Functions and graphs

Class 12 Mathematics, unit 1, book pages 7–42

A function is a rule that gives exactly one output for each allowed input. This unit first finds the domain, codomain and range of functions, sorts them into types (into, onto, one-one, bijective) and builds inverse functions. It then draws and reads the graphs of linear, quadratic, power, modulus, exponential, logarithmic and trigonometric functions, uses graphs to solve equations and daily-life problems, and ends with moving and resizing graphs by shifts and scaling. These skills are used throughout the calculus units that follow.

Learning outcomes (18)

Recall the definition of a function, and find its domain, codomain, range and type.
Find the inverse of a function and demonstrate its domain and range with examples.
Know linear, quadratic and square root functions.
Sketch graphs of linear and non-linear functions.
Plot the graph of a function of the type $y = x^{n}$ when (i) $n$ is a positive or negative integer and $x \neq = 0$ , (ii) $n$ is a rational number and $x > 0$ .
Plot the graph of a quadratic function of the form $y = a x^{2} + b x + c$ , where $a, b, c$ are integers and $a \neq = 0$ .
Draw graphs using factors and predict functions from their graphs.
Find intersecting points graphically when the intersection occurs between (i) a linear function and the coordinate axes, (ii) two linear functions, (iii) a linear function and a quadratic function.
Draw the graph of modulus functions.
Solve appropriate problems from daily life graphically.
Classify algebraic and transcendental functions, and describe trigonometric, inverse trigonometric, logarithmic and exponential functions.
Define logarithm, and derive and apply the laws of logarithm.
Graph and analyse exponential and logarithmic functions.
Apply the concept of exponential functions to find compound interest.
Solve problems involving exponential and logarithmic equations.
Identify the domain and range of transcendental functions through graphs.
Interpret the relation between a one-one function and its inverse through a graph.
Demonstrate the transformation of a graph through horizontal shift, vertical shift and scaling.

1.1Function

A function is a rule that links every input to exactly one output. Think of a vending machine: each button gives one fixed item. Pressing the same button never gives two different items, but two different buttons may give the same item.

Function: A function $f$ from a set A to a set B assigns to each element of A exactly one element of B. It is written $f : A \to B$ and read as “ $f$ is a function from A to B”.
Domain and co-domain: The set A is the domain of the function and the set B is its co-domain.
Image and pre-image: If $(x, y) \in f$ , then $y$ is the image of $x$ and $x$ is the pre-image of $y$ . In $(2, 4)$ , 4 is the image of 2 and 2 is the pre-image of 4.

Example. Let $A = {- 1, 1, 2}$ and $B = {1, 4, 9}$ . Then $f = {(- 1, 1), (1, 1), (2, 4)}$ is a function from A to B, because each element of A is used exactly once as a first element. No first element is repeated.

Each element of A has exactly one arrow leaving it. Two arrows may end at the same element (here

- 1

and

1

both go to 1), and some elements of B (here 9) may receive no arrow.

Function notation. If $x$ is the independent variable (the input) and $y$ is the dependent variable (the output), a function from A to B is written $y = f (x)$ . In the example above $f (- 1) = 1$ , $f (1) = 1$ and $f (2) = 4$ .

A function in daily life

A tree grows

15 cm

every year. Write its height as a function of its age and find the height after 10 years.

The height depends on the age, so age is the input and height is the output.
$h (age) = 15 \times age$
Put the input age $= 10$ years into the rule.
$h (10) = 15 \times 10 = 150 cm$
So the function relates the input 10 to the output 150. We can write $10 \to 150$ .

h (10) = 150 cm

Real valued function: A function whose inputs and outputs are real numbers. All functions in this unit are real valued functions.

1.1.1 to 1.1.3Domain, range and codomainOften examined

Domain of a function (1.1.1)

The domain is the set of all possible values of the independent variable that can be used as inputs. In the example $f = {(- 1, 1), (1, 1), (2, 4)}$ , $Dom f = {- 1, 1, 2}$ .

When a function is given only by a formula, its domain is every real number $x$ for which the formula gives a real value. Domain restrictions are the values of $x$ for which the function cannot be defined. Two operations cause trouble with real numbers: dividing by zero, and taking the square root (or any even root) of a negative number. A logarithm also needs a positive input.

How to find the domain

The book writes $R - {2}$ for “all real numbers except 2”.
Form of $f (x)$	Condition	Example	Domain
Polynomial	none	$3 x + 8$ , $x^{3} + 2 x - 5$	$R$ (all real numbers)
Fraction $\frac{p ( x )}{q ( x )}$	$q (x) \neq = 0$	$\frac{1}{x - 2}$	$R - {2}$
Square root $u$	$u \geq 0$	$x - 3$	$[3, \infty)$
Root in a denominator $\frac{1}{u}$	$u > 0$	$\frac{1}{x}$	$(0, \infty)$
Logarithm $ln u$ or $lo g u$	$u > 0$	$ln (x + 5)$	$(- 5, \infty)$
$e^{x}$ , $sin x$ , $cos x$ , $∣ x ∣$	none	$e^{x - 2} + 3$	$R$
Interval given with $f$	stated	$f (x) = x + 1, 0 < x < 10$	$(0, 10)$

Range of a function (1.1.2)

The range is the set of all outputs the function actually produces when $x$ takes every value of the domain. In $y = f (x)$ , the range is the spread of all values of $y$ from the minimum to the maximum. In the example above $Rang f = {1, 4}$ .

How to find the range

Find the domain first and remove the values of $x$ for which the function is not defined.
Use facts about the building blocks to find the smallest and largest values of $y$ : $x^{2} \geq 0$ , $u \geq 0$ , $∣ u ∣ \geq 0$ , $- 1 \leq sin x \leq 1$ and $e^{x} > 0$ .
For fractions, solve $y = f (x)$ for $x$ . Any value of $y$ that makes this impossible (for example a zero denominator) is not in the range.
Check with a quick sketch: squash the graph sideways onto the $y$ -axis. The part of the $y$ -axis it covers is the range.

Codomain of a function (1.1.3)

The codomain is the set of all possible outcomes that the function is allowed to give, the set B in $f : A \to B$ . The range is always a subset of the codomain. Sometimes the range equals the codomain. In the example above, $Codom f = {1, 4, 9}$ while $Rang f = {1, 4}$ .

The domain is

{a, b, c}

, the codomain is

{1, 2, 3, 4, 5}

and the range (the elements actually hit) is

{1, 2, 3}

Example 1: domain and range of

f (x) = 2 x^{2} - 4

Find the domain and range of the function

f (x) = 2 x^{2} - 4

Domain. $f$ is a polynomial. No value of $x$ makes it fail.
$Dom f = (- \infty, \infty) = R$
Range. A square is never negative, so the smallest value of $2 x^{2}$ is 0 (at $x = 0$ ).
$x^{2} \geq 0 \Rightarrow 2 x^{2} - 4 \geq - 4$
At $x = 0$ the output is $- 4$ . For every other real $x$ the output is greater than $- 4$ , and it grows without limit. Every $y \geq - 4$ is reached, because $y = 2 x^{2} - 4$ gives $x = \pm (y + 4) /2$ , which is real when $y \geq - 4$ .
$Rang f = [- 4, \infty)$

−3

−2

−1

−5

−4

−3

−2

−1

x

y

y = - 4

(0, - 4)

f (x) = 2 x^{2} - 4

The lowest point is

(0, - 4)

, so the graph covers

y \geq - 4

Dom f = R, Rang f = [- 4, \infty)

Example 2: domain and range of

g (x) = x - 3

Find the domain and range of the function

g (x) = x - 3

Domain. The expression under the square root must not be negative.
$x - 3 \geq 0 \Rightarrow x \geq 3, Dom g = [3, \infty)$
Range. At $x = 3$ , $g (3) = 0 = 0$ . For every $x > 3$ the output is positive, and it keeps growing as $x$ grows.
$Rang g = [0, \infty)$

−1

x

y

(4, 1)

(7, 2)

g (x) = x - 3

The graph starts at

(3, 0)

(filled dot, included) and rises slowly to the right.

Dom g = [3, \infty), Rang g = [0, \infty)

1.2Types of functionsOften examined

Functions are sorted into types by looking at how the elements of A are matched with the elements of B. Two checks decide the type: whether every element of B is used (onto) or not (into), and whether different inputs always give different outputs (one-one) or not (many-one).

Into function (1.2.1)

A function $f : A \to B$ is an into function if at least one element of B has no pre-image in A. In other words, not every element of the codomain is matched with an element of the domain, so $range (f) \neq = B$ .

f = {(a, 1), (b, 3), (c, 3)}

is into: the elements 2 and 4 of B have no pre-image.

$f (x) = sin x$ with $f : R \to R$ is into, because its outputs lie only in $[- 1, 1]$ , not in all of $R$ .
$g (x) = x^{2}$ with $g : R \to R$ is into, because it never gives a negative real number.
$h (x) = e^{x}$ with $h : R \to [0, \infty)$ is into, because $e^{x}$ is never 0.

Onto (surjective) function (1.2.2)

For non-empty sets A and B, $f : A \to B$ is onto if every element of B is the image of some element of A. That is, for every $y \in B$ there is an $x \in A$ with $f (x) = y$ , so $range (f) = B$ .

f = {(a, 2), (b, 1), (c, 3)}

is onto: every element of B receives an arrow.

$f (x) = x$ (identity function), $f : R \to R$ .
$g (x) = e^{x}$ (exponential function) when $g : R \to R^{+}$ .
$h (x) = x^{2}$ (square function) when $h : R \to [0, \infty)$ .
$m (x) = x^{3}$ (cubic function), $m : R \to R$ .
$p (x) = c$ (constant function) when its codomain is ${c}$ .

One-one (injective) function (1.2.3)

$f : A \to B$ is one-one if distinct elements of A have distinct images in B. Equivalently, equal outputs can come only from equal inputs:

f (x_{1}) = f (x_{2}) ⟹ x_{1} = x_{2}

f = {(a, 2), (b, 1), (c, 3)}

into

B = {1, 2, 3, 4}

is one-one: no element of B receives two arrows.

$f (x) = x$ (identity function)
$g (x) = 2 x + 3$ (linear function)
$h (x) = e^{x}$ (exponential function)
$m (x) = x$ (square root function, defined for $x \geq 0$ )

A function that is not one-one is called many-one: two or more elements of A share the same image. For example $x^{2}$ is many-one because $(- 2)^{2} = 2^{2} = 4$ .

Injective function (1.2.4)

The book defines an injective function as a function which is both into and one-one (a one-one into function).

Bijective function (1.2.5)

A function which is both onto and one-one is a bijective function. A bijective function sets up a one-one correspondence between the elements of the two sets: every element of A is paired with exactly one element of B and every element of B with exactly one element of A. Only bijective functions have inverse functions (section 1.3).

Summary of the types of functions.
Type	Condition	In an arrow diagram
Into	$range \neq = B$	some element of B gets no arrow
Onto (surjective)	$range = B$	every element of B gets at least one arrow
One-one	$f (x_{1}) = f (x_{2}) \Rightarrow x_{1} = x_{2}$	no element of B gets two arrows
Many-one	$f (x_{1}) = f (x_{2})$ for some $x_{1} \neq = x_{2}$	some element of B gets two or more arrows
Injective (book)	one-one and into	one arrow at most per element of B, some get none
Bijective	one-one and onto	exactly one arrow into every element of B

Key Facts table from the book: various types of functions.
Based on	Types of functions
elements	one-one, many-one, onto, bijective, into, constant function
the equation	identity, linear, quadratic, cubic, polynomial functions
the range	modulus, rational, even and odd, periodic, greatest and smallest integer, inverse, composite functions
the domain	algebraic, trigonometric, logarithmic, exponential functions

How to test the type of a function

One-one test (formula). Start from $f (x_{1}) = f (x_{2})$ and simplify. If you can reach $x_{1} = x_{2}$ , the function is one-one. If you reach something like $x_{1}^{2} = x_{2}^{2}$ , find two numbers with the same image to prove it is not one-one.
Onto test (formula). Put $y = f (x)$ and solve for $x$ . If the $x$ you get is defined and lies in the domain for every $y$ in the codomain, the function is onto.
Finite sets. List all images. Repeated image means not one-one. If the set of images equals the codomain the function is onto, otherwise into.

Example 3

Check whether the function

f (x) = 2 x + 3

is one-to-one or not if the domain is

{0.5, 1, 2}

and the codomain is

{4, 5, 7}

Find the image of each element of the domain.
$f (0.5) = 2 (0.5) + 3 = 4, f (1) = 5, f (2) = 7$
The three inputs give three different outputs, so no output is repeated. The function is one-one.
The images ${4, 5, 7}$ also fill the whole codomain, so the function is onto as well. Hence it is in fact bijective.

f is one-one (in fact bijective)

Example 4

Check whether the function

f (x) = 2 x^{2} + 1

is one-to-one or not.

Assume two inputs give the same output.
$f (x_{1}) = f (x_{2}) \Rightarrow 2 x_{1}^{2} + 1 = 2 x_{2}^{2} + 1 \Rightarrow x_{1}^{2} = x_{2}^{2}$
This gives $x_{1} = \pm x_{2}$ , not only $x_{1} = x_{2}$ . A counterexample confirms it.
$f (1) = 3 = f (- 1), but 1 \neq = - 1$

f (x) = 2 x^{2} + 1 is not one-one

Example 5

Check the type of the function

f (x) = x^{2} - 1

Dom f = {1, - 1, 2, - 2}

and

Codom f = {0, 3, - 3}

Substitute each element of the domain.
$f (1) = 0, f (- 1) = 0, f (2) = 3, f (- 2) = 3$
So $Rang f = {0, 3}$ . The element $- 3$ of the codomain has no pre-image.
$Rang f = {0, 3} \neq = {0, 3, - 3} = Codom f$
Hence $f$ is an into function. It is also many-one, since $1$ and $- 1$ have the same image.

f is a many-one into function

Example 6

Find the type of the function

f (x) = 3 x + 2

defined on

f : R \to R

Onto. Let $y$ be any real number and solve $y = 3 x + 2$ for $x$ .
$x = \frac{y - 2}{3} \in R$
Substitute this $x$ back into $f$ . We get $y$ back, so every real $y$ has a pre-image.
$f (\frac{y - 2}{3}) = 3 (\frac{y - 2}{3}) + 2 = y - 2 + 2 = y$
One-one. $3 x_{1} + 2 = 3 x_{2} + 2$ gives $x_{1} = x_{2}$ . So $f$ is one-one as well.
$f is onto and one-one, that is, bijective$

f (x) = 3 x + 2 is onto (in fact bijective)

1.2Even and odd functions

The book's Key Facts table lists even and odd functions among the types of functions. They were also asked in the 2026 old-syllabus paper, so the idea is summarised here. The test is to replace x by -x and compare with f(x).

Even function: f(-x)=f(x) for every x in the domain. The graph is symmetric about the y-axis. Examples: x^2, x^4, \cos x, |x|.
Odd function: f(-x)=-f(x) for every x in the domain. The graph is symmetric about the origin (a half turn about O leaves it unchanged). Examples: x, x^3, \sin x, x^3-x.

The even function is a mirror image in the y-axis. The odd function looks the same after a half turn about the origin.

Testing a function

Decide whether each function is even, odd or neither: (a) x^3-x, (b) x^2+x.

(a) Replace x by -x.
f(-x)=(-x)^3-(-x)=-x^3+x=-(x^3-x)=-f(x)\;\Rightarrow\;\text{odd}
(b) Replace x by -x and compare with both f(x) and -f(x).
g(-x)=x^2-x,\quad g(x)=x^2+x,\quad -g(x)=-x^2-x
g(-x) equals neither, for example g(1)=2 but g(-1)=0. So g is neither even nor odd.

x^3-x\ \text{is odd};\quad x^2+x\ \text{is neither}

1.3Inverse functionOften examined

The inverse of a function f is a function, written f^{-1}, which reverses the effect of f: it undoes what f does. If f takes the input x to the output y, then f^{-1} takes y back to x. If f:X\to Y, then f^{-1}:Y\to X.

f carries x in X to y in Y; f^{-1} carries y back to x. The two sets swap roles.

\text{Dom}\,f=\text{Rang}\,f^{-1}\qquad\text{and}\qquad\text{Rang}\,f=\text{Dom}\,f^{-1}

So the domain of the given function becomes the range of the inverse function, and the range of the given function becomes the domain of the inverse function. This is the fastest way to write the range of f^{-1}.

When does an inverse exist

Not every function has an inverse. If f took the same value twice, say f(1)=f(-1)=5, then f^{-1}(5) would have to be both 1 and -1, which a function cannot do. So the inverse exists only when f is one-one, and it is defined on the whole of Y only when f is onto. Hence f^{-1} exists exactly when f is bijective.

The book asks whether the inverse function is also bijective. Yes: f^{-1} is one-one (different y values come from different x values) and onto (every x in X is the image of f(x)), so f^{-1} is bijective too.

Composing a function with its inverse gives back the input:

f\circ f^{-1}(x)=f^{-1}\circ f(x)=x

If y=f(x) is a bijective function, then x=f^{-1}(y).
If f\circ g(x)=g\circ f(x)=x, then g=f^{-1} and f=g^{-1}.
(f^{-1})^{-1}=f: the inverse of the inverse is the original function.

Steps to find an inverse function (1.3.1)

Derivation Inverse of the linear function f(x)=ax+b

Aim. Find f^{-1}(x) for f(x)=ax+b, a\neq0.

Replace f(x) with y, so the rule reads as an equation.
y=ax+b
Solve this equation for x. Subtract b, then divide by a (allowed because a\neq0).
x=\dfrac{y-b}{a}
This x is the input that gives the output y, so it is f^{-1}(y).
f^{-1}(y)=\dfrac{y-b}{a}
Interchange y with x, because we normally use x for the input of any function.
f^{-1}(x)=\dfrac{x-b}{a}

Resultf(x)=ax+b\;\Longrightarrow\;f^{-1}(x)=\dfrac{x-b}{a}

The inverse undoes the steps of f in reverse order: f multiplies by a then adds b, so f^{-1} subtracts b then divides by a. For example, f(x)=3x-4 gives f^{-1}(x)=\dfrac{x+4}{3} and f^{-1}(5)=3.

Graph of an inverse function (1.3.2)

If (x,y) lies on the graph of f, then (y,x) lies on the graph of f^{-1}. Swapping the coordinates of every point reflects the graph in the line y=x. So the graphs of a function and its inverse are symmetric with respect to the line y=x, and two functions whose graphs are symmetric about y=x are inverses of each other.

y=2^{x} and its inverse y=\log_2x are mirror images in the dashed line y=x: (1,2) on one curve matches (2,1) on the other.

Example 7

Find the inverse function of f(x)=\dfrac{x}{x-2}. (i) Find the domain and range of the function and its inverse. (ii) Prove that f\circ f^{-1}(x)=f^{-1}\circ f(x)=x.

(i) Inverse, domain and range

The denominator cannot be zero, so remove x=2.
\text{Dom}\,f=\mathbb{R}-\{2\}
Put y=f(x) and clear the fraction.
y=\dfrac{x}{x-2}\;\Rightarrow\;y(x-2)=x\;\Rightarrow\;xy-2y=x
Collect the x terms on one side and factor out x.
xy-x=2y\;\Rightarrow\;x(y-1)=2y\;\Rightarrow\;x=\dfrac{2y}{y-1}
This x is f^{-1}(y). Replace y with x.
f^{-1}(y)=\dfrac{2y}{y-1}\;\Rightarrow\;f^{-1}(x)=\dfrac{2x}{x-1}
The inverse is not defined only at x=1. Its domain is the range of f, and its range is the domain of f.
\text{Dom}\,f^{-1}=\mathbb{R}-\{1\}=\text{Rang}\,f,\qquad \text{Rang}\,f^{-1}=\mathbb{R}-\{2\}=\text{Dom}\,f

(ii) Proof

Put f^{-1}(x) into f and simplify the compound fraction by multiplying top and bottom by (x-1).
f\circ f^{-1}(x)=f\!\left(\dfrac{2x}{x-1}\right)=\dfrac{\frac{2x}{x-1}}{\frac{2x}{x-1}-2}=\dfrac{2x}{2x-2x+2}=\dfrac{2x}{2}=x
Put f(x) into f^{-1} and simplify by multiplying top and bottom by (x-2).
f^{-1}\circ f(x)=f^{-1}\!\left(\dfrac{x}{x-2}\right)=\dfrac{2\left(\frac{x}{x-2}\right)}{\frac{x}{x-2}-1}=\dfrac{2x}{x-x+2}=\dfrac{2x}{2}=x
Both compositions give x.
f\circ f^{-1}(x)=f^{-1}\circ f(x)=x

Asymptotes of f: x=2 and y=1. Those of f^{-1} are x=1 and y=2 (not drawn). The two curves are reflections in y=x and meet on it at (0,0) and (3,3).

f^{-1}(x)=\dfrac{2x}{x-1};\quad \text{Dom}\,f^{-1}=\mathbb{R}-\{1\},\ \text{Rang}\,f^{-1}=\mathbb{R}-\{2\}

1.3Composite functions

Section 1.3 uses the symbol f\circ f^{-1}. It means “apply f^{-1} first, then apply f to the result”. In general the composite function f\circ g (read “f of g”) is defined by

(f\circ g)(x)=f\big(g(x)\big)

The function written nearest to x acts first. The output of g becomes the input of f, so g(x) must lie in the domain of f.

Composing two functions

If f(x)=x+2 and g(x)=x^2, find f(g(x)) and g(f(x)).

For f(g(x)), replace the x in f by g(x)=x^2.
f(g(x))=f(x^2)=x^2+2
For g(f(x)), replace the x in g by f(x)=x+2.
g(f(x))=g(x+2)=(x+2)^2=x^2+4x+4
The two answers are different, so the order of composition matters.

f(g(x))=x^2+2,\qquad g(f(x))=x^2+4x+4

1.4Linear, quadratic and square root functions

Linear function (1.4.1)

A function of the form

y=ax+b;\qquad a,b\in\mathbb{R},\ a\neq0

is called a linear function. It is a polynomial function of degree one. Here a is the slope (the rise in y for each unit step in x) and b is the y-intercept (the value of y at x=0). For example, f(x)=x+2 and g(x)=2x-1 are linear functions.

The graph of a linear function is a straight line. The slope between any two points of the line is the same, so two points are enough to draw it. The domain and the range of a linear function are both \mathbb{R}.

m=\dfrac{y_2-y_1}{x_2-x_1}

Slope of the line through (x_1,y_1) and (x_2,y_2). For y=ax+b this always equals a.

g has the larger slope (2 against 1), so it is steeper. The lines cross at (3,5), where x+2=2x-1.

Non-linear function (1.4.2)

A function that is not linear is called a non-linear function. Its graph is a curve, and the slope between two points changes from one pair of points to another. Quadratic, cubic, square root and exponential functions are non-linear. Here we recall the shapes of the quadratic and square root graphs.

(a) Quadratic function

A function of the form

y=ax^2+bx+c;\qquad a,b,c\in\mathbb{R},\ a\neq0

is called quadratic. It is a polynomial function of degree two. For example, f(x)=x^2+3x-2 and g(x)=5-2x^2 are quadratic functions. The graph of a quadratic function is a U-shaped curve called a parabola. It opens upward when a>0 and downward when a<0.

f opens upward (a=1>0) with lowest point (-1.5,-4.25); g opens downward (a=-2<0) with highest point (0,5).

(b) Square root function

The function y=\sqrt{x}, where x\ge0, is called a square root function. More generally \sqrt{u} with u an expression in x is a square root function; for example f(x)=\sqrt{x+2} and g(x)=\sqrt{x^2-3}. The domain of a square root function depends on its formula: the expression under the root must be \ge0. Its outputs are never negative.

Function	Condition	Domain	Range
y=\sqrt{x}	x\ge0	[0,\infty)	[0,\infty)
f(x)=\sqrt{x+2}	x+2\ge0	[-2,\infty)	[0,\infty)
g(x)=\sqrt{x^2-3}	x^2\ge3	(-\infty,-\sqrt3\,]\cup[\sqrt3,\infty)	[0,\infty)

g has two branches because x^2-3\ge0 holds for x\le-\sqrt3 and for x\ge\sqrt3. Far from the origin its branches approach the lines y=\pm x.

1.5.1 and 1.5.2Graphs of functions of the type y = x^n

Graph of y=x^n, n\in\mathbb{Z}, x\neq0 (1.5.1)

To plot y=x^n we take different integral values of n and make a table of values for each. For example, the table for y=\pm x^3 is:

x	0	0.5	1	2
f(x)=x^3	0	0.125	1	8
g(x)=-x^3	0	-0.125	-1	-8

x	f(x)=x^3	g(x)=-x^3
0	0	0
0.5	0.125	-0.125
1	1	-1
2	8	-8

The graphs of y=x^3 and y=-x^3 have the same shape but opposite behaviour: each is the mirror image of the other in the x-axis.

The graphs of y=x^n for n=-2,-1,0,1,2 are shown below in two groups.

n=0,1,2. The line y=x^0=1 has a hole at x=0, because 0^0 is not defined.

n=-1,-2. Both curves come very close to the axes but never touch them.

From these graphs we observe:

the graph of y=x^0 is a horizontal line through y=1;
the graph of y=x^1 is a straight line bisecting the first and third quadrants;
the graph of y=x^{-1} is a hyperbola with branches in the first and third quadrants;
the graph of y=x^2 is a parabola starting from the origin and opening upwards;
the graph of y=x^{-2} has two branches, both above the x-axis (first and second quadrants), rising very steeply near the y-axis;
all the graphs of y=x^n pass through (1,1).

How n decides the shape of y=x^n.
Type of n	Shape	Symmetry	Example
positive even	U-shape through O	about the y-axis (even function)	x^2, x^4
positive odd	rises through O from quadrant 3 to 1	about the origin (odd function)	x, x^3
negative odd	hyperbola in quadrants 1 and 3	about the origin	x^{-1}
negative even	two branches in quadrants 1 and 2	about the y-axis	x^{-2}

Example 8: graph of y=x^4

Draw the graph of y=x^4.

Make a table of values. The power is even, so x and -x give the same y.
\begin{array}{c|ccccccc} x & -1.5 & -1 & -0.5 & 0 & 0.5 & 1 & 1.5 \\ \hline y & 5.06 & 1 & 0.06 & 0 & 0.06 & 1 & 5.06 \end{array}
Plot the points and join them with a smooth curve.

The graph of y=x^4 is U-shaped and opens upward, starting from the origin. The value of y increases slowly for -0.5<x<0.5 (the curve is very flat near O), and after that it increases rapidly.

y=x^4:\ \text{U-shaped, symmetric about the } y\text{-axis, minimum at } (0,0)

Graph of y=x^n, n\in\mathbb{Q}, x>0 (1.5.2)

When n is a fraction, x^{n} may involve an even root, so we take x>0. For example, x^{3/2}=(\sqrt{x})^{3}. The table for y=x^{3/2} is:

x	0.5	1	4	6
f(x)=x^{3/2}	0.35	1	8	14.7

x	f(x)=x^{3/2}
0.5	0.35
1	1
4	8
6	14.7

y=x^{3/2} rises faster and faster as x grows.

The graphs of y=x^n for n=-\tfrac92,\ -\tfrac23,\ \tfrac23,\ \tfrac12 are shown below. (For n=0 the graph is the line y=1.)

All four curves pass through (1,1). Negative powers fall towards the x-axis; positive powers rise.

From these graphs we observe:

the graph of y=x^{-2/3} is closer to the y-axis;
the graph of y=x^{-9/2} moves away from the y-axis as compared with the graph of y=x^{-2/3} (for 0<x<1 it lies above it), and beyond x=1 it drops to the x-axis much faster;
the graph of y=x^{1/2} is closer to the x-axis (for x>1);
the graph of y=x^{2/3} moves away from the x-axis as compared with the graph of y=x^{1/2} (for x>1);
the graphs of y=x^n pass through (1,1).

1.5.3Graph of a quadratic functionOften examined

The polynomial function of degree two is called a quadratic function:

f(x)=ax^2+bx+c;\qquad a,b,c\in\mathbb{R}\ \text{and}\ a\neq0

Its graph is a parabola. For example, y=x^2+2x+1 and y=2-3x^2 are quadratic functions.

Understanding the graph

a: the coefficient a affects the direction and width of the parabola. If a>0 the parabola opens upwards; if a<0 it opens downwards. The larger the absolute value of a, the narrower the parabola.
b: together with a it fixes the horizontal position of the vertex (left or right), through x=-\dfrac{b}{2a}. It is also the slope of the curve at the point where it crosses the y-axis.
c: this is the y-intercept. So the point (0,c) is on the parabola.
x=-\dfrac{b}{2a} is the equation of the axis of symmetry, and it is also the x-coordinate of the vertex (the turning point).

Derivation Vertex of y=ax^2+bx+c by completing the square

Aim. Show that the vertex is at x=-\dfrac{b}{2a} and find its y-coordinate.

Take a out of the first two terms so that x^2 has coefficient 1.
y=a\left(x^2+\dfrac{b}{a}x\right)+c
Add and subtract \left(\dfrac{b}{2a}\right)^2 inside the bracket (half the coefficient of x, squared). This makes a perfect square without changing the value.
y=a\left(x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}\right)-a\cdot\dfrac{b^2}{4a^2}+c
Write the bracket as a square and combine the constants.
y=a\left(x+\dfrac{b}{2a}\right)^2+\dfrac{4ac-b^2}{4a}
The square is never negative and is zero only when x=-\dfrac{b}{2a}. If a>0, y is smallest there; if a<0, y is largest there. So that point is the vertex, and the graph is symmetric about the vertical line through it.
x=-\dfrac{b}{2a},\qquad y=\dfrac{4ac-b^2}{4a}

Result\text{Vertex}=\left(-\dfrac{b}{2a},\ \dfrac{4ac-b^2}{4a}\right),\qquad \text{axis of symmetry: } x=-\dfrac{b}{2a}

The vertex is the lowest point when a>0 and the highest point when a<0. So the range is \left[\frac{4ac-b^2}{4a},\infty\right) for a>0 and \left(-\infty,\frac{4ac-b^2}{4a}\right] for a<0. In practice, find x=-\frac{b}{2a} and substitute it into y.

Steps to draw the graph

Compare with y=ax^2+bx+c and decide whether the parabola opens upwards (a>0) or downwards (a<0).
Find and draw the axis of symmetry x=-\dfrac{b}{2a}.
Find and plot the vertex: substitute x=-\dfrac{b}{2a} into the equation to get its y-coordinate.
Find some more points (the y-intercept (0,c), the x-intercepts if any, and points on both sides of the axis), plot them and join with a smooth curve.

Example 9: graph of y=2x^2+3x-2

Draw the graph of y=2x^2+3x-2.

Compare with y=ax^2+bx+c.
a=2,\quad b=3,\quad c=-2
Step 1. a=2>0, so the parabola opens upwards.
Step 2. Axis of symmetry.
x=-\dfrac{b}{2a}=-\dfrac{3}{2(2)}=-\dfrac34
Step 3. The vertex has x=-\tfrac34. Substitute to get y.
\begin{aligned}y&=2\left(-\tfrac34\right)^2+3\left(-\tfrac34\right)-2\\ &=\dfrac98-\dfrac94-2=-\dfrac{25}{8}\\ \text{vertex}&=\left(-\tfrac34,-\tfrac{25}{8}\right)=(-0.75,-3.125)\end{aligned}
Step 4. More points from a table. The x-intercepts come from 2x^2+3x-2=(2x-1)(x+2)=0, so x=\tfrac12 or x=-2.
\begin{array}{c|ccccc} x & -2 & -1 & 0 & 0.5 & 1 \\ \hline y & 0 & -3 & -2 & 0 & 3 \end{array}

From the figure it is clear that the graph of y=2x^2+3x-2 is a parabola, symmetric about x=-\tfrac34.

\text{Vertex}\left(-\tfrac34,-\tfrac{25}{8}\right),\ \text{axis } x=-\tfrac34,\ \text{opens upward}

Example 10: graph of y=4-2x^2

Draw the graph of y=4-2x^2.

Write it in standard order and compare.
y=-2x^2+0x+4:\qquad a=-2,\ b=0,\ c=4
Step 1. a=-2<0, so the parabola opens downwards.
Step 2. Axis of symmetry.
x=-\dfrac{b}{2a}=-\dfrac{0}{2(-2)}=0
Step 3. Vertex: at x=0, y=-2(0)^2+0+4=4.
\text{vertex}=(0,4)
Step 4. Table of values. The x-intercepts are where 4-2x^2=0, that is x=\pm\sqrt2\approx\pm1.41.
\begin{array}{c|ccccc} x & -2 & -1 & 0 & 1 & 2 \\ \hline y & -4 & 2 & 4 & 2 & -4 \end{array}

\text{Vertex }(0,4),\ \text{axis } x=0,\ \text{opens downward}

1.6Drawing a graph using factorsOften examined

A parabola y=ax^2+bx+c (a\neq0) can be tricky to sketch by plotting many points. Factorising the quadratic gives all the information needed for a quick, accurate sketch.

Procedure

The points where the parabola meets the x-axis are the solutions of

ax^2+bx+c=0\qquad(\text{i})

If equation (i) can be factorised as a(x-x_1)(x-x_2)=0, then by the zero product property (if a product is zero, one factor is zero) we get x=x_1 and x=x_2. So the graph cuts the x-axis at (x_1,0) and (x_2,0): these are the x-intercepts.

**x-intercepts:** put y=0 and factorise; the roots x_1,x_2 give (x_1,0) and (x_2,0).
**y-intercept:** put x=0; the constant term gives the point (0,c).
Direction: if a is positive the parabola opens upward; if negative, downward.
Vertex: its x-coordinate is the arithmetic mean of the two x-intercepts, \dfrac{x_1+x_2}{2}. Substitute it into the function to get the y-coordinate.
Plot these four points (both x-intercepts, the y-intercept and the vertex), add the mirror image of the y-intercept in the axis, and sketch a smooth symmetric curve.

Derivation Why the vertex lies midway between the roots

Aim. Show that the axis of symmetry is x=\dfrac{x_1+x_2}{2}.

Expand the factorised form.
y=a(x-x_1)(x-x_2)=ax^2-a(x_1+x_2)x+ax_1x_2
Compare with y=ax^2+bx+c: the coefficients of x must be equal.
b=-a(x_1+x_2)
Substitute into the axis formula x=-\dfrac{b}{2a}.
x=-\dfrac{-a(x_1+x_2)}{2a}=\dfrac{x_1+x_2}{2}

Resultx_{\text{vertex}}=\dfrac{x_1+x_2}{2}

The parabola is symmetric, so its two x-intercepts are equally far from the axis. When the two roots are equal (a repeated root), the vertex lies on the x-axis and the parabola only touches it.

Example 10 (second example with this number in the book): y=x^2-8x+12

Draw the graph of y=x^2-8x+12 using factors.

x-intercepts: put y=0 and factorise.
x^2-8x+12=0\;\Rightarrow\;(x-2)(x-6)=0\;\Rightarrow\;x_1=2,\ x_2=6
So the graph intersects the x-axis at (2,0) and (6,0).
y-intercept: put x=0, which gives y=12. So the y-intercept is (0,12).
Vertex: the x-coordinate is the mean of the roots; substitute to get y.
x=\dfrac{2+6}{2}=4,\qquad y=4^2-8(4)+12=-4,\qquad \text{vertex}=(4,-4)
a=1>0, so the parabola opens upward and is symmetric about x=4. The mirror image of (0,12) in x=4 is (8,12).

x\text{-intercepts } (2,0),(6,0);\ y\text{-intercept }(0,12);\ \text{vertex }(4,-4)

Example 11: y=-x^2+4x-4

Draw the graph of y=-x^2+4x-4 using factors.

x-intercepts: put y=0. Take out -1 first to make factorising easier.
-x^2+4x-4=0\;\Rightarrow\;-(x^2-4x+4)=0\;\Rightarrow\;(x-2)^2=0
We get repeated roots x_1=x_2=2. The graph meets the x-axis only at (2,0), which is the vertex of the parabola (it touches the axis there).
y-intercept: put x=0, which gives y=-4. So the y-intercept is (0,-4).
a=-1<0, so the parabola opens downwards. It is symmetric about x=2, so it also passes through (4,-4), the mirror image of (0,-4).

\text{vertex }(2,0),\ y\text{-intercept }(0,-4),\ \text{opens downward}

Check point: y=2x^2+6x+4

Draw the graph of y=2x^2+6x+4 using factors.

Factorise after taking out the common factor 2.
2x^2+6x+4=2(x^2+3x+2)=2(x+1)(x+2)
x-intercepts: 2(x+1)(x+2)=0 gives x=-1 or x=-2, so (-1,0) and (-2,0).
y-intercept: x=0 gives y=4, so (0,4).
Vertex: mean of the roots, then substitute.
x=\dfrac{-1+(-2)}{2}=-1.5,\qquad y=2(-1.5)^2+6(-1.5)+4=-0.5
a=2>0, so it opens upward. The mirror image of (0,4) in x=-1.5 is (-3,4).

x\text{-intercepts } (-2,0),(-1,0);\ y\text{-intercept }(0,4);\ \text{vertex }(-1.5,-0.5)

1.7Predicting functions from their graphs

Here we work backwards: we are given a graph and must write its equation. First decide the type of function from the shape (straight line or parabola), then read clear points off the graph and use them to find the unknown constants.

Graph	Form to use	Points needed
straight line	y-y_1=m(x-x_1) with m=\dfrac{y_2-y_1}{x_2-x_1}	any two points
parabola cutting the x-axis at p and q	y=a(x-p)(x-q)	the two x-intercepts and one more point
parabola through (p,r) and (q,r)	y-r=a(x-p)(x-q)	two points at the same height and one more point
parabola with vertex (h,k)	y=a(x-h)^2+k	the vertex and one more point

Example 12

Predict the function from the graph: a straight line passing through (2,0) and (0,-2).

The given graph.

The graph is a straight line, so the function is linear. Find the slope from the two points.
m=\dfrac{0-(-2)}{2-0}=\dfrac{2}{2}=1
Use the point slope form of the line with the point (2,0).
y-0=1(x-2)
Simplify.
y=x-2

f(x)=x-2

Example 13

Predict the function from the graph: a parabola opening upwards and passing through (2,0), (-1,0) and (0,-4).

The given graph.

The graph is a parabola, so the function is quadratic. It cuts the x-axis at p=2 and q=-1, so it has the form below, with a>0 because it opens upwards.
y=a(x-p)(x-q)=a(x-2)(x+1)\qquad(\text{i})
The parabola also passes through (0,-4). Substitute x=0, y=-4 into (i) to find a.
-4=a(0-2)(0+1)\;\Rightarrow\;-4=-2a\;\Rightarrow\;a=2
Put a=2 back into (i).
y=2(x-2)(x+1)=2(x^2-x-2)=2x^2-2x-4

f(x)=2(x^2-x-2)=2x^2-2x-4

Derivation Parabola through two points at the same height

Aim. Show that a parabola through (p,r) and (q,r), p\neq q, has an equation y-r=a(x-p)(x-q).

Move the whole graph down by r units. The new function is Y=y-r, and the two points become (p,0) and (q,0).
Y=y-r
The moved parabola cuts the x-axis at p and q, so it has the factor form.
Y=a(x-p)(x-q)
Replace Y by y-r to move back up.
y-r=a(x-p)(x-q)

Resulty-r=a(x-p)(x-q)

This is the book's Key Facts result. The book adds that p, q and r are positive, but the formula works for any real numbers with p\neq q. The third point on the graph fixes a.

1.8Graph of modulus functions

A modulus function (absolute value function) gives the size of a number regardless of its sign. If x is a real number, the modulus function is

y=|x|\quad\text{or}\quad f(x)=|x|=\begin{cases}\ \ x, & x\ge0\\ -x, & x<0\end{cases}\qquad x\in\mathbb{R}

It takes the actual value of x if x is more than or equal to 0, and minus the value of x if x is less than 0. For example:

If x=-5, then y=f(x)=-(-5)=5, since x is less than zero.
If x=6, then y=f(x)=6, since x is greater than zero.
If x=0, then y=f(x)=0, since x is equal to zero.

Domain and range of the modulus function (1.8.1)

The modulus of any real number can be found, so the domain of the modulus function is \mathbb{R}. The output is never negative, so its range is the set of non-negative real numbers, [0,\infty).

Graph of the modulus function (1.8.2)

x	-3	-2	-1	0	1	2	3
f(x)	3	2	1	0	1	2	3

x	f(x)
-3	3
-2	2
-1	1
0	0
1	1
2	2
3	3

A V-shape with its corner (vertex) at the origin: the arm y=-x on the left and the arm y=x on the right.

For all possible values of x, f(x) is never negative: it is positive for x\neq0 and zero at x=0.

Sketching other modulus graphs

For y=|x-h|+k the corner of the V is where the expression inside the modulus is zero, at x=h, and it is at height k. Make a table with x-values on both sides of the corner.

Graph of y=|x+1|-2

Sketch y=|x+1|-2 and state its domain and range.

The corner is where x+1=0, that is x=-1; there y=|0|-2=-2.
\text{vertex}=(-1,-2)
Table of values on both sides of x=-1.
\begin{array}{c|ccccccc} x & -4 & -3 & -2 & -1 & 0 & 1 & 2 \\ \hline y & 1 & 0 & -1 & -2 & -1 & 0 & 1 \end{array}
x-intercepts: |x+1|=2 gives x+1=\pm2, so x=1 or x=-3. The y-intercept is (0,-1).
Every real x can be used, and the lowest output is -2 at the corner.
D=\mathbb{R},\qquad R=[-2,\infty)

D=\mathbb{R},\qquad R=[-2,\infty)

1.9Finding intersecting points graphicallyOften examined

A point that lies on two graphs satisfies both equations at the same time. So drawing two graphs on the same axes and reading where they meet solves the two equations together. This is called a graphical solution.

Intersection of a linear function and the coordinate axes (1.9.1)

The graph of a linear function is a straight line. Its points of intersection with the axes are called intercepts. The x-intercept is found by putting y=0; the y-intercept by putting x=0.

Example 14

Find the x-intercept and y-intercept of the function f(x)=\dfrac{12-2x}{3}.

Make a table of values. Choosing x=6 and x=0 gives the intercepts directly.
\begin{array}{c|ccc} x & 6 & 0 & 3 \\ \hline f(x) & 0 & 4 & 2 \end{array}
The line crosses the x-axis at (6,0), so its x-intercept is 6. It crosses the y-axis at (0,4), so its y-intercept is 4.

x\text{-intercept}=6,\qquad y\text{-intercept}=4

Check point

The x- and y-intercepts of a line are -3 and -5 respectively. Find the points of intersection of the line with the axes.

An x-intercept of -3 means the line meets the x-axis where x=-3 and y=0.
(-3,0)
A y-intercept of -5 means the line meets the y-axis where x=0 and y=-5.
(0,-5)
The line itself follows from the two-intercept form \dfrac{x}{a}+\dfrac{y}{b}=1.
\dfrac{x}{-3}+\dfrac{y}{-5}=1\;\Rightarrow\;5x+3y=-15

(-3,0)\ \text{and}\ (0,-5)

Intersection of two linear functions (1.9.2)

While solving two linear functions graphically, keep in mind the following points.

Draw each linear function on the same set of axes, using the same scale.
Find the coordinates of the point where the lines intersect.

Example 15

Find the graphical solution of f(x)=\dfrac{6-x}{2} and g(x)=x-3.

Table of values for f(x)=\dfrac{6-x}{2}.
\begin{array}{c|ccc} x & 6 & 0 & 4 \\ \hline f(x) & 0 & 3 & 1 \end{array}
Table of values for g(x)=x-3.
\begin{array}{c|ccc} x & 3 & 0 & 5 \\ \hline g(x) & 0 & -3 & 2 \end{array}
Plot both lines on the same axes. They intersect at the point (4,1).
Check algebraically: set f(x)=g(x).
\dfrac{6-x}{2}=x-3\;\Rightarrow\;6-x=2x-6\;\Rightarrow\;x=4,\ y=4-3=1

\text{Solution: the point } (4,1)

Intersection of a linear function and a quadratic function (1.9.3)

The graph of a quadratic function is a curve (a parabola). A point of intersection of a linear function and a quadratic function is a point where both graphs meet. A line can meet a parabola at two points, at one point (when it just touches, as a tangent) or not at all.

Example 16

Solve f(x)=3x+4 and g(x)=5+3x-2x^2 graphically.

Table of values for the line f(x)=3x+4.
\begin{array}{c|ccc} x & 0 & -1 & 1 \\ \hline f(x) & 4 & 1 & 7 \end{array}
Compare g(x)=5+3x-2x^2 with y=ax^2+bx+c: a=-2, b=3, c=5. Since a=-2<0, the curve opens downward.
Table of values for g(x).
\begin{array}{c|cccccc} x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline g(x) & -9 & 0 & 5 & 6 & 3 & -4 \end{array}
Plot both graphs on the same axes. They intersect at approximately A(-0.7,\,1.9) and B(0.7,\,6.1).
Check algebraically: 3x+4=5+3x-2x^2 gives 2x^2=1.
x=\pm\dfrac{1}{\sqrt2}=\pm0.71,\qquad y=3(\pm0.71)+4=1.88\ \text{or}\ 6.12

\text{Solution set}=\{(-0.7,\,1.9),\ (0.7,\,6.1)\}

Solving problems graphically from daily life (1.9.4)

Example 17

Two airplanes are moving along the paths represented by f(x)=x+2 and g(x)=2x-4 respectively. Draw the graph of the paths of both planes and find the point where both planes pass.

Table of values for the first path f(x)=x+2.
\begin{array}{c|ccc} x & 0 & 2 & 6 \\ \hline f(x) & 2 & 4 & 8 \end{array}
Table of values for the second path g(x)=2x-4.
\begin{array}{c|ccc} x & 2 & 4 & 6 \\ \hline g(x) & 0 & 4 & 8 \end{array}
Draw both lines on the same axes. They meet at (6,8).
Check: x+2=2x-4 gives x=6, and then y=6+2=8.

\text{Both planes pass through the point } (6,8)

1.10.1 and 1.10.2Algebraic and transcendental functions

Algebraic function (1.10.1)

An algebraic function is a function that involves only algebraic operations: addition, subtraction, multiplication, division, and raising to a power (including roots, which are fractional powers).

Main types of algebraic functions

(i) Polynomial functions. A function of the form

p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0

where a_0,a_1,\dots,a_n are constants and n is a non-negative integer, is called a polynomial function. The highest power n (with a_n\neq0) is its degree. Some examples are:

f(x)=3x+7 (linear function, degree 1)
f(x)=x^2-2x+5 (quadratic function, degree 2)
f(x)=x^3-7x+7 (cubic function, degree 3)
f(x)=x^4-5x^2+2x-8 (biquadratic function, degree 4)
f(x)=x^5-7x+3 (quintic function, degree 5)

(ii) Rational functions. A function written as a fraction of two polynomials, f(x)=\dfrac{p(x)}{q(x)} with q(x)\neq0, is called a rational function. Some examples are:

f(x)=\dfrac{x-4}{2x+3},\qquad g(x)=\dfrac{7}{x^2+5x+1}

(iii) Power functions. Functions of the form f(x)=kx^{a}, where k and a are real numbers, are power functions. The exponent a can be an integer or a rational number. Examples: f(x)=x^2, f(x)=x^{-1} (the reciprocal function) and f(x)=\sqrt{x-2}=(x-2)^{1/2}.

Properties

Algebraic functions are closed under addition, subtraction, multiplication, division and composition: combining two algebraic functions in these ways gives another algebraic function.
Algebraic functions are easy to solve, differentiate and integrate.

Applications

Physics and engineering: simple mechanical systems, and the motion of objects under constant acceleration (s=ut+\tfrac12at^2 is a quadratic function of t).
Geometry: many curves such as circles and ellipses are described by algebraic equations.

Transcendental functions (1.10.2)

Functions which are not algebraic are called transcendental functions. They cannot be built from x using a finite number of algebraic operations; they can be expressed only through infinite series. Some examples are:

Exponential functions: f(x)=e^{x}, g(x)=a^{3x}
Logarithmic functions: f(x)=\log_a x, g(x)=\ln x, where the base a is a positive constant
Trigonometric functions: f(x)=\sin x, g(x)=\cos x, h(x)=\tan x
Inverse trigonometric functions: f(x)=\sin^{-1}x, g(x)=\cos^{-1}x
Hyperbolic functions: f(x)=\sinh x, g(x)=\cosh x, h(x)=\tanh x
Inverse hyperbolic functions: f(x)=\sinh^{-1}x, g(x)=\cosh^{-1}x
Special functions: Bessel functions, Gamma functions, error functions and others

Properties

They are not expressible in terms of a finite combination of the algebraic operations of addition, subtraction, multiplication, division, raising to a power and extracting a root.
They often show more complex behaviour, such as periodicity (trigonometric functions) and rapid growth (exponential functions).

Applications

Science and engineering: exponential and logarithmic functions model growth, decay and oscillation in natural systems.
Signal processing: trigonometric functions are used to analyse waves, sounds and signals.
Mathematical analysis: many problems in calculus, differential equations and complex analysis involve transcendental functions.

Classifying functions.
Function	Type	Reason
x^3-7x+7	algebraic (polynomial)	only powers and sums
\dfrac{x-4}{2x+3}	algebraic (rational)	polynomial divided by polynomial
\sqrt{x-2}	algebraic (power)	a root is a fractional power
3^{x}\times5	transcendental (exponential)	the variable is in the exponent
\ln(x+3)	transcendental (logarithmic)	logarithm of the variable
\sin 2x	transcendental (trigonometric)	trigonometric ratio of the variable

Hyperbolic functions (for reference)

The hyperbolic functions in the list above are built from e^{x} and e^{-x}:

\sinh x=\dfrac{e^{x}-e^{-x}}{2},\qquad \cosh x=\dfrac{e^{x}+e^{-x}}{2},\qquad \tanh x=\dfrac{\sinh x}{\cosh x}

\operatorname{csch}x=\dfrac{1}{\sinh x},\qquad \operatorname{sech}x=\dfrac{1}{\cosh x},\qquad \coth x=\dfrac{\cosh x}{\sinh x}

Derivation The identity \cosh^2x-\sinh^2x=1

Aim. Prove \cosh^2x-\sinh^2x=1 from the definitions.

Square \cosh x. Use (e^{x})^2=e^{2x} and e^{x}e^{-x}=1.
\cosh^2x=\dfrac{e^{2x}+2+e^{-2x}}{4}
Square \sinh x in the same way.
\sinh^2x=\dfrac{e^{2x}-2+e^{-2x}}{4}
Subtract. The e^{2x} and e^{-2x} terms cancel.
\cosh^2x-\sinh^2x=\dfrac{2-(-2)}{4}=\dfrac44=1

Result\cosh^2x-\sinh^2x=1

This is the hyperbolic partner of \cos^2x+\sin^2x=1. Dividing it by \sinh^2x (x\neq0) gives \coth^2x-1=\operatorname{csch}^2x, and dividing by \cosh^2x gives 1-\tanh^2x=\operatorname{sech}^2x.

1.10.3Logarithmic functions and the laws of logarithmsOften examined

Logarithmic functions form a fundamental class of transcendental functions. They are the inverses of exponential functions. A logarithm is a power: since 2^3=8, the logarithm of 8 to the base 2 is 3, the power to which 2 must be raised to give 8.

Logarithmic function: If y=a^{x} where a>0 and a\neq1, then x=\log_a(y). Replacing y with x (and x with y) gives the logarithmic function y=\log_a(x), read as “logarithm of x to the base a”.

y=a^{x}\iff x=\log_a y\qquad(a>0,\ a\neq1,\ y>0)

Base of the logarithm

Natural logarithm: \log_e(x)=\ln x, where e=2.71828\ldots is called Euler's number.
Common logarithm: \log_{10}(x), base a=10. On calculators and in this book \log x means \log_{10}x.
Binary logarithm: \log_2(x), base a=2.

Properties

The logarithm is the inverse of the exponential: if y=a^{x}, then x=\log_a(y). This means \log_a(a^{x})=x and a^{\log_a(y)}=y.
The domain of \log_a(x) is x>0, because we cannot take the logarithm of zero or of a negative real number (a^{x} is always positive).
\log_a(1)=0 because a^{0}=1, and \log_a(a)=1 because a^{1}=a.

Laws of logarithms

Here x>0, y>0, a>0, a\neq1.
Law	Statement
Product rule	\log_a(xy)=\log_a x+\log_a y
Quotient rule	\log_a\!\left(\dfrac{x}{y}\right)=\log_a x-\log_a y
Power rule	\log_a(x^{n})=n\log_a x
Change of base	\log_a x=\dfrac{\log_b x}{\log_b a} for positive bases a\neq1, b\neq1

Each law comes from a law of indices. In every proof below, let m=\log_a x and n=\log_a y, so that x=a^{m} and y=a^{n}.

Derivation Product rule

Aim. Prove \log_a(xy)=\log_a x+\log_a y.

Write x and y as powers of a using the definition of a logarithm.
m=\log_a x\Rightarrow x=a^{m},\qquad n=\log_a y\Rightarrow y=a^{n}
Multiply. When powers of the same base are multiplied, the indices are added.
xy=a^{m}\cdot a^{n}=a^{m+n}
Change back to logarithmic form: the power of a that gives xy is m+n.
\log_a(xy)=m+n=\log_a x+\log_a y

Result\log_a(xy)=\log_a x+\log_a y

The log of a product is the sum of the logs. This is why logarithms turn multiplication into addition.

Derivation Quotient rule

Aim. Prove \log_a\!\left(\dfrac{x}{y}\right)=\log_a x-\log_a y.

As before, x=a^{m} and y=a^{n}.
x=a^{m},\qquad y=a^{n}
Divide. When powers of the same base are divided, the indices are subtracted.
\dfrac{x}{y}=\dfrac{a^{m}}{a^{n}}=a^{m-n}
Change back to logarithmic form.
\log_a\!\left(\dfrac{x}{y}\right)=m-n=\log_a x-\log_a y

Result\log_a\!\left(\dfrac{x}{y}\right)=\log_a x-\log_a y

The log of a quotient is the log of the numerator minus the log of the denominator.

Derivation Power rule

Aim. Prove \log_a(x^{k})=k\log_a x for any real number k.

Let m=\log_a x, so x=a^{m}.
x=a^{m}
Raise both sides to the power k. A power of a power multiplies the indices.
x^{k}=(a^{m})^{k}=a^{mk}
Change back to logarithmic form.
\log_a(x^{k})=mk=k\log_a x

Result\log_a(x^{n})=n\log_a x

The power comes down in front of the log. This is the key step for solving equations with the unknown in the exponent.

Derivation Change of base rule

Aim. Prove \log_a x=\dfrac{\log_b x}{\log_b a}.

Let y=\log_a x and write it in exponential form.
a^{y}=x
Take logarithms to the new base b of both sides.
\log_b(a^{y})=\log_b x
Use the power rule on the left.
y\log_b a=\log_b x
Divide by \log_b a, which is not zero because a\neq1.
y=\dfrac{\log_b x}{\log_b a}

Result\log_a x=\dfrac{\log_b x}{\log_b a}

A calculator has only \log (base 10) and \ln (base e) keys. This rule lets us find a log to any base, for example \log_2 10=\dfrac{\ln10}{\ln2}=\dfrac{2.3026}{0.6931}=3.32.

Using the laws

(a) Find \log_2 32. (b) Expand \log\dfrac{x^2y}{z}. (c) Write 2\log3+\log4-\log6 as a single logarithm.

(a) 32=2^{5}, so the power of 2 that gives 32 is 5.
\log_2 32=\log_2 2^{5}=5\log_2 2=5
(b) Quotient rule, then product rule, then power rule.
\log\dfrac{x^2y}{z}=\log(x^2y)-\log z=2\log x+\log y-\log z
(c) Power rule first, then combine with the product and quotient rules.
2\log3+\log4-\log6=\log9+\log4-\log6=\log\dfrac{9\times4}{6}=\log6

\text{(a) }5\qquad\text{(b) }2\log x+\log y-\log z\qquad\text{(c) }\log6

Exponential and logarithmic equations

When the unknown is in an exponent, take the logarithm of both sides and use the power rule to bring it down. When the unknown is inside logarithms, combine them into one logarithm and change to exponential form. Always check that every logarithm in the original equation has a positive input.

Solving equations

Solve (a) 3^{x+1}=20, (b) e^{2x}=7, (c) \log_2(x+1)+\log_2(x-1)=3.

(a) Take \ln of both sides and use the power rule.
(x+1)\ln3=\ln20\;\Rightarrow\;x=\dfrac{\ln20}{\ln3}-1=\dfrac{2.9957}{1.0986}-1=1.727
(b) \ln undoes e.
2x=\ln7\;\Rightarrow\;x=\dfrac{1.9459}{2}=0.973
(c) Product rule, then exponential form.
\log_2\big[(x+1)(x-1)\big]=3\;\Rightarrow\;x^2-1=2^{3}=8\;\Rightarrow\;x=\pm3
Check: x=-3 makes x+1 and x-1 negative, so their logarithms are not defined. Reject it.
x=3

\text{(a) }x\approx1.727\qquad\text{(b) }x\approx0.973\qquad\text{(c) }x=3

1.10.3Graphs of exponential and logarithmic functionsOften examined

Graph of the exponential function y=a^{x}

If the base a is greater than 1, the function increases exponentially at a growth rate of a. This is known as exponential growth.
If the base a is less than 1 (but greater than 0), the function decreases exponentially. This is known as exponential decay.
If the base a is equal to 1, the function trivially becomes y=1 (a horizontal line). That is why the base must satisfy a\neq1.
Since a^{0}=1, every exponential function passes through (0,1). The points (0,1) and (1,a) are always on the graph of y=a^{x}.
An exponential function takes only positive values, and its graph never touches the x-axis: the x-axis (y=0) is a horizontal asymptote.
The domain of the exponential function is the set of all real numbers, \mathbb{R}, whereas its range is the set of positive real numbers, (0,\infty).

Both curves pass through (0,1), stay above the x-axis and approach it on one side. y=0.5^{x} is the reflection of y=2^{x} in the y-axis.

Graph of the logarithmic function y=\log_a x

The logarithmic graph is the reflection of the exponential graph y=a^{x} in the line y=x, because the two functions are inverses. For a>1 its shape is similar to the square root graph: it rises quickly at first and then more and more slowly.
The logarithmic function always passes through the point (1,0) because \log_a(1)=0.
The curve approaches the y-axis but never touches it: x=0 is a vertical asymptote.
The domain of the logarithmic function is the set of all positive real numbers, (0,\infty), whereas its range is the set of all real numbers, \mathbb{R}.
For a>1 the value of the function increases as x increases. For 0<a<1 it decreases as x increases.

y=2^{x} and y=\log_2x are mirror images in y=x. The domain and range swap: \mathbb{R}\to(0,\infty) for the exponential and (0,\infty)\to\mathbb{R} for the logarithm.

Exponential and logarithmic functions compared.
	y=a^{x}	y=\log_a x
domain	\mathbb{R}	(0,\infty)
range	(0,\infty)	\mathbb{R}
fixed point	(0,1)	(1,0)
asymptote	y=0 (the x-axis)	x=0 (the y-axis)
a>1	increasing (growth)	increasing
0<a<1	decreasing (decay)	decreasing

Example 18

Draw the graph of f(x)=e^{-0.5x}.

Make a table of values with a calculator.
\begin{array}{c|ccccccc} x & -5 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline f(x) & 12.18 & 2.72 & 1.65 & 1 & 0.61 & 0.37 & 0.22 \end{array}
Plot the points and join them smoothly. The base is e^{-0.5}\approx0.61<1, so this is exponential decay.

The curve falls from left to right and approaches the x-axis without touching it.

\text{Decay curve through }(0,1),\ \text{domain }\mathbb{R},\ \text{range }(0,\infty)

Example 19

Draw the graphs of (i) f(x)=\ln x and (ii) g(x)=\ln(x+3).

Table of values. \ln x is not defined at x=0.
\begin{array}{c|cccccc} x & 0 & 0.1 & 0.5 & 1 & 4 & 10 \\ \hline f(x)=\ln x & \text{--} & -2.30 & -0.69 & 0 & 1.39 & 2.30 \\ g(x)=\ln(x+3) & 1.10 & 1.13 & 1.25 & 1.39 & 1.95 & 2.56 \end{array}
g(x)=\ln(x+3) is the graph of \ln x moved 3 units to the left. Its domain is x>-3 and its vertical asymptote is x=-3; it crosses the x-axis where x+3=1, at x=-2.

Both curves rise slowly to the right. \ln x hugs the y-axis; \ln(x+3) hugs the line x=-3.

\ln x:\ D=(0,\infty),\ R=\mathbb{R};\qquad \ln(x+3):\ D=(-3,\infty),\ R=\mathbb{R}

Applications of logarithms

Growth and decay models: logarithms describe quantities that grow rapidly at first and then slow down, such as population growth or the spread of diseases.
pH in chemistry: the pH of a solution is the logarithmic measure of the hydrogen ion concentration, \text{pH}=-\log_{10}[\ce{H+}]. For [\ce{H+}]=10^{-3}\,\mathrm{mol\,dm^{-3}}, \text{pH}=3.
Sound intensity (decibels): the decibel scale is logarithmic, \text{decibels}=10\log_{10}\!\left(\dfrac{I}{I_0}\right), where I is the intensity of the sound and I_0 is the reference intensity. A sound 1000 times the reference intensity is 10\log_{10}1000=30 decibels.
Information theory: logarithms measure information content and entropy.
Financial models: logarithms are used in finance, for the time value of money and compound interest.
Computer science: logarithms appear in the analysis of algorithms and data structures (for example binary search needs about \log_2 n steps).

Conclusion. Logarithmic functions are powerful tools for dealing with exponential growth and decay, and for measuring and comparing quantities on vastly different scales. Most of their applications are in engineering and computer technology.

1.10.3Compound interest, exponential growth and decayOften examined

When a quantity grows by the same percentage in each equal time interval, it is multiplied by the same factor each time. Repeated multiplication by a fixed factor is exactly what an exponential function does, so money in a bank, populations and radioactive substances all follow exponential functions.

Derivation Compound interest formula

Aim. Find the amount A(t) after t years when a principal P is invested at an annual interest rate r (as a decimal), compounded annually.

After 1 year the interest Pr is added to the principal.
A(1)=P+Pr=P(1+r)
In year 2 the interest is earned on the new amount P(1+r), so we multiply by (1+r) again.
A(2)=P(1+r)(1+r)=P(1+r)^{2}
Each further year multiplies the amount by (1+r) once more. After t years:
A(t)=P(1+r)^{t}

ResultA(t)=P(1+r)^{t}

The amount is an exponential function of time with base 1+r>1, so it shows exponential growth. For Rs. 30,000 at 8%, A(t)=30000(1.08)^{t}.

For populations the book uses the continuous growth model P(t)=P_0e^{rt}, where P_0 is the initial value (at t=0), r is the growth rate per unit time as a decimal and t is the time.

Exponential growth model: P(t)=P_0e^{rt}, r>0: the quantity increases.
Exponential decay model: P(t)=P_0e^{-rt}, r>0: the quantity decreases.

Derivation Doubling time

Aim. Find the time T for a quantity to double.

Annual compounding. Double the principal means A(T)=2P.
P(1+r)^{T}=2P\;\Rightarrow\;(1+r)^{T}=2
Take the natural log of both sides and use the power rule.
T\ln(1+r)=\ln2\;\Rightarrow\;T=\dfrac{\ln2}{\ln(1+r)}
Continuous growth. Put P(T)=2P_0 in P_0e^{rT} and take \ln (since \ln e=1).
e^{rT}=2\;\Rightarrow\;rT=\ln2\;\Rightarrow\;T=\dfrac{\ln2}{r}

ResultT=\dfrac{\ln2}{\ln(1+r)}\ \ \text{(annual)},\qquad T=\dfrac{\ln2}{r}\ \ \text{(continuous)}

The doubling time does not depend on the starting amount. Since \ln2\approx0.693, a quantity growing continuously at p\% per year doubles in about 70/p years (the “rule of 70”).

Method for solving growth problems

Write the model with the given numbers.
Put the target value equal to the model.
Divide to isolate the power.
Take \ln of both sides and bring the exponent down (power rule).
Divide, evaluate, and answer in words with units.

Example 20

Suppose that Rs. 30,000 is invested at 8% interest compounded annually. In t years it will grow to the amount A(t)=30{,}000\,(1.08)^{t}. (i) How long will it take until there is Rs. 150,000 in the account? (ii) Let T be the time it takes for Rs. 30,000 to double. Find T.

(i) Set A(t)=150{,}000 and isolate the power.
150000=30000(1.08)^{t}\;\Rightarrow\;(1.08)^{t}=\dfrac{150000}{30000}=5
Take the natural log of both sides and use the power rule.
t\ln(1.08)=\ln5\;\Rightarrow\;t=\dfrac{\ln5}{\ln1.08}=\dfrac{1.6094}{0.07696}\approx20.9
So it will take almost 20.9 years for Rs. 30,000 to grow to Rs. 150,000.
(ii) Doubling means A(T)=60{,}000.
60000=30000(1.08)^{T}\;\Rightarrow\;(1.08)^{T}=2
Take logs.
T\ln1.08=\ln2\;\Rightarrow\;T=\dfrac{\ln2}{\ln1.08}=\dfrac{0.6931}{0.07696}\approx9
So the doubling time is about 9 years.

The account grows slowly at first and faster later; it doubles about every 9 years. Amounts are in thousands of rupees.

\text{(i) } t\approx20.9\ \text{years}\qquad\text{(ii) } T\approx9\ \text{years}

Example 21

In 2020, the population of a country was 249 million and the exponential growth rate was 0.9% per year. If P(t)=P_0e^{rt} is the exponential growth function, then (i) find the exponential growth function for the given data; (ii) what would you expect the population to be in the year 2028?

(i) Take t=0 in 2020, so P_0=249 (million). The rate is r=0.9\%=0.009.
P(t)=249\,e^{0.009t}\qquad(\text{a})
(ii) The year 2028 is 8 years after 2020, so t=8. Substitute into (a).
P(8)=249\,e^{0.009\times8}=249\,e^{0.072}\approx249\times1.0747\approx267.6
So the population of the country in 2028 will be about 267.6 million.

P(t)=249e^{0.009t};\qquad P(8)\approx267.6\ \text{million}

1.11.1 to 1.11.3Domain and range of trigonometric functions through graphsOften examined

If a weight attached to a spring is pushed up or pulled down and released, it rises and falls again and again. The weight is said to be oscillating in harmonic motion. If the position y of the weight is graphed against time, the result is a sine or cosine curve. Graphs like this let us read the domain and range of trigonometric functions directly.

Graphs of y=\sin\theta and y=\cos\theta (1.11.1)

The horizontal axis shows \theta (in degrees or radians) and the vertical axis shows \sin\theta or \cos\theta. The points plotted are (\theta,\sin\theta) or (\theta,\cos\theta).

Values to two decimal places. The book's table uses steps of 15^\circ.
\theta	0^\circ	30^\circ	60^\circ	90^\circ	120^\circ	150^\circ	180^\circ	210^\circ	240^\circ	270^\circ	300^\circ	330^\circ	360^\circ
\sin\theta	0	0.5	0.87	1	0.87	0.5	0	-0.5	-0.87	-1	-0.87	-0.5	0
\cos\theta	1	0.87	0.5	0	-0.5	-0.87	-1	-0.87	-0.5	0	0.5	0.87	1

Values to two decimal places. The book's table uses steps of 15^\circ.
\theta	\sin\theta	\cos\theta
0^\circ	0	1
30^\circ	0.5	0.87
60^\circ	0.87	0.5
90^\circ	1	0
120^\circ	0.87	-0.5
150^\circ	0.5	-0.87
180^\circ	0	-1
210^\circ	-0.5	-0.87
240^\circ	-0.87	-0.5
270^\circ	-1	0
300^\circ	-0.87	0.5
330^\circ	-0.5	0.87
360^\circ	0	1

Both waves repeat every 2\pi (360^\circ) and stay between -1 and 1. The cosine curve is the sine curve moved \tfrac{\pi}{2} to the left.

Function	Domain	Range
y=\sin\theta	\mathbb{R}, that is -\infty<\theta<\infty	[-1,1], that is -1\le y\le1
y=\cos\theta	\mathbb{R}, that is -\infty<\theta<\infty	[-1,1], that is -1\le y\le1

Graphs of y=\tan\theta and y=\cot\theta (1.11.2)

\tan\theta=\dfrac{\sin\theta}{\cos\theta} is not defined where \cos\theta=0, that is at odd multiples of \dfrac{\pi}{2}. There the graph has vertical asymptotes. \cot\theta=\dfrac{\cos\theta}{\sin\theta} is not defined where \sin\theta=0, at multiples of \pi.

y=\tan\theta: asymptotes (dashed) at \theta=\pm\tfrac{\pi}{2},\pm\tfrac{3\pi}{2},\dots; the curve takes every real value between them.

y=\cot\theta: asymptotes at \theta=0,\pm\pi,\pm2\pi,\dots (the y-axis is one of them).

Function	Domain	Range
y=\tan\theta	\theta\neq(2n+1)\dfrac{\pi}{2};\ n\in\mathbb{Z}	\mathbb{R}
y=\cot\theta	\theta\neq n\pi;\ n\in\mathbb{Z}	\mathbb{R}

Graphs of y=\sec\theta and y=\operatorname{cosec}\theta (1.11.3)

\sec\theta=\dfrac{1}{\cos\theta} and \operatorname{cosec}\theta=\dfrac{1}{\sin\theta}. Since |\cos\theta|\le1, its reciprocal satisfies |\sec\theta|\ge1, so the graph never enters the band -1<y<1. The same holds for \operatorname{cosec}\theta.

y=\sec\theta: U-shaped pieces touch y=1 and y=-1; asymptotes where \cos\theta=0.

y=\operatorname{cosec}\theta: asymptotes where \sin\theta=0, at \theta=0,\pm\pi,\pm2\pi,\dots

Function	Domain	Range
y=\sec\theta	\theta\neq(2n+1)\dfrac{\pi}{2};\ n\in\mathbb{Z}	y\le-1 or y\ge1, that is (-\infty,-1]\cup[1,\infty)
y=\operatorname{cosec}\theta	\theta\neq n\pi;\ n\in\mathbb{Z}	y\le-1 or y\ge1, that is (-\infty,-1]\cup[1,\infty)

Summary of the six trigonometric functions (n\in\mathbb{Z}).
Function	Domain	Range	Period
\sin\theta	\mathbb{R}	[-1,1]	2\pi
\cos\theta	\mathbb{R}	[-1,1]	2\pi
\tan\theta	\mathbb{R}-\{(2n+1)\tfrac{\pi}{2}\}	\mathbb{R}	\pi
\cot\theta	\mathbb{R}-\{n\pi\}	\mathbb{R}	\pi
\sec\theta	\mathbb{R}-\{(2n+1)\tfrac{\pi}{2}\}	(-\infty,-1]\cup[1,\infty)	2\pi
\operatorname{cosec}\theta	\mathbb{R}-\{n\pi\}	(-\infty,-1]\cup[1,\infty)	2\pi

1.11.4Inverse trigonometric functions

Inverse trigonometric functions, also known as arc functions or anti-trigonometric functions, are the inverses of the six trigonometric functions. They are used to find the angle when the trigonometric ratio is known. For example, \arcsin(x) finds the angle whose sine is x.

A trigonometric function repeats its values, so it is not one-one and has no inverse on its whole domain. We therefore restrict each function to one interval (the principal value interval) on which it is one-one and takes every value of its range once. The inverse is taken on that interval.

Notation, domain and range of the inverse trigonometric functions.
Name	Usual notation	Definition	Domain of x for real result	Range of principal value (radians)	Range (degrees)
arcsine	y=\sin^{-1}(x)	x=\sin(y)	-1\le x\le1	-\tfrac{\pi}{2}\le y\le\tfrac{\pi}{2}	-90^\circ\le y\le90^\circ
arccosine	y=\cos^{-1}(x)	x=\cos(y)	-1\le x\le1	0\le y\le\pi	0^\circ\le y\le180^\circ
arctangent	y=\tan^{-1}(x)	x=\tan(y)	\mathbb{R}	-\tfrac{\pi}{2}<y<\tfrac{\pi}{2}	-90^\circ<y<90^\circ
arccotangent	y=\cot^{-1}(x)	x=\cot(y)	\mathbb{R}	0<y<\pi	0^\circ<y<180^\circ
arcsecant	y=\sec^{-1}(x)	x=\sec(y)	\|x\|\ge1	0\le y<\tfrac{\pi}{2} or \tfrac{\pi}{2}<y\le\pi	0^\circ\le y<90^\circ or 90^\circ<y\le180^\circ
arccosecant	y=\csc^{-1}(x)	x=\csc(y)	\|x\|\ge1	-\tfrac{\pi}{2}\le y<0 or 0<y\le\tfrac{\pi}{2}	-90^\circ\le y<0^\circ or 0^\circ<y\le90^\circ

\sin^{-1}x runs from -\tfrac{\pi}{2} to \tfrac{\pi}{2} and \cos^{-1}x from \pi down to 0, both only for -1\le x\le1.

1.12One-one functions and their inverses through graphsOften examined

One-one function and its graph (1.12.1)

A one-one function is a special function in which every element of the range comes from exactly one element of the domain, that is, the outputs never repeat.

The function f(x)=x-3 is a one-to-one function, since it produces a different answer for every input.
The function g(x)=x^2-1 is not a one-to-one function, since it produces the same output 0 for the two inputs 1 and -1.

Horizontal line test (1.12.2)

The horizontal line test is used to decide whether a function is one-one when its graph is given. Imagine a horizontal line (a horizontal stick) and slide it up and down across the whole graph.

If no horizontal line passes through more than one point of the graph, the function is one-one.
If some horizontal line passes through more than one point of the graph, the function is not one-one.

f(x)=x-3 passes the horizontal line test: every horizontal line meets it exactly once. So f is one-one.

g(x)=x^2-1 fails the test: the line y=3 meets it twice, at x=-2 and x=2. So g is not one-one.

Check point

By using the horizontal line test, check whether the function y=x^3 is a 1-1 function or not.

The graph of y=x^3 rises all the time from left to right. Every horizontal line y=k meets it exactly once, at x=\sqrt[3]{k}.
So it passes the horizontal line test. Algebraically, x_1^3=x_2^3 gives x_1=x_2.
y=x^3\ \text{is a one-one function}

y=x^3\ \text{is one-one}

Inverse of a one-one function (1.12.3)

Suppose f:X\to Y is a one-one function. Since every element y of Y corresponds to precisely one element x of X, the function f determines a “reverse function” g:Y\to X whose domain is Y and range is X. Then f and g imply that:

f(x)=y\ \text{ and }\ g(y)=x,\qquad f(g(y))=y\ \text{ and }\ g(f(x))=x

The function g is given the formal name “inverse of f”. From the above it is clear that \text{Dom}\,f=\text{Rang}\,g and \text{Rang}\,f=\text{Dom}\,g.

Inverse function: Let f be a one-to-one function with domain X and range Y. The inverse of f is the function g with domain Y and range X for which f(g(y))=y for every y in Y and g(f(x))=x for every x in X. It is denoted by f^{-1}, so g(x)=f^{-1}(x).

f\big(f^{-1}(x)\big)=f^{-1}\big(f(x)\big)=x

Properties of the inverse of a one-to-one function (1.12.4)

The function f has an inverse function if and only if f is a one-to-one function.
If the functions f and g are inverses of each other, then both functions are one-to-one.
f and g are inverses of each other if and only if f(g(x))=x for x in the domain of g, and g(f(x))=x for x in the domain of f.
If f and g are inverses of each other, then the domain of f is equal to the range of g, and the range of g is equal to the domain of f.
If f and g are inverses of each other, then their graphs are reflections of each other in the line y=x.
If the point (a,b) is on the graph of f, then the point (b,a) is on the graph of f^{-1}.

Derivation Why the graph of f^{-1} is the reflection of f in y=x

Aim. Show that (b,a) is the mirror image of (a,b) in the line y=x.

The segment joining (a,b) and (b,a) is cut at right angles by y=x, at its midpoint M.

Link the two points through the definition of the inverse.
(a,b)\in f\iff b=f(a)\iff a=f^{-1}(b)\iff(b,a)\in f^{-1}
Find the midpoint M of the segment joining (a,b) and (b,a). Its two coordinates are equal, so M lies on y=x.
M=\left(\dfrac{a+b}{2},\dfrac{b+a}{2}\right)
Find the slope of the segment (for a\neq b). The line y=x has slope 1 and (-1)(1)=-1, so the segment is perpendicular to y=x.
m=\dfrac{a-b}{b-a}=-1
The line y=x cuts the segment at its midpoint and at right angles, so it is the perpendicular bisector. Hence (b,a) is the mirror image of (a,b) in y=x.

Result\text{graph of } f^{-1}=\text{reflection of the graph of } f \text{ in } y=x

To sketch f^{-1}, reflect the graph of f in y=x: swap the coordinates of a few key points (intercepts, end points) and swap the asymptotes (a vertical asymptote x=k of f becomes a horizontal asymptote y=k of f^{-1}).

Example 22

Find the inverse of f(x)=\dfrac{1}{2x-3};\ x\neq\dfrac32, then represent f and f^{-1} graphically.

Since f is a one-to-one function, it has an inverse. Replace x with f^{-1}(x) in f(f^{-1}(x))=x.
f\big(f^{-1}(x)\big)=\dfrac{1}{2f^{-1}(x)-3}=x
Solve for f^{-1}(x): take reciprocals of both sides.
2f^{-1}(x)-3=\dfrac{1}{x}\;\Rightarrow\;2f^{-1}(x)=\dfrac1x+3=\dfrac{1+3x}{x}
Divide by 2.
f^{-1}(x)=\dfrac{1+3x}{2x},\qquad x\neq0
Domains and ranges swap: \text{Dom}\,f=\mathbb{R}-\{\tfrac32\}=\text{Rang}\,f^{-1} and \text{Rang}\,f=\mathbb{R}-\{0\}=\text{Dom}\,f^{-1}.

f has asymptotes x=\tfrac32 and y=0; f^{-1} has asymptotes x=0 and y=\tfrac32. The point (2,1) on f matches (1,2) on f^{-1}.

From the graph it is clear that if any point (a,b) is on the graph of f(x), then the point (b,a) is on the graph of f^{-1}(x).

f^{-1}(x)=\dfrac{1+3x}{2x},\quad x\neq0

Example 23

Given that f(x)=3-4x is one to one, find its inverse and represent f and f^{-1} graphically.

Put y=3-4x and solve for x.
4x=3-y\;\Rightarrow\;x=\dfrac{3-y}{4}
So f^{-1}(y)=\dfrac{3-y}{4}. Replace y with x.
f^{-1}(x)=\dfrac{3-x}{4}
Key points: f(0)=3, so (0,3) is on the graph of f(x). Swapping gives (3,0) on the graph of f^{-1}(x). The two lines meet on y=x where 3-4x=x, at (0.6,0.6).

The steep line is f and the gentle line is f^{-1}; they are reflections of each other in y=x.

Therefore both graphs are reflections of each other in the line y=x.

f^{-1}(x)=\dfrac{3-x}{4}

1.13.1Vertical and horizontal shiftsOften examined

A transformation changes a graph by a rule. A shift is a rigid translation: it does not change the shape or size of the graph, only its location. In this section c is a positive constant added to or subtracted from f or its independent variable x.

Vertical shift: Adds or subtracts a positive constant to or from every y-coordinate while leaving the x-coordinates unchanged: y=f(x)\pm c.
Horizontal shift: Adds or subtracts a positive constant to or from every x-coordinate while leaving the y-coordinates unchanged: y=f(x\pm c).

Table 1.1: vertical and horizontal shifts (c>0).
Original function	Add c to f(x)	Subtract c from f(x)	Add c to x	Subtract c from x
y=f(x)	y=f(x)+c	y=f(x)-c	y=f(x+c)	y=f(x-c)
geometric effect	shifts the graph c units up	shifts the graph c units down	shifts the graph c units left	shifts the graph c units right

Vertical shifts: x^2+2 is 2 units up and x^2-3 is 3 units down.

Horizontal shifts: (x+3)^2 is 3 units left and (x-3)^2 is 3 units right.

Example 24

Sketch the graphs of (a) y=\sqrt{x}, (b) y=\sqrt{x-2}, (c) y=\sqrt{x+2}. Which kind of shift did you observe after sketching the graphs?

(a) y=\sqrt{x} starts at (0,0) and passes through (1,1) and (4,2).
(b) In \sqrt{x-2} the constant 2 is grouped with x and subtracted: shift 2 units right. It starts at (2,0).
(c) In \sqrt{x+2} the constant is added to x: shift 2 units left. It starts at (-2,0).

The graphs show a horizontal shift. The graph of y=\sqrt{x-2} is obtained by moving the graph of y=\sqrt{x} 2 units to the right, while the graph of y=\sqrt{x+2} is obtained by moving it 2 units to the left.

\sqrt{x-2}:\ 2\ \text{units right};\qquad \sqrt{x+2}:\ 2\ \text{units left}

Example 25

Draw the graph of y=|x| and then sketch the graphs of (a) y=|x|-1, (b) y=|x|+1, (c) y=|x-1|, (d) y=|x+1|, (e) y=|x-1|-1, (f) y=|x+1|-1. Which kind of shift did you observe after sketching the graphs?

Table of values for y=|x|.
\begin{array}{c|ccccccc} x & 0 & \pm1 & \pm2 & \pm3 & \pm4 & \pm5 & \pm6 \\ \hline f(x) & 0 & 1 & 2 & 3 & 4 & 5 & 6 \end{array}
Each other graph is the V of y=|x| moved so that its corner (vertex) goes to a new point. Find the corner by making the inside of the modulus zero.

(a) vertical shift 1 unit down; (b) vertical shift 1 unit up.

(e) 1 unit right and 1 unit down; (f) 1 unit left and 1 unit down.

Graph	Corner (vertex)	Shift of y=\|x\|
(a) \|x\|-1	(0,-1)	vertical, 1 unit down
(b) \|x\|+1	(0,1)	vertical, 1 unit up
(c) \|x-1\|	(1,0)	horizontal, 1 unit right
(d) \|x+1\|	(-1,0)	horizontal, 1 unit left
(e) \|x-1\|-1	(1,-1)	1 unit right, 1 unit down
(f) \|x+1\|-1	(-1,-1)	1 unit left, 1 unit down

\text{(a),(b) vertical; (c),(d) horizontal; (e),(f) both}

1.13.2Scaling: stretching and compressing

Scaling is a non-rigid transformation in which the shape and size of the graph are altered. A scale multiplies or divides coordinates, which changes the appearance of the graph as well as its location.

Vertical scaling: Multiplies or divides every y-coordinate by a constant while leaving the x-coordinates unchanged: y=cf(x) or y=\dfrac{f(x)}{c}.
Horizontal scaling: Multiplies or divides every x-coordinate by a constant while leaving the y-coordinates unchanged: y=f(cx) or y=f\!\left(\dfrac{x}{c}\right).

The vertical and horizontal scaling can be combined into one expression. Below, c is a constant that multiplies or divides f or its independent variable x; the book's examples use c=2 or c=-2.

Table 1.2: vertical scaling.
Rule	What happens to the points	Standard name	Example
y=cf(x),\ c>0	every y-value is multiplied by c	vertical stretch by factor c (for c>1); the curve looks narrower	y=2x^2
y=cf(x),\ c<0	y-values multiplied by \|c\| and their signs changed	vertical stretch and reflection in the x-axis	y=-2x^2
y=\dfrac{f(x)}{c},\ c>0	every y-value is divided by c	vertical compression (for c>1); the curve looks wider	y=\dfrac{x^2}{2}
y=\dfrac{f(x)}{c},\ c<0	y-values divided by \|c\| and their signs changed	vertical compression and reflection in the x-axis	y=\dfrac{x^2}{-2}

Multiplying by 2 doubles every height, so (1,1) moves to (1,2); dividing by 2 halves it, so (2,4) moves to (2,2).

A negative constant also turns the graph upside down (reflection in the x-axis).

Table 1.3: horizontal scaling.
Rule	What happens to the points	Book's description	Example
y=f(cx),\ c>0	every x-value is divided by c	compressed horizontally (for c>1)	y=(2x)^3
y=f(cx),\ c<0	x-values divided by \|c\| and signs changed	compressed, opposite direction (reflection in the y-axis)	y=(-2x)^3
y=f\!\left(\dfrac{x}{c}\right),\ c>0	every x-value is multiplied by c	stretched horizontally (for c>1)	y=\left(\dfrac{x}{2}\right)^3
y=f\!\left(\dfrac{x}{c}\right),\ c<0	x-values multiplied by \|c\| and signs changed	stretched, opposite direction (reflection in the y-axis)	y=\left(-\dfrac{x}{2}\right)^3

(1,1) on y=x^3 moves to (0.5,1) on y=(2x)^3 (squeezed towards the y-axis) and to (2,1) on y=(x/2)^3 (pulled away from it).

A negative constant inside also reflects the graph in the y-axis.

Example 26

Draw the graph of y=|x| and then sketch the graphs of (a) y=|1.5x| and y=|-1.5x|, (b) y=\dfrac{|x|}{1.5}, (c) y=\dfrac{|x|}{-1.5}. Which kind of scaling did you observe after sketching the graphs?

Table of values for y=|x|: x=0,\pm1,\pm2,\pm3,\pm4,\pm5,\pm6 give y=0,1,2,3,4,5,6. The graph is a V with its corner at the origin.
(a) |-1.5x|=|1.5x| because the modulus removes the sign. Every x-value is divided by 1.5, so the V is compressed horizontally (it looks narrower). Both give the same graph, with arms of slope \pm1.5.
(b) Every y-value is divided by 1.5. The V becomes wider, with arms of slope \pm\tfrac23.
(c) Dividing by -1.5 does the same and also turns the V upside down (reflection in the x-axis): y=-\tfrac23|x|\le0.

\text{(a) horizontal compression by }1.5;\ \text{(b), (c) vertical scaling by }1.5\text{, (c) also reflected}

The book summarises Example 26 as: in (a) the figure is compressed by changing x-values by 1.5 in both cases; in (b) and (c) the figure is stretched by changing y-values by 1.5 in both cases, but with opposite behaviour.

All transformations of this unit at a glance (c>0).
Equation	Effect on the graph of y=f(x)
y=f(x)+c / y=f(x)-c	shift up / down by c
y=f(x+c) / y=f(x-c)	shift left / right by c
y=cf(x), c>1	y-values multiplied by c (vertical stretch)
y=f(cx), c>1	x-values divided by c (horizontal compression)
y=-f(x)	reflection in the x-axis
y=f(-x)	reflection in the y-axis

Formula sheet

Function notation

f:A\to B,\qquad y=f(x)

A: domain, B: codomain, x: independent variable (input), y: dependent variable (output)

Range and codomain

\text{Rang}\,f\subseteq\text{Codom}\,f

The range (outputs actually produced) is a subset of the codomain; they are equal for an onto function

One-one test

f(x_1)=f(x_2)\;\Longrightarrow\;x_1=x_2

If equal outputs force equal inputs, f is one-one

Onto test

\forall\,y\in B\ \ \exists\,x\in A:\ f(x)=y

Solve y=f(x) for x; if x exists in A for every y in B, f is onto

Even and odd functions

\text{even: } f(-x)=f(x),\qquad \text{odd: } f(-x)=-f(x)

Even graphs are symmetric about the y-axis; odd graphs about the origin

Inverse function identities

f\big(f^{-1}(x)\big)=f^{-1}\big(f(x)\big)=x

f^{-1}: inverse of the bijective function f (not 1/f)

Domain and range of an inverse

\text{Dom}\,f^{-1}=\text{Rang}\,f,\qquad \text{Rang}\,f^{-1}=\text{Dom}\,f

The inverse swaps the roles of inputs and outputs

Inverse of a linear function

f(x)=ax+b\;\Rightarrow\;f^{-1}(x)=\dfrac{x-b}{a}

a\neq0

Inverse of a fractional linear function (check only)

f(x)=\dfrac{ax+b}{cx+d}\;\Rightarrow\;f^{-1}(x)=\dfrac{dx-b}{a-cx}

Domain of f^{-1}: x\neq\frac{a}{c}; range of f^{-1}: y\neq-\frac{d}{c}

Composite function

(f\circ g)(x)=f\big(g(x)\big)

Apply g first, then f; in general f\circ g\neq g\circ f

Linear function and slope

y=ax+b\ (a\neq0),\qquad m=\dfrac{y_2-y_1}{x_2-x_1}

a (or m): slope, b: y-intercept

Point slope form

y-y_1=m(x-x_1)

Line of slope m through (x_1,y_1)

Two-intercept form

\dfrac{x}{a}+\dfrac{y}{b}=1

Line with x-intercept a and y-intercept b

Quadratic function

y=ax^2+bx+c,\qquad a\neq0

a>0: opens upward; a<0: opens downward; (0,c): y-intercept

Axis of symmetry and vertex

x=-\dfrac{b}{2a},\qquad \text{vertex}=\left(-\dfrac{b}{2a},\ \dfrac{4ac-b^2}{4a}\right)

The vertex is the lowest point if a>0 and the highest if a<0

Vertex from the roots

x_{\text{vertex}}=\dfrac{x_1+x_2}{2}

x_1,x_2: x-intercepts from a(x-x_1)(x-x_2)=0

Parabola from points

y=a(x-p)(x-q),\qquad y-r=a(x-p)(x-q)

Through (p,0) and (q,0); or through (p,r) and (q,r)

Modulus function

|x|=\begin{cases}\ \ x, & x\ge0\\ -x, & x<0\end{cases}

Domain \mathbb{R}, range [0,\infty)

Power function

f(x)=kx^{a}

k,a real constants; every y=x^n passes through (1,1)

Definition of logarithm

y=a^{x}\iff x=\log_a y

a>0, a\neq1, y>0; \ln x=\log_e x with e=2.71828\ldots

Inverse properties of logarithms

\log_a(a^{x})=x,\qquad a^{\log_a x}=x,\qquad \log_a1=0,\qquad \log_a a=1

a>0, a\neq1, x>0 in the second

Product rule

\log_a(xy)=\log_a x+\log_a y

x,y>0

Quotient rule

\log_a\!\left(\dfrac{x}{y}\right)=\log_a x-\log_a y

x,y>0

Power rule

\log_a(x^{n})=n\log_a x

x>0, n any real number

Change of base

\log_a x=\dfrac{\log_b x}{\log_b a}

a,b>0, a\neq1, b\neq1

Exponential and logarithmic graphs

y=a^{x}:\ D=\mathbb{R},\ R=(0,\infty);\qquad y=\log_a x:\ D=(0,\infty),\ R=\mathbb{R}

a^{x} passes through (0,1) and (1,a); \log_a x through (1,0)

Compound interest (annual)

A(t)=P(1+r)^{t}

P: principal, r: annual rate as a decimal, t: years

Unit: A in rupees

Exponential growth and decay

P(t)=P_0e^{rt}\ \ \text{(growth)},\qquad P(t)=P_0e^{-rt}\ \ \text{(decay)}

P_0: value at t=0, r>0: rate as a decimal, t: time

Doubling time

T=\dfrac{\ln2}{\ln(1+r)}\ \ \text{(annual)},\qquad T=\dfrac{\ln2}{r}\ \ \text{(continuous)}

r: rate per year as a decimal

Unit: years

pH and decibels

\text{pH}=-\log_{10}[\ce{H+}],\qquad \text{dB}=10\log_{10}\!\left(\dfrac{I}{I_0}\right)

[\ce{H+}]: hydrogen ion concentration; I: sound intensity, I_0: reference intensity

Hyperbolic functions

\sinh x=\dfrac{e^{x}-e^{-x}}{2},\quad \cosh x=\dfrac{e^{x}+e^{-x}}{2},\quad \cosh^2x-\sinh^2x=1

\tanh x=\frac{\sinh x}{\cosh x}, \coth^2x-1=\operatorname{csch}^2x

Domains and ranges of trigonometric functions

\begin{aligned}\sin,\cos&:\ D=\mathbb{R},\ R=[-1,1]\\ \tan&:\ x\neq(2n+1)\tfrac{\pi}{2},\ R=\mathbb{R}\\ \sec&:\ R=(-\infty,-1]\cup[1,\infty)\end{aligned}

\cot and \operatorname{cosec} exclude x=n\pi; n\in\mathbb{Z}

Principal values of inverse sine, cosine, tangent

\sin^{-1}x\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right],\quad \cos^{-1}x\in[0,\pi],\quad \tan^{-1}x\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)

Domains: [-1,1], [-1,1] and \mathbb{R}

Shifts

y=f(x)\pm c\ \ \text{(up/down)},\qquad y=f(x\pm c)\ \ \text{(left/right)}

c>0; the sign inside the bracket is opposite to the direction of the shift

Scaling

y=cf(x)\ \ \text{(vertical)},\qquad y=f(cx)\ \ \text{(horizontal)}

cf(x) multiplies y-values by c; f(cx) divides x-values by c; a negative c also reflects the graph

Key terms

Function: A rule f:A\to B that assigns to each element of A exactly one element of B.
Domain: The set of all possible inputs of a function (the set A in f:A\to B).
Codomain: The set of all possible outcomes a function is allowed to give (the set B in f:A\to B).
Range: The set of outputs a function actually produces; a subset of the codomain.
Image and pre-image: If f(x)=y, then y is the image of x and x is the pre-image of y.
Real valued function: A function whose inputs and outputs are real numbers.
Domain restriction: A value of x for which a function cannot be defined, such as a zero denominator or a negative number under a square root.
Into function: A function in which at least one element of the codomain has no pre-image, so the range is not the whole codomain.
Onto (surjective) function: A function in which every element of the codomain is the image of some element of the domain; range = codomain.
One-one function: A function in which distinct inputs have distinct images: f(x_1)=f(x_2)\Rightarrow x_1=x_2.
Many-one function: A function in which two or more inputs have the same image.
Injective function: In the book, a function that is both one-one and into. In general usage, any one-one function.
Bijective function: A function that is both one-one and onto; it pairs the elements of two sets in a one-one correspondence.
Even function: A function with f(-x)=f(x); its graph is symmetric about the y-axis.
Odd function: A function with f(-x)=-f(x); its graph is symmetric about the origin.
Inverse function: The function f^{-1} that reverses f: if f(x)=y then f^{-1}(y)=x. It exists only when f is bijective.
Composite function: The function f\circ g defined by (f\circ g)(x)=f(g(x)).
Linear function: A polynomial function of degree one, y=ax+b with a\neq0; its graph is a straight line.
Non-linear function: A function whose graph is a curve, so its slope changes from point to point.
Quadratic function: A polynomial function of degree two, y=ax^2+bx+c with a\neq0; its graph is a parabola.
Parabola: The U-shaped graph of a quadratic function, symmetric about its axis.
Vertex: The turning point of a parabola (lowest if a>0, highest if a<0), at x=-\frac{b}{2a}.
Axis of symmetry: The vertical line x=-\frac{b}{2a} through the vertex about which a parabola is symmetric.
Square root function: A function of the form y=\sqrt{u}; the expression u must be non-negative.
Zero product property: If a product of factors is zero, at least one factor is zero; used to find x-intercepts from factors.
Intercepts: The points where a graph meets the axes: the x-intercept (put y=0) and the y-intercept (put x=0).
Graphical solution: The coordinates of the intersection point(s) of two graphs drawn on the same axes.
Modulus (absolute value) function: The function |x|, which gives the size of x regardless of its sign.
Algebraic function: A function built from x by addition, subtraction, multiplication, division, powers and roots only.
Polynomial function: A function a_nx^n+\cdots+a_1x+a_0 with constant coefficients and a non-negative integer n.
Rational function: A function \frac{p(x)}{q(x)} where p and q are polynomials and q(x)\neq0.
Power function: A function of the form f(x)=kx^{a} with real constants k and a.
Transcendental function: A function that is not algebraic, such as exponential, logarithmic, trigonometric and hyperbolic functions.
Exponential function: A function y=a^{x} with a>0, a\neq1; the variable is in the exponent.
Exponential growth and decay: Growth when the base a>1 (or P_0e^{rt}); decay when 0<a<1 (or P_0e^{-rt}).
Logarithm: \log_a y is the power to which a must be raised to give y: x=\log_a y\iff a^{x}=y.
Natural, common and binary logarithms: Logarithms to the bases e (\ln x), 10 (\log_{10}x) and 2 (\log_2x).
Euler's number: The constant e=2.71828\ldots, the base of natural logarithms.
Compound interest: Interest added to the principal each period so that later interest is earned on it too: A=P(1+r)^{t}.
Doubling time: The time a quantity growing exponentially takes to double; it does not depend on the starting amount.
Asymptote: A line that a graph approaches more and more closely without touching, such as y=0 for y=a^{x}.
Periodic function: A function whose values repeat at regular intervals, such as \sin x (period 2\pi) and \tan x (period \pi).
Inverse trigonometric function: An arc function such as \sin^{-1}x that gives the angle (principal value) whose trigonometric ratio is x.
Principal value: The value of an inverse trigonometric function taken from the restricted interval on which the trigonometric function is one-one.
Horizontal line test: A function is one-one if and only if no horizontal line meets its graph more than once.
Shift (translation): A rigid transformation that moves a graph without changing its shape: vertical f(x)\pm c, horizontal f(x\pm c).
Scaling: A non-rigid transformation that multiplies or divides the y-coordinates (cf(x)) or the x-coordinates (f(cx)) and so changes the shape of a graph.
Hyperbolic functions: \sinh x=\frac{e^{x}-e^{-x}}{2}, \cosh x=\frac{e^{x}+e^{-x}}{2} and the functions built from them, such as \tanh, \coth and \operatorname{csch}.

Exercise 1.1p. 14–15

Domain, range, types of functions and inverse functions. For a formula, remove the inputs that make it fail (zero denominator, negative under a square root); for the type, test one-one with f(x_1)=f(x_2) and onto by solving y=f(x) for x.

Find the domain of following functions.

Solution

(i) f(x)=x^2-6

f is a polynomial. There is no denominator and no root, so every real number can be substituted.
So the function is defined for all real x.
\text{Dom}\,f=\mathbb{R}=(-\infty,\infty)

The parabola extends without break to the left and right: every x is allowed.

\text{Dom}\,f=\mathbb{R}

(ii) g(x)=\dfrac{x}{x+3}

A fraction is not defined when its denominator is zero.
x+3=0\;\Rightarrow\;x=-3
Remove this value from the real numbers.
\text{Dom}\,g=\mathbb{R}-\{-3\}

The graph breaks at the vertical asymptote x=-3, the excluded value.

\text{Dom}\,g=\mathbb{R}-\{-3\}

(iii) h(x)=\dfrac{x+4}{x^2-9}

Set the denominator equal to zero and factorise it (difference of squares).
x^2-9=0\;\Rightarrow\;(x-3)(x+3)=0\;\Rightarrow\;x=3\ \text{or}\ x=-3
Both values make the denominator zero, so both are removed.
\text{Dom}\,h=\mathbb{R}-\{-3,3\}

Two vertical asymptotes, at x=-3 and x=3.

\text{Dom}\,h=\mathbb{R}-\{-3,3\}

(iv) i(x)=\dfrac{x}{5x+2}

The denominator must not be zero.
5x+2=0\;\Rightarrow\;x=-\dfrac25
Remove this value.
\text{Dom}\,i=\mathbb{R}-\left\{-\dfrac25\right\}

\text{Dom}\,i=\mathbb{R}-\left\{-\tfrac25\right\}

(v) j(x)=\dfrac{x}{x^2+4}

Check whether the denominator can be zero. A square is never negative, so x^2+4\ge4 for every real x.
x^2+4=0\;\Rightarrow\;x^2=-4,\ \text{which has no real solution}
The denominator is never zero, so no real number is excluded.
\text{Dom}\,j=\mathbb{R}

A smooth curve with no break: the denominator never vanishes.

\text{Dom}\,j=\mathbb{R}

(vi) k(x)=\sqrt{x+1}

The expression under the square root must be non-negative.
x+1\ge0\;\Rightarrow\;x\ge-1
Write the answer in interval notation. -1 is included (square bracket).
\text{Dom}\,k=[-1,\infty)

The graph starts at (-1,0) and exists only to its right.

\text{Dom}\,k=[-1,\infty)

Find the domain and range of the functions.

Solution

(i) f(x)=x+7

Domain. A linear function is defined for every real x.
D_f=\mathbb{R}
Range. For any real y, the input x=y-7 gives f(x)=y. So every real number is an output.
R_f=\mathbb{R}

D_f=\mathbb{R},\qquad R_f=\mathbb{R}

(ii) f(x)=2x^2+1

Domain. A polynomial is defined for every real x.
D_f=\mathbb{R}
Range. x^2\ge0, so 2x^2\ge0 and 2x^2+1\ge1. The value 1 is reached at x=0, and f grows without limit.
R_f=[1,\infty)

The lowest point is the vertex (0,1).

D_f=\mathbb{R},\qquad R_f=[1,\infty)

(iii) f(x)=2\sqrt{x-5}

Domain. The expression under the root must be non-negative.
x-5\ge0\;\Rightarrow\;x\ge5,\qquad D_f=[5,\infty)
Range. \sqrt{x-5}\ge0, so 2\sqrt{x-5}\ge0. It equals 0 at x=5 and grows without limit.
R_f=[0,\infty)

D_f=[5,\infty),\qquad R_f=[0,\infty)

(iv) f(x)=|x-2|-3

Domain. The modulus of any real number exists.
D_f=\mathbb{R}
Range. |x-2|\ge0, so |x-2|-3\ge-3. The minimum -3 occurs at x=2 (the corner of the V).
R_f=[-3,\infty)
Graph. V-shape with corner (2,-3); it cuts the x-axis where |x-2|=3, at x=-1 and x=5.

D_f=\mathbb{R},\qquad R_f=[-3,\infty)

(v) f(x)=1+\sin x

Domain. \sin x is defined for every real x.
D_f=\mathbb{R}
Range. Add 1 to every part of -1\le\sin x\le1.
0\le1+\sin x\le2\;\Rightarrow\;R_f=[0,2]

The sine wave lifted 1 unit up: it oscillates between 0 and 2.

D_f=\mathbb{R},\qquad R_f=[0,2]

(vi) f(x)=3+\sqrt{x-2}

Domain. The expression under the root must be non-negative.
x-2\ge0\;\Rightarrow\;x\ge2,\qquad D_f=[2,\infty)
Range. \sqrt{x-2}\ge0, so 3+\sqrt{x-2}\ge3. The value 3 occurs at x=2.
R_f=[3,\infty)

D_f=[2,\infty),\qquad R_f=[3,\infty)

(vii) f(x)=\dfrac{3e^{x}}{7}

Domain. e^{x} is defined for every real x.
D_f=\mathbb{R}
Range. e^{x}>0 for all x and takes every positive value, so \dfrac{3}{7}e^{x} also takes every positive value but never 0.
R_f=(0,\infty)

The curve approaches the x-axis on the left but never touches it.

D_f=\mathbb{R},\qquad R_f=(0,\infty)

(viii) f(x)=\dfrac{x^2-16}{x+4}

Domain. The denominator must not be zero.
x+4=0\;\Rightarrow\;x=-4,\qquad D_f=\mathbb{R}-\{-4\}
Simplify. Factorise the numerator. For x\neq-4 the factor (x+4) cancels.
f(x)=\dfrac{(x-4)(x+4)}{x+4}=x-4,\qquad x\neq-4
Range. The line y=x-4 gives every real number except the value at the removed point x=-4, which would be -4-4=-8.
R_f=\mathbb{R}-\{-8\}

A straight line with a hole (hollow dot) at (-4,-8).

D_f=\mathbb{R}-\{-4\},\qquad R_f=\mathbb{R}-\{-8\}

(ix) f(x)=(x-1)^2+1

Domain. A polynomial, so all real numbers.
D_f=\mathbb{R}
Range. (x-1)^2\ge0, so f(x)\ge1. The minimum 1 occurs at x=1.
R_f=[1,\infty)

The parabola y=x^2 moved 1 unit right and 1 unit up; vertex (1,1).

D_f=\mathbb{R},\qquad R_f=[1,\infty)

(x) f(x)=\dfrac{1}{x-1}

Domain. The denominator must not be zero.
x-1=0\;\Rightarrow\;x=1,\qquad D_f=\mathbb{R}-\{1\}
Range. Solve y=\dfrac{1}{x-1} for x. The result exists for every y except y=0 (a fraction with numerator 1 is never 0).
x=1+\dfrac1y,\ y\neq0\;\Rightarrow\;R_f=\mathbb{R}-\{0\}

Vertical asymptote x=1 (excluded input) and horizontal asymptote y=0 (excluded output).

D_f=\mathbb{R}-\{1\},\qquad R_f=\mathbb{R}-\{0\}

(xi) f(x)=\dfrac{x-2}{x+3}

Domain. The denominator must not be zero.
x+3=0\;\Rightarrow\;x=-3,\qquad D_f=\mathbb{R}-\{-3\}
Range. Put y=\dfrac{x-2}{x+3} and solve for x.
y(x+3)=x-2\;\Rightarrow\;xy-x=-2-3y\;\Rightarrow\;x=\dfrac{3y+2}{1-y}
This x exists for every y except y=1.
R_f=\mathbb{R}-\{1\}

D_f=\mathbb{R}-\{-3\},\qquad R_f=\mathbb{R}-\{1\}

(xii) f(x)=\dfrac{x^2-x-6}{x-3}

Domain. The denominator must not be zero.
x-3=0\;\Rightarrow\;x=3,\qquad D_f=\mathbb{R}-\{3\}
Simplify. Factorise the numerator and cancel (x-3) for x\neq3.
f(x)=\dfrac{(x-3)(x+2)}{x-3}=x+2,\qquad x\neq3
Range. The line y=x+2 takes every real value except the one at x=3, which would be 3+2=5.
R_f=\mathbb{R}-\{5\}

A straight line with a hole at (3,5).

D_f=\mathbb{R}-\{3\},\qquad R_f=\mathbb{R}-\{5\}

Given that A=\{0,1,2,3\}, B=\{p,q,r,s\} and f=\{(0,p),(1,q),(2,r),(3,s)\}. Check whether the function is one to one, onto and/or into.

Solution

Each element of A goes to a different element of B, and every element of B is reached.

One-one. The images of 0,1,2,3 are p,q,r,s. No two elements of A share an image, so distinct elements have distinct images. Hence f is one-one.
Onto or into. The range is the set of images.
\text{Rang}\,f=\{p,q,r,s\}=B
Every element of B has a pre-image, so f is onto. It is therefore not into.
A function that is one-one and onto is bijective.

f\ \text{is one-one and onto (bijective); it is not into}

A=\{2,3,4,5\}, B=\{b,c,d,e\}. The function is defined as f=\{(2,b),(3,c),(4,e),(5,e)\}. Check whether the function is one to one, into or onto.

Solution

Two arrows end at e, and no arrow reaches d.

One-one. f(4)=e and f(5)=e: two different elements of A have the same image. So f is not one-one (it is many-one).
Onto or into. Collect the images.
\text{Rang}\,f=\{b,c,e\}\neq\{b,c,d,e\}=B
The element d of B has no pre-image, so f is not onto. It is an into function.

f\ \text{is into (not onto) and not one-one: a many-one into function}

Check whether the functions are one-to-one or not.

Solution

(i) f(x)=4x-7

Assume two inputs give the same output and simplify.
f(x_1)=f(x_2)\;\Rightarrow\;4x_1-7=4x_2-7\;\Rightarrow\;4x_1=4x_2\;\Rightarrow\;x_1=x_2
Equal outputs force equal inputs, so f is one-one. On the graph every horizontal line meets the line once.

f(x)=4x-7\ \text{is one-one}

(ii) f(x)=6x^2+2

Assume f(x_1)=f(x_2).
6x_1^2+2=6x_2^2+2\;\Rightarrow\;x_1^2=x_2^2\;\Rightarrow\;x_1=\pm x_2
This does not force x_1=x_2. A counterexample proves it.
f(1)=6(1)^2+2=8=f(-1),\quad \text{but } 1\neq-1
So f is not one-one. The horizontal line y=8 meets the graph twice.

f(x)=6x^2+2\ \text{is not one-one}

(iii) f(x)=\dfrac{x^3-1}{2}

Assume f(x_1)=f(x_2) and simplify.
\dfrac{x_1^3-1}{2}=\dfrac{x_2^3-1}{2}\;\Rightarrow\;x_1^3=x_2^3
A real number has only one real cube root, so the cubes can be equal only if the numbers are equal.
x_1=x_2
Hence f is one-one. The graph always rises, so it passes the horizontal line test.

f(x)=\dfrac{x^3-1}{2}\ \text{is one-one}

Check the type of function g(x)=2x^2+3x+1 if \text{Dom}\,g(x)=\{0,1,2,3\} and \text{Rang}\,g(x)=\{1,6,15,28,35\}.

Solution

Find the image of each element of the domain.
\begin{aligned}g(0)&=1\\ g(1)&=2+3+1=6\\ g(2)&=8+6+1=15\\ g(3)&=18+9+1=28\end{aligned}
One-one. The four images 1,6,15,28 are all different, so g is one-one.
Into. The set of images is \{1,6,15,28\}. The element 35 of the given set has no pre-image.
\{1,6,15,28\}\neq\{1,6,15,28,35\}
So g is not onto; it is into. A function that is one-one and into is called injective in the book (1.2.4).

Each input has its own image; 35 receives no arrow.

g\ \text{is one-one and into, that is, injective}

Note. The book calls \{1,6,15,28,35\} the range, but the actual range of g is \{1,6,15,28\}. The given set must be the codomain, and the answer is based on that.

Find the type of the function h(x)=2x+1 defined on h:\mathbb{R}\to\mathbb{R}.

Solution

One-one. Assume h(x_1)=h(x_2).
2x_1+1=2x_2+1\;\Rightarrow\;x_1=x_2
So h is one-one.
Onto. Take any y\in\mathbb{R} (the codomain) and solve y=2x+1 for x.
x=\dfrac{y-1}{2}\in\mathbb{R}
Check that this input really gives y.
h\!\left(\dfrac{y-1}{2}\right)=2\cdot\dfrac{y-1}{2}+1=y-1+1=y
Every real y has a pre-image, so h is onto. Being one-one and onto, h is bijective.

A line that is not horizontal covers every y exactly once.

h\ \text{is one-one and onto, that is, bijective}

If f:A\to B is defined by f(x)=\dfrac{x-2}{x-3} for all x\in A where A=\mathbb{R}-\{3\} and B=\mathbb{R}-\{1\}. Then show that the function f is bijective.

Solution

One-one

Assume f(x_1)=f(x_2) for x_1,x_2\in A and cross-multiply.
\dfrac{x_1-2}{x_1-3}=\dfrac{x_2-2}{x_2-3}\;\Rightarrow\;(x_1-2)(x_2-3)=(x_2-2)(x_1-3)
Expand both sides.
x_1x_2-3x_1-2x_2+6=x_1x_2-3x_2-2x_1+6
Cancel x_1x_2 and 6, then collect terms.
-3x_1-2x_2=-3x_2-2x_1\;\Rightarrow\;-x_1=-x_2\;\Rightarrow\;x_1=x_2
So f is one-one.

Onto

Take any y\in B, so y\neq1. Solve y=\dfrac{x-2}{x-3} for x.
y(x-3)=x-2\;\Rightarrow\;xy-x=3y-2\;\Rightarrow\;x=\dfrac{3y-2}{y-1}
This x exists because y\neq1. It also lies in A: if it were 3, then 3y-2=3y-3, that is -2=-3, which is false. So x\neq3.
Check that this x is mapped to y.
f\!\left(\dfrac{3y-2}{y-1}\right)=\dfrac{\frac{3y-2}{y-1}-2}{\frac{3y-2}{y-1}-3}=\dfrac{3y-2-2y+2}{3y-2-3y+3}=\dfrac{y}{1}=y
Every element of B has a pre-image in A, so f is onto.

Since f is both one-one and onto, f is bijective.

The graph misses exactly x=3 (not in A) and y=1 (not in B), and every horizontal line y\neq1 meets it once.

f\ \text{is one-one and onto, hence bijective}

92026 Annual I, Set B, Q.2 (i)2026 Annual I, Set B, Q.2 (xii) OR2026 Annual I, Set D, Q.2 (i)

Find the domain and range of inverse functions when: (i) f(x)=4x-3 (ii) f(x)=\dfrac{x}{x-5} (iii) f(x)=\dfrac{x+2}{x-1} (iv) f(x)=\sqrt{x+2} (v) f(x)=x^2+6 (vi) f(x)=\dfrac{2x-1}{x+4}. Also prove that f(f^{-1}(x))=f^{-1}(f(x))=x.

Solution

(i) f(x)=4x-3

Inverse. Put y=4x-3 and solve for x.
4x=y+3\;\Rightarrow\;x=\dfrac{y+3}{4}\;\Rightarrow\;f^{-1}(x)=\dfrac{x+3}{4}
Domain and range. f^{-1} is linear, so it is defined for all real x and takes all real values. (Also \text{Rang}\,f^{-1}=\text{Dom}\,f=\mathbb{R}.)
\text{Dom}\,f^{-1}=\mathbb{R},\qquad \text{Rang}\,f^{-1}=\mathbb{R}
Proof. Substitute each function into the other.
f\big(f^{-1}(x)\big)=4\left(\dfrac{x+3}{4}\right)-3=x+3-3=x
f^{-1}\big(f(x)\big)=\dfrac{(4x-3)+3}{4}=\dfrac{4x}{4}=x
Hence f(f^{-1}(x))=f^{-1}(f(x))=x.

The two lines are reflections in y=x and meet on it at (1,1).

f^{-1}(x)=\dfrac{x+3}{4};\ \ \text{Dom}\,f^{-1}=\mathbb{R},\ \text{Rang}\,f^{-1}=\mathbb{R}

(ii) f(x)=\dfrac{x}{x-5}

\text{Dom}\,f=\mathbb{R}-\{5\}. Inverse. Put y=\dfrac{x}{x-5}, clear the fraction and collect the x terms.
y(x-5)=x\;\Rightarrow\;xy-x=5y\;\Rightarrow\;x(y-1)=5y\;\Rightarrow\;x=\dfrac{5y}{y-1}
Replace y with x.
f^{-1}(x)=\dfrac{5x}{x-1}
Domain and range. f^{-1} is not defined at x=1. Its range is the domain of f.
\text{Dom}\,f^{-1}=\mathbb{R}-\{1\},\qquad \text{Rang}\,f^{-1}=\mathbb{R}-\{5\}
Proof. Multiply the top and bottom of each compound fraction by the small denominator.
f\big(f^{-1}(x)\big)=\dfrac{\frac{5x}{x-1}}{\frac{5x}{x-1}-5}=\dfrac{5x}{5x-5(x-1)}=\dfrac{5x}{5}=x
f^{-1}\big(f(x)\big)=\dfrac{5\cdot\frac{x}{x-5}}{\frac{x}{x-5}-1}=\dfrac{5x}{x-(x-5)}=\dfrac{5x}{5}=x

f: asymptotes x=5, y=1. f^{-1}: asymptotes x=1, y=5. They are mirror images in y=x.

f^{-1}(x)=\dfrac{5x}{x-1};\ \ \text{Dom}\,f^{-1}=\mathbb{R}-\{1\},\ \text{Rang}\,f^{-1}=\mathbb{R}-\{5\}

(iii) f(x)=\dfrac{x+2}{x-1}

\text{Dom}\,f=\mathbb{R}-\{1\}. Inverse. Put y=\dfrac{x+2}{x-1} and solve for x.
y(x-1)=x+2\;\Rightarrow\;xy-x=y+2\;\Rightarrow\;x(y-1)=y+2\;\Rightarrow\;x=\dfrac{y+2}{y-1}
Replace y with x. The inverse has the same formula as f: this function is its own inverse.
f^{-1}(x)=\dfrac{x+2}{x-1}
Domain and range.
\text{Dom}\,f^{-1}=\mathbb{R}-\{1\},\qquad \text{Rang}\,f^{-1}=\text{Dom}\,f=\mathbb{R}-\{1\}
Proof. Since f^{-1}=f, both compositions are f(f(x)). Multiply top and bottom by (x-1).
f\big(f^{-1}(x)\big)=f^{-1}\big(f(x)\big)=\dfrac{\frac{x+2}{x-1}+2}{\frac{x+2}{x-1}-1}=\dfrac{x+2+2x-2}{x+2-x+1}=\dfrac{3x}{3}=x

The graph is symmetric about y=x itself (for example (-2,0) and (0,-2) both lie on it), which is why f=f^{-1}.

f^{-1}(x)=\dfrac{x+2}{x-1};\ \ \text{Dom}\,f^{-1}=\mathbb{R}-\{1\},\ \text{Rang}\,f^{-1}=\mathbb{R}-\{1\}

(iv) f(x)=\sqrt{x+2}

For f: x+2\ge0, so \text{Dom}\,f=[-2,\infty), and a square root is never negative, so \text{Rang}\,f=[0,\infty).
Inverse. Put y=\sqrt{x+2} (with y\ge0) and square both sides.
y^2=x+2\;\Rightarrow\;x=y^2-2\;\Rightarrow\;f^{-1}(x)=x^2-2,\quad x\ge0
Domain and range. The domain of f^{-1} is the range of f, and its range is the domain of f. (On x\ge0, x^2-2 runs from -2 upwards.)
\text{Dom}\,f^{-1}=[0,\infty),\qquad \text{Rang}\,f^{-1}=[-2,\infty)
Proof. For x\ge0, \sqrt{x^2}=|x|=x.
f\big(f^{-1}(x)\big)=\sqrt{(x^2-2)+2}=\sqrt{x^2}=x
For x\ge-2:
f^{-1}\big(f(x)\big)=\big(\sqrt{x+2}\big)^2-2=x+2-2=x

f starts at (-2,0) and f^{-1} at (0,-2): swapped coordinates. They meet at (2,2) on y=x.

f^{-1}(x)=x^2-2\ (x\ge0);\ \ \text{Dom}\,f^{-1}=[0,\infty),\ \text{Rang}\,f^{-1}=[-2,\infty)

(v) f(x)=x^2+6

On all of \mathbb{R} this function is not one-one (f(1)=f(-1)=7), so it has no inverse there. Restrict the domain to x\ge0, where it is one-one. Then \text{Dom}\,f=[0,\infty) and \text{Rang}\,f=[6,\infty).
Inverse. Put y=x^2+6 and solve for x, taking the non-negative root because x\ge0.
x^2=y-6\;\Rightarrow\;x=\sqrt{y-6}\;\Rightarrow\;f^{-1}(x)=\sqrt{x-6}
Domain and range. x-6\ge0 gives the domain; a square root is never negative.
\text{Dom}\,f^{-1}=[6,\infty),\qquad \text{Rang}\,f^{-1}=[0,\infty)
Proof. For x\ge6:
f\big(f^{-1}(x)\big)=\big(\sqrt{x-6}\big)^2+6=x-6+6=x
For x\ge0, \sqrt{x^2}=x:
f^{-1}\big(f(x)\big)=\sqrt{(x^2+6)-6}=\sqrt{x^2}=x

The right half of the parabola and its reflection in y=x. They never meet, since x^2+6>x for all x.

f^{-1}(x)=\sqrt{x-6};\ \ \text{Dom}\,f^{-1}=[6,\infty),\ \text{Rang}\,f^{-1}=[0,\infty)

(vi) f(x)=\dfrac{2x-1}{x+4}

\text{Dom}\,f=\mathbb{R}-\{-4\}. Inverse. Put y=\dfrac{2x-1}{x+4} and solve for x.
y(x+4)=2x-1\;\Rightarrow\;xy-2x=-1-4y\;\Rightarrow\;x(y-2)=-(4y+1)\;\Rightarrow\;x=\dfrac{4y+1}{2-y}
Replace y with x.
f^{-1}(x)=\dfrac{4x+1}{2-x}
Domain and range.
\text{Dom}\,f^{-1}=\mathbb{R}-\{2\},\qquad \text{Rang}\,f^{-1}=\text{Dom}\,f=\mathbb{R}-\{-4\}
Proof. Multiply top and bottom by (2-x).
f\big(f^{-1}(x)\big)=\dfrac{2\cdot\frac{4x+1}{2-x}-1}{\frac{4x+1}{2-x}+4}=\dfrac{8x+2-2+x}{4x+1+8-4x}=\dfrac{9x}{9}=x
Multiply top and bottom by (x+4).
f^{-1}\big(f(x)\big)=\dfrac{4\cdot\frac{2x-1}{x+4}+1}{2-\frac{2x-1}{x+4}}=\dfrac{8x-4+x+4}{2x+8-2x+1}=\dfrac{9x}{9}=x

f: asymptotes x=-4, y=2. f^{-1}: asymptotes x=2, y=-4. Both curves touch y=x at (-1,-1).

f^{-1}(x)=\dfrac{4x+1}{2-x};\ \ \text{Dom}\,f^{-1}=\mathbb{R}-\{2\},\ \text{Rang}\,f^{-1}=\mathbb{R}-\{-4\}

Note. Part (v): f(x)=x^2+6 is not one-one on \mathbb{R}, so its domain is restricted to x\ge0 (the usual choice). With x\le0 instead, f^{-1}(x)=-\sqrt{x-6} and \text{Rang}\,f^{-1}=(-\infty,0].

Exercise 1.2p. 24–25

Plotting linear, quadratic, power and modulus graphs, and reading intersections. Always make a table of values, mark intercepts and vertices, and draw every pair of functions on the same axes.

Plot the graph of the functions:

Solution

(i) f(x)=3x-2

A linear function, so the graph is a straight line with slope 3 and y-intercept -2. Make a table of values.
\begin{array}{c|cccc} x & -1 & 0 & 1 & 2 \\ \hline f(x) & -5 & -2 & 1 & 4 \end{array}
Intercepts: x=0 gives (0,-2); 3x-2=0 gives x=\tfrac23, so \left(\tfrac23,0\right).
Plot the points and draw a straight line through them.

\text{Straight line through }(0,-2)\text{ and }\left(\tfrac23,0\right)

(ii) f(x)=3x

A linear function with b=0, so the line passes through the origin with slope 3.
\begin{array}{c|ccccc} x & -2 & -1 & 0 & 1 & 2 \\ \hline f(x) & -6 & -3 & 0 & 3 & 6 \end{array}
Plot the points and join them with a straight line.

\text{Straight line through the origin with slope }3

(iii) f(x)=1-2x

A linear function with slope -2 (the line falls to the right) and y-intercept 1.
\begin{array}{c|cccc} x & -1 & 0 & 1 & 2 \\ \hline f(x) & 3 & 1 & -1 & -3 \end{array}
Intercepts: (0,1), and 1-2x=0 gives \left(\tfrac12,0\right).

\text{Straight line through }(0,1)\text{ and }\left(\tfrac12,0\right)

(iv) g(x)=x^2+4

A quadratic with a=1>0, b=0, c=4: a parabola opening upward, axis x=-\tfrac{b}{2a}=0, vertex (0,4).
Table of values (symmetric about x=0).
\begin{array}{c|ccccccc} x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline g(x) & 13 & 8 & 5 & 4 & 5 & 8 & 13 \end{array}
x^2+4\ge4>0, so the graph never meets the x-axis (no x-intercepts).

\text{Parabola opening upward, vertex }(0,4)

(v) g(x)=x^2-x-6

a=1>0, so the parabola opens upward. Axis and vertex:
x=-\dfrac{-1}{2(1)}=\dfrac12,\qquad g\!\left(\tfrac12\right)=\tfrac14-\tfrac12-6=-\tfrac{25}{4}
x-intercepts from x^2-x-6=(x-3)(x+2)=0: x=3, x=-2. The y-intercept is (0,-6).
Table of values.
\begin{array}{c|cccccccc} x & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline g(x) & 6 & 0 & -4 & -6 & -6 & -4 & 0 & 6 \end{array}

\text{Vertex }\left(\tfrac12,-\tfrac{25}{4}\right),\ x\text{-intercepts }-2,\,3,\ y\text{-intercept }-6

(vi) g(x)=\sqrt{2x+1}

The expression under the root must be \ge0: 2x+1\ge0, so x\ge-\tfrac12. The graph starts at \left(-\tfrac12,0\right).
Choose x-values that make 2x+1 a perfect square.
\begin{array}{c|ccccc} x & -0.5 & 0 & 1.5 & 4 & 7.5 \\ \hline g(x) & 0 & 1 & 2 & 3 & 4 \end{array}
Join the points with a smooth curve rising slowly to the right. Domain \left[-\tfrac12,\infty\right), range [0,\infty).

\text{Half-parabola starting at }\left(-\tfrac12,0\right)

Plot the graph of following functions.

Solution

(i) f(x)=-x^2+1

a=-1<0, so the parabola opens downward. Axis x=0, vertex (0,1).
x-intercepts: 1-x^2=0 gives x=\pm1.
\begin{array}{c|ccccc} x & -2 & -1 & 0 & 1 & 2 \\ \hline f(x) & -3 & 0 & 1 & 0 & -3 \end{array}

\text{Parabola opening downward, vertex }(0,1)

(ii) f(x)=2x^3

A power function with an odd power, so the graph passes through the origin and is symmetric about it: f(-x)=-f(x).
\begin{array}{c|ccccccccc} x & -2 & -1.5 & -1 & -0.5 & 0 & 0.5 & 1 & 1.5 & 2 \\ \hline f(x) & -16 & -6.75 & -2 & -0.25 & 0 & 0.25 & 2 & 6.75 & 16 \end{array}
The curve is flat near O and rises steeply for larger |x|.

\text{S-shaped cubic through }(0,0),\ (1,2),\ (2,16)

(iii) f(x)=1+x^{-2}

f(x)=1+\dfrac{1}{x^2} is not defined at x=0. Since \dfrac{1}{x^2}>0, every output is greater than 1.
The power is even, so f(-x)=f(x): the graph is symmetric about the y-axis.
\begin{array}{c|cccccccc} x & -3 & -2 & -1 & -0.5 & 0.5 & 1 & 2 & 3 \\ \hline f(x) & 1.11 & 1.25 & 2 & 5 & 5 & 2 & 1.25 & 1.11 \end{array}
As x\to0 the values grow without limit (vertical asymptote x=0); as |x| grows they approach 1 (horizontal asymptote y=1).

\text{Two branches above }y=1;\ \text{asymptotes } x=0,\ y=1

(iv) f(x)=3x^{\frac12}

f(x)=3\sqrt{x} is defined for x\ge0. Choose perfect squares for x.
\begin{array}{c|cccc} x & 0 & 1 & 4 & 9 \\ \hline f(x) & 0 & 3 & 6 & 9 \end{array}
Plot and join with a smooth curve that starts at the origin and rises more and more slowly.

\text{Curve from }(0,0)\text{ through }(1,3),(4,6),(9,9)

(v) f(x)=2-x^{-\frac12}

f(x)=2-\dfrac{1}{\sqrt{x}} needs x>0 (a square root in a denominator).
Table of values.
\begin{array}{c|ccccc} x & 0.0625 & 0.25 & 1 & 4 & 9 \\ \hline f(x) & -2 & 0 & 1 & 1.5 & 1.67 \end{array}
The curve crosses the x-axis where \dfrac{1}{\sqrt x}=2, at x=\tfrac14. As x\to0^{+} it falls without limit (asymptote x=0); as x grows it rises towards y=2 but stays below it.

\text{Increasing curve for }x>0;\ \text{asymptotes } x=0,\ y=2

(vi) f(x)=x^{\frac52}

x^{5/2}=(\sqrt{x})^{5} needs x\ge0.
\begin{array}{c|ccccc} x & 0 & 0.5 & 1 & 1.5 & 2 \\ \hline f(x) & 0 & 0.18 & 1 & 2.76 & 5.66 \end{array}
The curve starts at O, passes through (1,1) and then rises very steeply.

\text{Curve from }(0,0)\text{ through }(1,1),\ (2,5.66)

Find possible x-intercept, y-intercept and vertex of the following functions and then plot.

Solution

(i) f(x)=x^2+2x+1

x-intercept: put f(x)=0.
x^2+2x+1=(x+1)^2=0\;\Rightarrow\;x=-1\ \text{(repeated)}
y-intercept: put x=0, so f(0)=1.
Vertex: a=1, b=2.
x=-\dfrac{2}{2(1)}=-1,\qquad f(-1)=0,\qquad \text{vertex}=(-1,0)
a>0, so the parabola opens upward and just touches the x-axis at its vertex.
\begin{array}{c|ccccc} x & -3 & -2 & -1 & 0 & 1 \\ \hline f(x) & 4 & 1 & 0 & 1 & 4 \end{array}

x\text{-int }(-1,0),\ y\text{-int }(0,1),\ \text{vertex }(-1,0)

(ii) f(x)=x^2-4x+4

x-intercept:
x^2-4x+4=(x-2)^2=0\;\Rightarrow\;x=2\ \text{(repeated)}
y-intercept: f(0)=4, so (0,4).
Vertex:
x=-\dfrac{-4}{2(1)}=2,\qquad f(2)=0,\qquad \text{vertex}=(2,0)
The parabola opens upward and touches the x-axis at (2,0).
\begin{array}{c|ccccc} x & 0 & 1 & 2 & 3 & 4 \\ \hline f(x) & 4 & 1 & 0 & 1 & 4 \end{array}

x\text{-int }(2,0),\ y\text{-int }(0,4),\ \text{vertex }(2,0)

(iii) f(x)=x^2+2x

x-intercepts:
x^2+2x=x(x+2)=0\;\Rightarrow\;x=0\ \text{or}\ x=-2
y-intercept: f(0)=0, so the graph passes through the origin.
Vertex:
x=-\dfrac{2}{2(1)}=-1,\qquad f(-1)=1-2=-1,\qquad \text{vertex}=(-1,-1)
Table of values.
\begin{array}{c|ccccc} x & -3 & -2 & -1 & 0 & 1 \\ \hline f(x) & 3 & 0 & -1 & 0 & 3 \end{array}

x\text{-int }(-2,0),(0,0),\ y\text{-int }(0,0),\ \text{vertex }(-1,-1)

(iv) f(x)=9-x^2

x-intercepts:
9-x^2=(3-x)(3+x)=0\;\Rightarrow\;x=\pm3
y-intercept: f(0)=9.
Vertex: a=-1, b=0.
x=-\dfrac{0}{2(-1)}=0,\qquad f(0)=9,\qquad \text{vertex}=(0,9)
a<0, so the parabola opens downward.
\begin{array}{c|ccccccccc} x & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline f(x) & -7 & 0 & 5 & 8 & 9 & 8 & 5 & 0 & -7 \end{array}

x\text{-int }(\pm3,0),\ y\text{-int }(0,9),\ \text{vertex }(0,9)

Draw the graph of following function using factors.

Solution

(i) f(x)=x^2-2x+1

Factorise and put f(x)=0.
x^2-2x+1=(x-1)^2=0\;\Rightarrow\;x_1=x_2=1
A repeated root: the graph touches the x-axis at (1,0), which is the vertex.
y-intercept (0,1). The axis is x=1, so the mirror point of (0,1) is (2,1). a=1>0: opens upward.

\text{Vertex }(1,0),\ y\text{-int }(0,1)

(ii) f(x)=x^2-7x+12

Factorise and put f(x)=0.
x^2-7x+12=(x-3)(x-4)=0\;\Rightarrow\;x_1=3,\ x_2=4
x-intercepts (3,0) and (4,0); y-intercept (0,12).
Vertex: mean of the roots, then substitute.
x=\dfrac{3+4}{2}=3.5,\qquad f(3.5)=12.25-24.5+12=-0.25
a>0: opens upward, vertex (3.5,-0.25). The mirror point of (0,12) in x=3.5 is (7,12).

Zoomed view near the vertex: the curve dips just below the x-axis between the roots 3 and 4.

x\text{-int }(3,0),(4,0);\ y\text{-int }(0,12);\ \text{vertex }(3.5,-0.25)

(iii) f(x)=x^2-2x

Factorise and put f(x)=0.
x^2-2x=x(x-2)=0\;\Rightarrow\;x_1=0,\ x_2=2
x-intercepts (0,0) and (2,0); the y-intercept is also (0,0).
Vertex:
x=\dfrac{0+2}{2}=1,\qquad f(1)=1-2=-1,\qquad \text{vertex}=(1,-1)
a>0: opens upward.

x\text{-int }(0,0),(2,0);\ \text{vertex }(1,-1)

(iv) f(x)=-2x^2+x+3

Take out -1 and factorise.
-2x^2+x+3=-(2x^2-x-3)=-(2x-3)(x+1)
Put f(x)=0: x_1=\tfrac32, x_2=-1. x-intercepts \left(\tfrac32,0\right) and (-1,0); y-intercept (0,3).
Vertex:
x=\dfrac{\frac32+(-1)}{2}=\dfrac14,\qquad f\!\left(\tfrac14\right)=-2\left(\tfrac1{16}\right)+\tfrac14+3=\dfrac{25}{8}=3.125
a=-2<0: opens downward. The mirror point of (0,3) in x=\tfrac14 is \left(\tfrac12,3\right).

x\text{-int }(-1,0),\left(\tfrac32,0\right);\ y\text{-int }(0,3);\ \text{vertex }\left(\tfrac14,\tfrac{25}{8}\right)

(v) f(x)=4x^2-4x

Take out the common factor 4x.
4x^2-4x=4x(x-1)=0\;\Rightarrow\;x_1=0,\ x_2=1
x-intercepts (0,0) and (1,0); y-intercept (0,0).
Vertex:
x=\dfrac{0+1}{2}=\dfrac12,\qquad f\!\left(\tfrac12\right)=1-2=-1,\qquad \text{vertex}=\left(\tfrac12,-1\right)
a=4>0: opens upward, and it is narrow because |a| is large.

x\text{-int }(0,0),(1,0);\ \text{vertex }\left(\tfrac12,-1\right)

(vi) f(x)=6-x^2-x

Write in standard order, take out -1 and factorise.
6-x^2-x=-(x^2+x-6)=-(x+3)(x-2)
Put f(x)=0: x_1=-3, x_2=2. x-intercepts (-3,0) and (2,0); y-intercept (0,6).
Vertex:
x=\dfrac{-3+2}{2}=-\dfrac12,\qquad f\!\left(-\tfrac12\right)=6-\tfrac14+\tfrac12=\dfrac{25}{4}=6.25
a=-1<0: opens downward. The mirror point of (0,6) in x=-\tfrac12 is (-1,6).

x\text{-int }(-3,0),(2,0);\ y\text{-int }(0,6);\ \text{vertex }\left(-\tfrac12,\tfrac{25}{4}\right)

Note. The book prints the last part as a second (iv); it is part (vi) here.

Predict algebraic functions from the following graphs.

Solution

(i) A straight line through the marked points (-1,-1) and (0,1).

The given graph (redrawn).

The graph is a straight line, so the function is linear. Slope from the two marked points:
m=\dfrac{1-(-1)}{0-(-1)}=\dfrac21=2
The line crosses the y-axis at (0,1), so the y-intercept is c=1. Use y=mx+c.
y=2x+1
Check with (-1,-1): 2(-1)+1=-1. Correct.

f(x)=2x+1

(ii) A parabola opening upward with lowest point (0,-1), passing through (-1,1) and (1,1).

The given graph (redrawn).

The graph is a parabola with vertex (0,-1) on the y-axis, so it has the form below.
y=a(x-0)^2+(-1)=ax^2-1
It passes through (1,1). Substitute to find a.
1=a(1)^2-1\;\Rightarrow\;a=2
So the function is y=2x^2-1. Check with (-1,1): 2(1)-1=1. Correct.

f(x)=2x^2-1

(iii) A parabola opening downward with highest point (0,3), passing through (-2,-1) and (2,-1).

The given graph (redrawn).

The vertex is (0,3), so the parabola has the form below (with a<0 since it opens downward).
y=ax^2+3
It passes through (2,-1).
-1=a(2)^2+3\;\Rightarrow\;4a=-4\;\Rightarrow\;a=-1
So y=3-x^2. Check with (-2,-1): 3-4=-1. Correct. (Using the key fact with (p,r)=(-2,-1) and (q,r)=(2,-1) gives the same: y+1=a(x+2)(x-2) with a=-1.)

f(x)=3-x^2

(iv) A parabola with a=1, touching the x-axis at (-1,0) and passing through (0,1).

The given graph (redrawn).

With a=1 the function is y=x^2+bx+c. Use the two marked points.
Through (0,1):
1=0+0+c\;\Rightarrow\;c=1
Through (-1,0):
0=1-b+1\;\Rightarrow\;b=2
So y=x^2+2x+1=(x+1)^2. This has a repeated root at x=-1, which agrees with the graph just touching the x-axis at its vertex (-1,0).

f(x)=x^2+2x+1=(x+1)^2

Plot the graph of following and find point of intersection of function with axes.

Solution

(i) y=x+3

Table of values.
\begin{array}{c|cccc} x & -3 & -1 & 0 & 2 \\ \hline y & 0 & 2 & 3 & 5 \end{array}
x-axis: put y=0, so x=-3. y-axis: put x=0, so y=3.

\text{Meets the axes at }(-3,0)\text{ and }(0,3)

(ii) y=6-3x

Table of values.
\begin{array}{c|ccccc} x & -1 & 0 & 1 & 2 & 3 \\ \hline y & 9 & 6 & 3 & 0 & -3 \end{array}
x-axis: 6-3x=0 gives x=2. y-axis: x=0 gives y=6.

\text{Meets the axes at }(2,0)\text{ and }(0,6)

(iii) y=x^2-5x

A parabola opening upward. Vertex:
x=-\dfrac{-5}{2}=2.5,\qquad y=6.25-12.5=-6.25
Table of values.
\begin{array}{c|ccccccccc} x & -1 & 0 & 1 & 2 & 2.5 & 3 & 4 & 5 & 6 \\ \hline y & 6 & 0 & -4 & -6 & -6.25 & -6 & -4 & 0 & 6 \end{array}
x-axis: x^2-5x=x(x-5)=0 gives x=0 and x=5. y-axis: x=0 gives y=0.

\text{Meets the }x\text{-axis at }(0,0),(5,0)\text{ and the }y\text{-axis at }(0,0)

Find graphical solution of:

Solution

(i) f(x)=4-3x,\quad g(x)=-x+1

Tables of values for both lines.
\begin{array}{c|ccccc} x & -1 & 0 & 1 & 2 & 3 \\ \hline f(x) & 7 & 4 & 1 & -2 & -5 \\ g(x) & 2 & 1 & 0 & -1 & -2 \end{array}
Draw both lines on the same axes. They cross at (1.5,-0.5).
Check: 4-3x=1-x gives 3=2x, x=1.5 and y=1-1.5=-0.5.

(1.5,\,-0.5)

(ii) f(x)=2(2+x),\quad g(x)=x^2+1

f(x)=2x+4 is a line; g(x)=x^2+1 is a parabola opening upward with vertex (0,1).
\begin{array}{c|cccccccc} x & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline f(x) & -2 & 0 & 2 & 4 & 6 & 8 & 10 & 12 \\ g(x) & 10 & 5 & 2 & 1 & 2 & 5 & 10 & 17 \end{array}
Draw both graphs. They meet at (-1,2) and (3,10); both points appear in the table.
Check: x^2+1=2x+4 gives x^2-2x-3=(x-3)(x+1)=0, so x=3 or x=-1.

(-1,\,2)\ \text{and}\ (3,\,10)

(iii) f(x)=5+3x,\quad g(x)=-x^2+5

f is a line; g is a parabola opening downward with vertex (0,5).
\begin{array}{c|ccccccc} x & -4 & -3 & -2 & -1 & 0 & 1 & 2 \\ \hline f(x) & -7 & -4 & -1 & 2 & 5 & 8 & 11 \\ g(x) & -11 & -4 & 1 & 4 & 5 & 4 & 1 \end{array}
The graphs meet at (-3,-4) and (0,5).
Check: 5+3x=5-x^2 gives x^2+3x=x(x+3)=0, so x=0 or x=-3.

(-3,\,-4)\ \text{and}\ (0,\,5)

(iv) f(x)=1,\quad g(x)=-2x^2+2x+5

f(x)=1 is a horizontal line. g opens downward; its vertex is at x=-\dfrac{2}{2(-2)}=\tfrac12, g\!\left(\tfrac12\right)=5.5.
\begin{array}{c|ccccccc} x & -2 & -1 & 0 & 0.5 & 1 & 2 & 3 \\ \hline g(x) & -7 & 1 & 5 & 5.5 & 5 & 1 & -7 \end{array}
The line y=1 cuts the parabola at (-1,1) and (2,1).
Check: -2x^2+2x+5=1 gives x^2-x-2=(x-2)(x+1)=0.

(-1,\,1)\ \text{and}\ (2,\,1)

(v) f(x)=2+3x+x^2,\quad g(x)=5+3x-2x^2

f opens upward (vertex (-1.5,-0.25)); g opens downward (vertex (0.75,6.125)).
\begin{array}{c|cccccc} x & -3 & -2 & -1 & 0 & 1 & 2 \\ \hline f(x) & 2 & 0 & 0 & 2 & 6 & 12 \\ g(x) & -22 & -9 & 0 & 5 & 6 & 3 \end{array}
The two parabolas meet at (-1,0) and (1,6).
Check: 2+3x+x^2=5+3x-2x^2 gives 3x^2=3, so x=\pm1; f(-1)=0 and f(1)=6.

(-1,\,0)\ \text{and}\ (1,\,6)

Draw the graph of following modulus functions.

Solution

(i) f(x)=-1.5|x|

For x\ge0, f(x)=-1.5x; for x<0, f(x)=1.5x. Both arms go downward from the origin.
\begin{array}{c|ccccccc} x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline f(x) & -4.5 & -3 & -1.5 & 0 & -1.5 & -3 & -4.5 \end{array}
The graph is an upside-down V with its corner at (0,0). Domain \mathbb{R}, range (-\infty,0].

\text{Inverted V, vertex }(0,0);\ D=\mathbb{R},\ R=(-\infty,0]

(ii) f(x)=1+2|x|

The V of y=|x| is stretched (arms of slope \pm2) and lifted 1 unit.
\begin{array}{c|ccccccc} x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline f(x) & 7 & 5 & 3 & 1 & 3 & 5 & 7 \end{array}
Corner at (0,1). Domain \mathbb{R}, range [1,\infty).

\text{V-shape, vertex }(0,1);\ D=\mathbb{R},\ R=[1,\infty)

(iii) f(x)=3|x|+x

Remove the modulus on each side of 0.
f(x)=\begin{cases}3x+x=4x, & x\ge0\\ -3x+x=-2x, & x<0\end{cases}
Table of values.
\begin{array}{c|ccccccc} x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline f(x) & 6 & 4 & 2 & 0 & 4 & 8 & 12 \end{array}
The graph is a lopsided V with corner (0,0): the right arm (slope 4) is steeper than the left arm (slope -2). Domain \mathbb{R}, range [0,\infty).

\text{V-shape with corner }(0,0);\ D=\mathbb{R},\ R=[0,\infty)

The equations for supply and demand are given by two linear equations. Supply equation: S(x)=2x+10, where x is quantity and S(x) is the price. Demand equation: D(x)=-3x+40, where x is quantity and D(x) is the price. Find the equilibrium point where the price of supply equals the price of demand by drawing the graphs of both equations.

Solution

Make tables of values for both lines.
\begin{array}{c|cccccc} x & 0 & 2 & 4 & 6 & 8 & 10 \\ \hline S(x) & 10 & 14 & 18 & 22 & 26 & 30 \\ D(x) & 40 & 34 & 28 & 22 & 16 & 10 \end{array}
Draw both lines on the same axes, with quantity x across and price up. The supply line rises and the demand line falls.
They cross at (6,22): both give the price 22 when the quantity is 6. This is the equilibrium point.
Check: 2x+10=-3x+40 gives 5x=30, so x=6 and S(6)=2(6)+10=22.

Quantity x across, price up. The lines cross at the equilibrium point (6,22).

At equilibrium the quantity is 6 units and the price is 22.

\text{Equilibrium point }(6,\,22):\ x=6,\ \text{price}=22

Suppose a ball is thrown into the air and its height h(t) after t seconds is given by the parabolic trajectory h(t)=-6t^2+10t+5. If this ball hits a wall 10 m high representing the equation h(t)=9t. By drawing the graphs, find out when and where the ball reaches the wall.

Solution

Table of values for the ball's height (a parabola opening downward, since a=-6<0).
\begin{array}{c|ccccccc} t & 0 & 0.25 & 0.5 & 0.75 & 1 & 1.5 & 2 \\ \hline h(t) & 5 & 7.125 & 8.5 & 9.125 & 9 & 6.5 & 1 \end{array}
The wall line h=9t passes through (0,0), (1,9) and (1.1,9.9).
Draw both graphs with t across and height up. They meet at t=1, where both heights are 9.
Check: -6t^2+10t+5=9t gives 6t^2-t-5=0, that is (6t+5)(t-1)=0. So t=1 (the value t=-\tfrac56 is rejected because time cannot be negative), and h(1)=9.

The ball's highest point is h=\tfrac{55}{6}\approx9.17\,\mathrm{m} at t=\tfrac56\,\mathrm{s}, then it comes down and meets the wall line at (1,9).

The ball reaches the wall after 1\,\mathrm{s}, at a height of 9\,\mathrm{m}. This is below the top of the 10\,\mathrm{m} wall, so the ball hits the wall.

t=1\,\mathrm{s},\qquad h=9\,\mathrm{m}\ \ \text{(the point }(1,9)\text{)}

Note. The book represents the wall by the line h(t)=9t; the meeting point of the two graphs is taken as the moment the ball reaches the wall.

Two asteroids are following the parabolic paths represented by f(x)=x^2-7x+12 and g(x)=x(x-3). By drawing the graphs of both trajectories, find out the place from where, both asteroids will pass.

Solution

f(x)=(x-3)(x-4) cuts the x-axis at 3 and 4, vertex (3.5,-0.25). g(x)=x^2-3x cuts it at 0 and 3, vertex (1.5,-2.25). Both open upward.
Tables of values.
\begin{array}{c|ccccccc} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline f(x) & 12 & 6 & 2 & 0 & 0 & 2 & 6 \\ g(x) & 0 & -2 & -2 & 0 & 4 & 10 & 18 \end{array}
Draw both parabolas on the same axes. They meet at only one point, (3,0).
Check: x^2-7x+12=x^2-3x gives -4x+12=0, so x=3 and f(3)=0.

Both asteroids pass through the point (3,0).

(3,\,0)

Exercise 1.3p. 30

Graphs of exponential and logarithmic functions (drawn from a table of values, with intercepts and asymptotes), and growth and decay problems solved by taking logarithms of both sides.

Draw the graphs of functions.

Solution

(i) f(x)=e^{2x}

Table of values. Substitute some values of x and use a calculator (e\approx2.718).
\begin{array}{c|cccccc} x & -2 & -1 & -0.5 & 0 & 0.5 & 1 \\ \hline f(x) & 0.02 & 0.14 & 0.37 & 1 & 2.72 & 7.39 \end{array}
Intercept. At x=0, f(0)=e^{0}=1, so the graph cuts the y-axis at (0,1).
Asymptote. A power of e is always positive, so the graph stays above the x-axis. As x\to-\infty, e^{2x}\to0, so the x-axis (y=0) is a horizontal asymptote.
Shape. e^{2x}=\left(e^{2}\right)^{x} has base e^{2}\approx7.39>1, so the graph shows exponential growth. It rises faster than y=e^{x} because x is doubled.
Domain and range. Any real x can be used, and the outputs are all the positive numbers.

Exponential growth: the curve comes in along the x-axis on the left and rises steeply on the right.

\text{Domain}=\mathbb{R},\qquad \text{Range}=(0,\infty)

(ii) g(x)=e^{0.5x}

Table of values. Even values of x make 0.5x a whole number.
\begin{array}{c|ccccc} x & -4 & -2 & 0 & 2 & 4 \\ \hline g(x) & 0.14 & 0.37 & 1 & 2.72 & 7.39 \end{array}
Intercept and asymptote. g(0)=e^{0}=1, so the graph passes through (0,1). As x\to-\infty, e^{0.5x}\to0, so y=0 is a horizontal asymptote.
Shape. The base is e^{0.5}\approx1.65>1, so this is exponential growth, but slower than y=e^{x} because x is halved: x must increase by 2 for g to be multiplied by e.

A gentler growth curve. It still passes through (0,1) and never touches the x-axis.

\text{Domain}=\mathbb{R},\qquad \text{Range}=(0,\infty)

(iii) h(x)=2-e^{x}

Idea. Start from y=e^{x}. The minus sign reflects it in the x-axis (y=-e^{x}), and adding 2 moves that curve 2 units up.
Table of values.
\begin{array}{c|ccccccc} x & -3 & -2 & -1 & 0 & 0.5 & 1 & 1.5 \\ \hline h(x) & 1.95 & 1.86 & 1.63 & 1 & 0.35 & -0.72 & -2.48 \end{array}
Intercepts. h(0)=2-1=1, so the y-intercept is (0,1). For the x-intercept put h(x)=0.
2-e^{x}=0\;\Longrightarrow\;e^{x}=2\;\Longrightarrow\;x=\ln2\approx0.69
Asymptote. As x\to-\infty, e^{x}\to0, so h(x)\to2. The line y=2 is a horizontal asymptote and the curve stays below it.
Range. e^{x}>0 for every x, so 2-e^{x}<2. Every value below 2 is reached.
e^{x}>0\;\Longrightarrow\;-e^{x}<0\;\Longrightarrow\;h(x)<2

A decreasing curve that levels off towards y=2 on the left and falls steeply on the right.

\text{Domain}=\mathbb{R},\qquad \text{Range}=(-\infty,\,2)

(iv) h(x)=1+e^{-2x}

Idea. e^{-2x}=\left(e^{-2}\right)^{x} has base e^{-2}\approx0.135, which lies between 0 and 1, so it is an exponential decay curve. Adding 1 lifts it 1 unit up.
Table of values.
\begin{array}{c|cccccc} x & -1 & -0.5 & 0 & 0.5 & 1 & 2 \\ \hline h(x) & 8.39 & 3.72 & 2 & 1.37 & 1.14 & 1.02 \end{array}
Intercept. h(0)=1+e^{0}=2, so the graph passes through (0,2).
Asymptote. As x\to\infty, e^{-2x}\to0, so h(x)\to1. The line y=1 is a horizontal asymptote on the right.
Range. e^{-2x}>0, so h(x)=1+e^{-2x}>1.

Exponential decay lifted up by 1: the curve falls and approaches y=1 without touching it.

\text{Domain}=\mathbb{R},\qquad \text{Range}=(1,\,\infty)

(v) f(x)=\ln(2x)

Domain. A logarithm needs a positive input.
2x>0\;\Longrightarrow\;x>0
Table of values.
\begin{array}{c|cccccc} x & 0.1 & 0.25 & 0.5 & 1 & 2 & 4 \\ \hline f(x) & -1.61 & -0.69 & 0 & 0.69 & 1.39 & 2.08 \end{array}
Intercept. \ln(2x)=0 when 2x=1, that is at x=\tfrac12. So the graph cuts the x-axis at \left(\tfrac12,0\right).
Asymptote. As x\to0^{+}, 2x\to0^{+} and \ln(2x)\to-\infty, so the y-axis (x=0) is a vertical asymptote.
Link with the basic log graph. By the product rule the graph is the curve y=\ln x moved up by \ln2\approx0.69.
\ln(2x)=\ln2+\ln x

\ln(2x) (solid) is \ln x (dashed) moved up by \ln 2. Both approach the y-axis but never touch it.

\text{Domain}=(0,\infty),\qquad \text{Range}=\mathbb{R}

(vi) g(x)=\log(x+1)

Base. \log without a base means the common logarithm, base 10.
Domain. The input of a logarithm must be positive.
x+1>0\;\Longrightarrow\;x>-1
Table of values. Choose x so that x+1 is easy: 0.1,\ 0.5,\ 1,\ 2,\ 5,\ 10.
\begin{array}{c|cccccc} x & -0.9 & -0.5 & 0 & 1 & 4 & 9 \\ \hline g(x) & -1 & -0.3 & 0 & 0.3 & 0.7 & 1 \end{array}
Key points. g(0)=\log1=0, so the graph passes through the origin. The line x=-1 is a vertical asymptote: as x\to-1^{+}, g(x)\to-\infty.
Shape. This is y=\log x moved 1 unit to the left. It rises slowly: x must grow from 0 to 9 for g to grow from 0 to 1.

The curve passes through the origin and approaches the asymptote x=-1.

\text{Domain}=(-1,\infty),\qquad \text{Range}=\mathbb{R}

(vii) h(x)=3+\log(x)

Domain. \log x (base 10) needs x>0.
Table of values. Powers of 10 give whole-number logarithms.
\begin{array}{c|cccccc} x & 0.01 & 0.1 & 0.5 & 1 & 5 & 10 \\ \hline h(x) & 1 & 2 & 2.7 & 3 & 3.7 & 4 \end{array}
Key points. h(1)=3+0=3 and h(10)=3+1=4. The x-intercept is where \log x=-3, that is x=10^{-3}=0.001, very close to the origin.
Asymptote and shape. The y-axis is a vertical asymptote. The graph is y=\log x moved 3 units up, so it keeps the same shape and range.

The curve of \log x lifted 3 units. It drops sharply near the y-axis.

\text{Domain}=(0,\infty),\qquad \text{Range}=\mathbb{R}

(viii) f(x)=e^{0.6x} and g(x)=\ln(0.6x)

Table of values for f.
\begin{array}{c|ccccccc} x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline f(x) & 0.17 & 0.3 & 0.55 & 1 & 1.82 & 3.32 & 6.05 \end{array}
Features. f has base e^{0.6}\approx1.82>1, so it shows exponential growth through (0,1), with the x-axis as horizontal asymptote. Domain \mathbb{R}, range (0,\infty).
Table of values for g, which needs 0.6x>0, that is x>0.
\begin{array}{c|cccccc} x & 0.5 & 1 & \tfrac53 & 3 & 5 & 8 \\ \hline g(x) & -1.2 & -0.51 & 0 & 0.59 & 1.1 & 1.57 \end{array}
Features. g(x)=0 when 0.6x=1, that is x=\tfrac53\approx1.67. The y-axis is a vertical asymptote. Domain (0,\infty), range \mathbb{R}.

Both graphs on one set of axes. They are not mirror images in the line y=x.

D_f=\mathbb{R},\ R_f=(0,\infty);\qquad D_g=(0,\infty),\ R_g=\mathbb{R}

The number of compact discs N (in million) purchased each year increasing exponentially is given by: N(t)=7.5(6)^{0.5t}, where t=0 corresponds to 2024, t=1 corresponds to 2025 and so on, t being the number of years after 2024.

Solution

a. After what amount of time will one billion compact discs be sold in a year?

Units. N is in millions and one billion =1000 million, so put N(t)=1000.
7.5(6)^{0.5t}=1000
Isolate the power. Divide both sides by 7.5.
6^{0.5t}=\dfrac{1000}{7.5}=133.33
Take natural logs. The unknown is in the exponent. The power rule \ln a^{n}=n\ln a brings it down.
0.5t\,\ln6=\ln133.33
Solve.
t=\dfrac{\ln133.33}{0.5\ln6}=\dfrac{4.8929}{0.5(1.7918)}=\dfrac{4.8929}{0.8959}\approx5.46
Interpret. t=5 is 2029 and t=6 is 2030. So yearly sales reach one billion discs about 5.46 years after 2024, that is, during 2029.

N in millions. The curve meets the level N=1000 at t\approx5.46.

t\approx5.46\ \text{years (during 2029)}

b. What is the doubling time on the sale of compact discs?

Meaning. The doubling time T is the time for yearly sales to become twice as large. At t=0 sales are 7.5 million, so find when they are 15 million.
7.5(6)^{0.5T}=15
Divide both sides by 7.5.
6^{0.5T}=2
Take natural logs and solve.
0.5T\ln6=\ln2\;\Longrightarrow\;T=\dfrac{\ln2}{0.5\ln6}=\dfrac{0.6931}{0.8959}\approx0.774
Convert to months. 0.774\times12\approx9.3 months. The same T works from any starting year, because N(t+T)=N(t)\,6^{0.5T}=2N(t).

T\approx0.77\ \text{year}\approx9.3\ \text{months}

Suppose that Rs. 50,000 is invested at 6% interest compounded annually. After t years, it grows to the amount A given by the function: A(t)=50{,}000(1.06)^{t}

Solution

a. After what amount of time will Rs. 50,000 grows to Rs. 450,000?

Put A(t)=450{,}000, as in Example 20.
450{,}000=50{,}000(1.06)^{t}
Divide both sides by 50,000.
(1.06)^{t}=9
Take natural logs of both sides and bring the power down.
t\ln(1.06)=\ln9
Solve for t.
t=\dfrac{\ln9}{\ln1.06}=\dfrac{2.1972}{0.05827}\approx37.7

A in thousands of rupees. The amount doubles to 100 thousand at t\approx11.9 and reaches 450 thousand at t\approx37.7.

t\approx37.7\ \text{years}

b. Find the doubling time.

The money doubles when A(T)=2\times50{,}000=100{,}000.
100{,}000=50{,}000(1.06)^{T}\;\Longrightarrow\;(1.06)^{T}=2
Take natural logs and solve.
T=\dfrac{\ln2}{\ln1.06}=\dfrac{0.6931}{0.05827}\approx11.9
Check. 9\approx2^{3.17}, so reaching 9 times the money needs about 3.17 doublings: 3.17\times11.9\approx37.7 years, which agrees with part a.

T\approx11.9\ \text{years}

The exponential growth rate of the population of the city is 1% per year. After how many years, the population will be doubled?

Solution

Model. As in Example 21, a population with exponential growth rate r is P(t)=P_0e^{rt}, where P_0 is the present population. Here r=1\%=0.01.
P(t)=P_0e^{0.01t}
Doubling. Put P(t)=2P_0. P_0 cancels, so the answer does not depend on the size of the population.
2P_0=P_0e^{0.01t}\;\Longrightarrow\;e^{0.01t}=2
Take natural logs. \ln e^{u}=u.
0.01t=\ln2\;\Longrightarrow\;t=\dfrac{\ln2}{0.01}=\dfrac{0.6931}{0.01}\approx69.3

t\approx69.3\ \text{years}

Note. The book's model for an exponential growth rate is P(t)=P_0e^{rt} (Example 21), which gives 69.3 years. If the 1% is instead added once a year (compound growth), (1.01)^{t}=2 gives t=\dfrac{\ln2}{\ln1.01}\approx69.7 years.

The population of the world was 5.2 billion in 1990. The exponential growth rate was 1.6% per year at that time.

Solution

a. Find the exponential growth function.

Use the growth model P(t)=P_0e^{rt} with t in years after 1990 and P in billions.
At t=0 (1990) the population is P_0=5.2, and r=1.6\%=0.016.
P(t)=5.2\,e^{0.016t}

P(t)=5.2\,e^{0.016t}\quad(t\text{ in years after 1990})

b. Find the population of the world in 2000.

2000 is 10 years after 1990, so put t=10.
P(10)=5.2\,e^{0.016\times10}=5.2\,e^{0.16}
Evaluate with a calculator.
P(10)=5.2\times1.1735\approx6.10

P(10)\approx6.10\ \text{billion}

c. In which year the world population was 8 billion?

Put P(t)=8 and divide by 5.2.
8=5.2\,e^{0.016t}\;\Longrightarrow\;e^{0.016t}=\dfrac{8}{5.2}=1.5385
Take natural logs.
0.016t=\ln1.5385=0.4308\;\Longrightarrow\;t=\dfrac{0.4308}{0.016}\approx26.9
Add to the starting year: 1990+26.9\approx2016.9. So the model gives 8 billion in the year 2017 (to the nearest year).

P in billions, t in years after 1990.

\text{In about }2017\ (t\approx26.9)

Students in a mathematics class took a final exam in monthly intervals thereafter. The average score S(t), after t months was given by: S(t)=68-20\log(t+1);\ t\ge0

Solution

a. What was the average score when they initially took the test (t=0)?

Put t=0. Since \log1=0, the log term vanishes.
S(0)=68-20\log(0+1)=68-20(0)=68

S(0)=68

b. What was the average score (i) after 4 months (ii) after 24 months?

(i) Put t=4, so t+1=5 and \log5=0.6990.
S(4)=68-20\log5=68-20(0.6990)=68-13.98=54.02\approx54
(ii) Put t=24, so t+1=25 and \log25=1.3979.
S(24)=68-20\log25=68-20(1.3979)=68-27.96=40.04\approx40

\text{(i) }S(4)\approx54,\qquad \text{(ii) }S(24)\approx40

c. Graph the function.

Table of values. Choose t so that t+1 is easy to take the log of.
\begin{array}{c|cccccc} t & 0 & 1 & 4 & 9 & 24 & 49 \\ \hline S(t) & 68 & 61.98 & 54.02 & 48 & 40.04 & 34.02 \end{array}
Features. The domain is t\ge0 and the graph starts at (0,68). Since \log(t+1) increases, S(t) decreases. It falls quickly in the first months and then more and more slowly.

The average score drops quickly at first and then slowly: students forget fast, then more slowly.

\text{Decreasing log curve from }(0,68)

d. After what time was the average score 50?

Put S(t)=50 and move the log term to one side.
50=68-20\log(t+1)\;\Longrightarrow\;20\log(t+1)=18
Divide by 20.
\log(t+1)=0.9
Antilog. \log u=0.9 means u=10^{0.9}.
t+1=10^{0.9}=7.943\;\Longrightarrow\;t\approx6.94
So the average score falls to 50 after about 7 months.

The graph meets the level S=50 at t\approx6.94 months.

t\approx6.94\approx7\ \text{months}

Note. Here \log is the common logarithm (base 10). The book's first sentence is shortened: the class took a final exam and then an equivalent test every month after it.

If P(t)=P_0e^{kt} denotes the growth function of oil and the exponential growth rate of the demand for oil is 10% per year, when will the demand be doubled?

Solution

Data. The growth rate is k=10\%=0.10, so the demand is P(t)=P_0e^{0.1t}, where P_0 is the present demand.
Doubling. Put P(t)=2P_0 and cancel P_0.
2P_0=P_0e^{0.1t}\;\Longrightarrow\;e^{0.1t}=2
Take natural logs and solve.
0.1t=\ln2\;\Longrightarrow\;t=\dfrac{\ln2}{0.1}=\dfrac{0.6931}{0.1}\approx6.93

So the demand for oil doubles in about 6.93 years, that is, in about 7 years.

t\approx6.93\ \text{years}

Approximately two third of all Aluminum cans distributed are recycled each year. A beverage company distributes 250,000 cans. The number still in use after t years is given by the function: N(t)=250{,}000\left(\dfrac{2}{3}\right)^{t}

Solution

a. After how many years will 60,000 cans be in use?

Put N(t)=60{,}000 and divide by 250,000.
60{,}000=250{,}000\left(\tfrac23\right)^{t}\;\Longrightarrow\;\left(\tfrac23\right)^{t}=0.24
Take natural logs and bring the power down.
t\ln\tfrac23=\ln0.24
Solve. Both logs are negative (their numbers are less than 1), so t comes out positive.
t=\dfrac{\ln0.24}{\ln(2/3)}=\dfrac{-1.4271}{-0.4055}\approx3.52

t\approx3.52\ \text{years}

b. After what amount of time will only 1,000 cans be in use?

Put N(t)=1000 and divide by 250,000.
1000=250{,}000\left(\tfrac23\right)^{t}\;\Longrightarrow\;\left(\tfrac23\right)^{t}=0.004
Take natural logs and solve.
t=\dfrac{\ln0.004}{\ln(2/3)}=\dfrac{-5.5215}{-0.4055}\approx13.6

N in thousands of cans. Exponential decay: the number in use falls by a factor \tfrac23 every year.

t\approx13.6\ \text{years}

Exercise 1.4p. 36

Reading the domain and range of trigonometric functions from their graphs, and using the horizontal line test to decide whether a function is one to one before finding its inverse and drawing it as the mirror image in y=x.

Find the domain and range of the functions graphically.

Solution

(i) f(x)=\sin\left(\dfrac{x}{2}\right)

Period. \sin u repeats when u increases by 2\pi. Here u=\tfrac{x}{2}, so x must increase by 4\pi.
\text{Period}=\dfrac{2\pi}{1/2}=4\pi
Table of values over one period.
\begin{array}{c|ccccc} x & 0 & \pi & 2\pi & 3\pi & 4\pi \\ \hline \tfrac{x}{2} & 0 & \tfrac{\pi}{2} & \pi & \tfrac{3\pi}{2} & 2\pi \\ \hline f(x) & 0 & 1 & 0 & -1 & 0 \end{array}
Domain from the graph. The curve has no breaks and goes on for ever to the left and right, because \sin is defined for every angle.
Range from the graph. The whole curve lies between the lines y=-1 and y=1 and touches both, so it covers every value from -1 to 1.

One full wave takes 4\pi. The graph stays between the dotted lines y=\pm1.

\text{Domain}=\mathbb{R},\qquad \text{Range}=[-1,\,1]

(ii) g(x)=3\cos\left(\dfrac{x}{3}\right)

Period. \cos u repeats every 2\pi; with u=\tfrac{x}{3} the period is \dfrac{2\pi}{1/3}=6\pi.
Table of values over one period.
\begin{array}{c|ccccc} x & 0 & \tfrac{3\pi}{2} & 3\pi & \tfrac{9\pi}{2} & 6\pi \\ \hline \tfrac{x}{3} & 0 & \tfrac{\pi}{2} & \pi & \tfrac{3\pi}{2} & 2\pi \\ \hline g(x) & 3 & 0 & -3 & 0 & 3 \end{array}
Domain. The graph is an unbroken wave for all x, so the domain is \mathbb{R}.
Range. \cos\tfrac{x}{3} takes every value from -1 to 1. Multiplying by 3 (the amplitude) stretches these to every value from -3 to 3; the graph touches y=3 and y=-3 and never goes beyond.
-1\le\cos\tfrac{x}{3}\le1\;\Longrightarrow\;-3\le g(x)\le3

A cosine wave of amplitude 3 and period 6\pi.

\text{Domain}=\mathbb{R},\qquad \text{Range}=[-3,\,3]

(iii) h(x)=2\tan x

Where it is not defined. \tan x=\dfrac{\sin x}{\cos x} does not exist when \cos x=0. At these values the graph has vertical asymptotes.
x=\pm\dfrac{\pi}{2},\ \pm\dfrac{3\pi}{2},\ \dots=(2n+1)\dfrac{\pi}{2}
Table of values between two asymptotes (\tan has period \pi).
\begin{array}{c|ccccc} x & -\tfrac{\pi}{3} & -\tfrac{\pi}{4} & 0 & \tfrac{\pi}{4} & \tfrac{\pi}{3} \\ \hline h(x) & -3.46 & -2 & 0 & 2 & 3.46 \end{array}
Domain. Every real number except the asymptote values.
Range. On each branch the curve rises from -\infty to +\infty, so every real number is an output. The factor 2 makes the curve steeper but does not change the range.

Dashed lines are the asymptotes x=\pm\tfrac{\pi}{2},\ \pm\tfrac{3\pi}{2}. Each branch covers all real y.

\begin{gathered}\text{Domain}=\left\{x:\ x\neq(2n+1)\tfrac{\pi}{2},\ n\in\mathbb{Z}\right\}\\ \text{Range}=\mathbb{R}\end{gathered}

(iv) y=\cot\left(\dfrac{x}{4}\right)

Where it is not defined. \cot u=\dfrac{\cos u}{\sin u} does not exist when \sin u=0, that is when u=n\pi.
\dfrac{x}{4}=n\pi\;\Longrightarrow\;x=4n\pi=0,\ \pm4\pi,\ \pm8\pi,\ \dots
Period. \cot has period \pi, so \tfrac{x}{4} must increase by \pi: the period is 4\pi.
Table of values on the branch 0<x<4\pi.
\begin{array}{c|ccc} x & \pi & 2\pi & 3\pi \\ \hline \tfrac{x}{4} & \tfrac{\pi}{4} & \tfrac{\pi}{2} & \tfrac{3\pi}{4} \\ \hline y & 1 & 0 & -1 \end{array}
Domain and range. The graph breaks only at x=4n\pi. Each branch falls from +\infty (just right of an asymptote) to -\infty (just left of the next one), so every real number is an output.

Asymptotes at x=0 (the y-axis) and x=\pm4\pi. Each branch is 4\pi wide.

\begin{gathered}\text{Domain}=\left\{x:\ x\neq4n\pi,\ n\in\mathbb{Z}\right\}\\ \text{Range}=\mathbb{R}\end{gathered}

(v) y=2\sec(2x)

Where it is not defined. \sec u=\dfrac{1}{\cos u} does not exist when \cos u=0, that is u=(2n+1)\tfrac{\pi}{2}.
2x=(2n+1)\dfrac{\pi}{2}\;\Longrightarrow\;x=(2n+1)\dfrac{\pi}{4}=\pm\dfrac{\pi}{4},\ \pm\dfrac{3\pi}{4},\ \dots
Period. \sec has period 2\pi, so \sec 2x has period \dfrac{2\pi}{2}=\pi.
Table of values.
\begin{array}{c|cccccc} x & -\tfrac{\pi}{6} & 0 & \tfrac{\pi}{6} & \tfrac{\pi}{3} & \tfrac{\pi}{2} & \tfrac{2\pi}{3} \\ \hline 2x & -\tfrac{\pi}{3} & 0 & \tfrac{\pi}{3} & \tfrac{2\pi}{3} & \pi & \tfrac{4\pi}{3} \\ \hline y & 4 & 2 & 4 & -4 & -2 & -4 \end{array}
Range. |\cos2x|\le1, so |\sec2x|\ge1 and |y|\ge2. The upper branches have lowest point y=2 and the lower branches have highest point y=-2. No part of the graph lies between y=-2 and y=2.
y\le-2\ \text{ or }\ y\ge2

Dashed lines: asymptotes x=\pm\tfrac{\pi}{4},\ \pm\tfrac{3\pi}{4}. The band between the dotted lines y=\pm2 is never reached.

\begin{gathered}\text{Domain}=\left\{x:\ x\neq(2n+1)\tfrac{\pi}{4},\ n\in\mathbb{Z}\right\}\\ \text{Range}=(-\infty,-2]\cup[2,\infty)\end{gathered}

(vi) y=\sin(2x)

Period. With u=2x, u increases by 2\pi when x increases by \pi, so the period is \pi.
Table of values over one period.
\begin{array}{c|ccccc} x & 0 & \tfrac{\pi}{4} & \tfrac{\pi}{2} & \tfrac{3\pi}{4} & \pi \\ \hline 2x & 0 & \tfrac{\pi}{2} & \pi & \tfrac{3\pi}{2} & 2\pi \\ \hline y & 0 & 1 & 0 & -1 & 0 \end{array}
Domain. No breaks in the graph: every real x is allowed.
Range. The graph is y=\sin x squeezed sideways by a factor of 2. It repeats twice as often, but its height is unchanged: it still stays between -1 and 1 and touches both.

Two full waves between -\pi and \pi: the period is \pi.

\text{Domain}=\mathbb{R},\qquad \text{Range}=[-1,\,1]

Determine whether the given function is one to one by examining its graph. If the function is one to one, find its inverse. Also draw the graphs of inverse function.

Solution

(i) f(x)=\dfrac{1}{3}x+3

Graph. A straight line with slope \tfrac13 and y-intercept 3. Two points: f(0)=3 and f(-9)=0.
Horizontal line test. A line with non-zero slope is not horizontal, so every horizontal line cuts it exactly once. Hence f is one to one. (Algebraically: \tfrac{x_1}{3}+3=\tfrac{x_2}{3}+3\Rightarrow x_1=x_2.)
Find the inverse. Put y=f(x) and solve for x.
y=\dfrac{x}{3}+3\;\Longrightarrow\;\dfrac{x}{3}=y-3\;\Longrightarrow\;x=3y-9
So f^{-1}(y)=3y-9. Interchange y with x.
f^{-1}(x)=3x-9
Check.
f\left(f^{-1}(x)\right)=\dfrac{3x-9}{3}+3=x-3+3=x
Graph of the inverse. Swap the coordinates of the points: (0,3)\to(3,0) and (-9,0)\to(0,-9). The two lines are mirror images in y=x and meet on it at (4.5,\,4.5).

f and f^{-1} are reflections of each other in the line y=x.

\text{One to one};\quad f^{-1}(x)=3x-9

(ii) g(x)=x(x-5)

Graph. g(x)=x^{2}-5x is a parabola opening upwards. It cuts the x-axis at x=0 and x=5, and its lowest point is at x=\tfrac52, where g\left(\tfrac52\right)=-\tfrac{25}{4}.
Horizontal line test. Draw the line y=-4. It cuts the parabola at two points.
x^{2}-5x=-4\;\Longrightarrow\;(x-1)(x-4)=0\;\Longrightarrow\;x=1,\ 4
So g(1)=g(4)=-4: two different inputs give the same output. The graph fails the test, so g is not one to one and has no inverse function on \mathbb{R}.

The horizontal line y=-4 meets the parabola twice, so g is not one to one.

\text{Not one to one; no inverse}

(iii) h(x)=(x+1)^{2}

Graph. This is y=x^{2} moved 1 unit left: a parabola opening upwards with vertex (-1,0).
Horizontal line test. The line y=1 cuts it twice.
(x+1)^{2}=1\;\Longrightarrow\;x+1=\pm1\;\Longrightarrow\;x=0,\ -2
h(0)=h(-2)=1, so h is not one to one and has no inverse on \mathbb{R}.

The line y=1 cuts the graph at two points.

\text{Not one to one; no inverse}

(iv) f(x)=x^{3}-8

Graph. This is y=x^{3} moved 8 units down. It passes through (0,-8) and (2,0) and rises all the time from left to right.
Horizontal line test. Because the curve is always increasing, every horizontal line cuts it exactly once. So f is one to one.
Find the inverse. Put y=f(x) and solve for x.
y=x^{3}-8\;\Longrightarrow\;x^{3}=y+8\;\Longrightarrow\;x=\sqrt[3]{y+8}
Interchange y with x.
f^{-1}(x)=\sqrt[3]{x+8}
Check.
f\left(f^{-1}(x)\right)=\left(\sqrt[3]{x+8}\right)^{3}-8=x+8-8=x
Graph of the inverse. Reflect the points in y=x: (2,0)\to(0,2) and (0,-8)\to(-8,0).

The cubic and the cube-root curve are mirror images in y=x.

\text{One to one};\quad f^{-1}(x)=\sqrt[3]{x+8}

(v) g(x)=\dfrac{4}{x}

Graph. A hyperbola in the first and third quadrants (x\neq0), with both axes as asymptotes. Points: (1,4),\ (2,2),\ (4,1) and (-1,-4),\ (-2,-2),\ (-4,-1).
Horizontal line test. A line y=k with k\neq0 cuts the graph only at x=\dfrac4k, and y=0 does not cut it at all. So g is one to one.
Find the inverse.
y=\dfrac4x\;\Longrightarrow\;xy=4\;\Longrightarrow\;x=\dfrac4y
Interchange y with x.
g^{-1}(x)=\dfrac4x,\quad x\neq0
Result. The inverse is the same function (a self-inverse function). Its graph is symmetric about y=x: the point (1,4) reflects to (4,1), which is also on the curve.

The graph of g is its own mirror image in y=x, so g and g^{-1} share one graph.

\text{One to one};\quad g^{-1}(x)=\dfrac{4}{x}\ \ (\text{same as }g)

(vi) h(x)=\dfrac{1}{3x+5}

Graph. The denominator is zero at x=-\tfrac53, which gives a vertical asymptote. The fraction is never 0, so y=0 is a horizontal asymptote. Points: h(0)=\tfrac15, h\left(-\tfrac43\right)=1, h(-2)=-1.
Horizontal line test. Each horizontal line y=k with k\neq0 cuts the graph once, and y=0 does not cut it. So h is one to one.
Find the inverse.
y=\dfrac{1}{3x+5}\;\Longrightarrow\;3x+5=\dfrac1y\;\Longrightarrow\;3x=\dfrac{1-5y}{y}\;\Longrightarrow\;x=\dfrac{1-5y}{3y}
Interchange y with x.
h^{-1}(x)=\dfrac{1-5x}{3x},\quad x\neq0
Check.
h\left(h^{-1}(x)\right)=\dfrac{1}{3\cdot\dfrac{1-5x}{3x}+5}=\dfrac{1}{\dfrac{1-5x+5x}{x}}=x
Domains and ranges swap. \mathrm{Dom}\,h=\mathbb{R}\setminus\{-\tfrac53\}=\mathrm{Rang}\,h^{-1} and \mathrm{Rang}\,h=\mathbb{R}\setminus\{0\}=\mathrm{Dom}\,h^{-1}. So h^{-1} has asymptotes x=0 and y=-\tfrac53.

The asymptotes swap too: x=-\tfrac53,\ y=0 for h become y=-\tfrac53,\ x=0 for h^{-1}.

\text{One to one};\quad h^{-1}(x)=\dfrac{1-5x}{3x},\ x\neq0

(vii) f(x)=x^{4}+2

Graph. x^{4}\ge0, so the lowest point is (0,2). The graph is U-shaped and symmetric about the y-axis, because (-x)^{4}=x^{4}.
Horizontal line test. The line y=3 cuts it twice.
x^{4}+2=3\;\Longrightarrow\;x^{4}=1\;\Longrightarrow\;x=\pm1
f(1)=f(-1)=3, so f is not one to one and has no inverse on \mathbb{R}.

Symmetric about the y-axis, so each output above 2 comes from two inputs.

\text{Not one to one; no inverse}

(viii) g(x)=5

Graph. A constant function: its graph is the horizontal line y=5.
Horizontal line test. The horizontal line y=5 lies along the graph, so it meets the graph at infinitely many points. For example g(-2)=g(0)=g(3)=5.
So g is not one to one (all inputs share one output) and it has no inverse.

Every input gives the same output 5.

\text{Not one to one; no inverse}

(ix) h(x)=|x|

Graph. A V shape with vertex at the origin: h(x)=x for x\ge0 and h(x)=-x for x<0.
Horizontal line test. The line y=2 cuts it at x=-2 and x=2.
h(-2)=h(2)=2, so h is not one to one and has no inverse on \mathbb{R}.

The two arms of the V give every positive output twice.

\text{Not one to one; no inverse}

Exercise 1.5p. 40

Draw the graphs of the given functions and then sketch the graphs of other functions using translation. Verify the results using graphical calculator. Rules (Section 1.13): y=f(x)\pm c moves the graph c units up or down; y=f(x+c) moves it c units left and y=f(x-c) moves it c units right; y=cf(x) scales it vertically, y=f(cx) scales it horizontally, and a negative factor also reflects it. In every graph below, the basic curve is dashed.

12026 Annual I, Old syllabus (Curriculum 2006), Q.2 (i), old syllabus

Draw the graph of y=|x| and then sketch the graphs of the other functions using translation.

Solution

Basic graph. Table of values for y=|x|.
\begin{array}{c|ccccccc} x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline y & 3 & 2 & 1 & 0 & 1 & 2 & 3 \end{array}
The graph is a V with its vertex (corner) at (0,0), symmetric about the y-axis. Domain \mathbb{R}, range [0,\infty).

The basic graph y=|x|.

(a) y=|x+2|

Rule. |x+2|=f(x+2) with f(x)=|x|. A constant added to x gives a horizontal shift: the graph moves 2 units left (Table 1.1).
Key point. The vertex is where the inside is zero: x+2=0, so the vertex moves from (0,0) to (-2,0).
Check. At x=0: y=|0+2|=2, so (0,2) is on the new graph. The range stays [0,\infty).

\text{Shift 2 units left; vertex }(-2,\,0)

(b) y=|x-2|

Rule. |x-2|=f(x-2): 2 is subtracted from x, so the graph moves 2 units right.
Key point. Vertex where x-2=0: it moves from (0,0) to (2,0).
Table of values to confirm the V shape.
\begin{array}{c|ccccccc} x & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline y & 3 & 2 & 1 & 0 & 1 & 2 & 3 \end{array}
Domain and range. |x-2| is defined for every real x, so the domain is \mathbb{R}. A modulus is never negative and |x-2|=0 at x=2, so the range is [0,\infty).

\begin{gathered}\text{Shift 2 units right; vertex }(2,\,0)\\ \text{Domain}=\mathbb{R}\\ \text{Range}=[0,\infty)\end{gathered}

Rule. |x|+2=f(x)+2: 2 is added to every y-coordinate, so the graph moves 2 units up.
Key point. The vertex moves from (0,0) to (0,2).
Check. At x=3: y=3+2=5. The range becomes [2,\infty).

\text{Shift 2 units up; vertex }(0,\,2)

(d) y=|x|-2

Rule. |x|-2=f(x)-2: 2 is subtracted from every y-coordinate, so the graph moves 2 units down.
Key points. The vertex moves to (0,-2). The graph now cuts the x-axis where |x|=2, that is at x=\pm2.
Range. [-2,\infty).

\text{Shift 2 units down; vertex }(0,\,-2)

Draw the graph of y=x^{2} and then sketch the graphs of the other functions using translation.

Solution

Basic graph. Table of values for y=x^{2}.
\begin{array}{c|ccccccc} x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline y & 9 & 4 & 1 & 0 & 1 & 4 & 9 \end{array}
A parabola opening upwards with vertex (0,0), symmetric about the y-axis. Domain \mathbb{R}, range [0,\infty).

The basic parabola y=x^{2}.

(a) y=x^{2}+4

Rule. x^{2}+4=f(x)+4: every y-coordinate increases by 4, so the parabola moves 4 units up.
Key point. Vertex (0,0)\to(0,4). The range becomes [4,\infty).
Check. At x=2: y=4+4=8.

\text{Shift 4 units up; vertex }(0,\,4)

(b) y=x^{2}-4

Rule. x^{2}-4=f(x)-4: the parabola moves 4 units down.
Key points. Vertex (0,0)\to(0,-4). It cuts the x-axis where x^{2}=4, that is at x=\pm2.
Range. [-4,\infty).

\text{Shift 4 units down; vertex }(0,\,-4)

Rule. (x-4)^{2}=f(x-4): 4 is subtracted from x, so the parabola moves 4 units right.
Key point. The vertex is where x-4=0: (0,0)\to(4,0). The axis of symmetry is now x=4.
Check. At x=2: y=(2-4)^{2}=4, so (2,4) is on the graph.

\text{Shift 4 units right; vertex }(4,\,0)

(d) y=(x+4)^{2}

Rule. (x+4)^{2}=f(x+4): 4 is added to x, so the parabola moves 4 units left.
Key point. Vertex where x+4=0: (0,0)\to(-4,0). Axis of symmetry x=-4.
Check. At x=-2: y=(-2+4)^{2}=4.

\text{Shift 4 units left; vertex }(-4,\,0)

Draw the graph of y=\sqrt{x} and then sketch the graphs of the other functions using translation.

Solution

Basic graph. Table of values for y=\sqrt{x} (perfect squares make it easy).
\begin{array}{c|cccc} x & 0 & 1 & 4 & 9 \\ \hline y & 0 & 1 & 2 & 3 \end{array}
The graph starts at (0,0) and rises more and more slowly. Domain [0,\infty), range [0,\infty).

The basic graph y=\sqrt{x} starts at the origin.

(a) y=\sqrt{x+3}

Rule. \sqrt{x+3}=f(x+3): the graph moves 3 units left, as in Example 24.
Starting point. The root needs x+3\ge0, so the curve starts where x=-3: (0,0)\to(-3,0). Domain [-3,\infty).
Check. At x=1: y=\sqrt4=2.

\text{Shift 3 units left; starts at }(-3,\,0)

(b) y=\sqrt{x-3}

Rule. \sqrt{x-3}=f(x-3): the graph moves 3 units right.
Starting point. x-3\ge0 gives x\ge3, so the curve starts at (3,0). Domain [3,\infty).
Check. At x=7: y=\sqrt4=2.

\text{Shift 3 units right; starts at }(3,\,0)

Rule. \sqrt{x}+3=f(x)+3: the graph moves 3 units up.
Starting point. (0,0)\to(0,3). The domain stays [0,\infty); the range becomes [3,\infty).
Check. At x=4: y=2+3=5.

\text{Shift 3 units up; starts at }(0,\,3)

(d) y=\sqrt{x}-3

Rule. \sqrt{x}-3=f(x)-3: the graph moves 3 units down.
Starting point. (0,0)\to(0,-3). Range [-3,\infty).
Intercept. \sqrt{x}-3=0 gives \sqrt{x}=3, so x=9: the curve crosses the x-axis at (9,0).

\text{Shift 3 units down; starts at }(0,\,-3)

Draw the graph of y=x and then sketch the graphs of the other functions.

Solution

Basic graph. Table of values for y=x.
\begin{array}{c|ccccc} x & -2 & -1 & 0 & 1 & 2 \\ \hline y & -2 & -1 & 0 & 1 & 2 \end{array}
A straight line through the origin with slope 1, making 45^\circ with the x-axis. Domain and range are both \mathbb{R}.

The identity line y=x.

(a) y=x+5

Rule. x+5=f(x)+5: every point moves 5 units up. The new line is parallel to y=x (same slope 1).
Intercepts. y-intercept (0,5); x-intercept where x+5=0, that is (-5,0).
For a straight line, moving 5 units up and moving 5 units left give the same line, since x+5=f(x+5) as well.

\text{Shift 5 units up; parallel line through }(0,\,5)

(b) y=x-5

Rule. x-5=f(x)-5: every point moves 5 units down. The slope stays 1.
Intercepts. y-intercept (0,-5); x-intercept (5,0).

\text{Shift 5 units down; parallel line through }(0,\,-5)

Rule. 5x=5f(x). This is not a translation but a scaling: every y-coordinate is multiplied by 5 while x stays the same (Table 1.2).
Effect. The line still passes through the origin, but its slope becomes 5, so it is much steeper: (1,1)\to(1,5).

\text{Vertical stretch by }5;\ \text{slope }5

(d) y=-5x

Rule. -5x=-5f(x): multiply every y-coordinate by -5. The factor 5 stretches the line and the minus sign reflects it in the x-axis.
Effect. A line through the origin with slope -5, falling steeply from left to right: (1,1)\to(1,-5) and (-1,-1)\to(-1,5).

\text{Stretch by }5\text{ and reflect in the }x\text{-axis; slope }-5

Draw the graph of y=x^{3} and then sketch the graphs of the other functions using translation.

Solution

Basic graph. Table of values for y=x^{3}.
\begin{array}{c|ccccc} x & -2 & -1 & 0 & 1 & 2 \\ \hline y & -8 & -1 & 0 & 1 & 8 \end{array}
The curve rises from left to right and is symmetric about the origin. It flattens out at its centre point (0,0), where it changes the way it bends. Domain and range are \mathbb{R}.

The basic cubic y=x^{3}.

(a) y=x^{3}+1

Rule. x^{3}+1=f(x)+1: the curve moves 1 unit up.
Key points. Centre (0,0)\to(0,1). It cuts the x-axis where x^{3}=-1, that is at (-1,0).

\text{Shift 1 unit up; centre }(0,\,1)

(b) y=x^{3}-1

Rule. x^{3}-1=f(x)-1: the curve moves 1 unit down.
Key points. Centre (0,0)\to(0,-1); x-intercept where x^{3}=1, that is (1,0).

\text{Shift 1 unit down; centre }(0,\,-1)

Rule. (x-1)^{3}=f(x-1): the curve moves 1 unit right.
Key points. Centre (0,0)\to(1,0); y-intercept y=(0-1)^{3}=-1, that is (0,-1).

\text{Shift 1 unit right; centre }(1,\,0)

(d) y=(x+1)^{3}

Rule. (x+1)^{3}=f(x+1): the curve moves 1 unit left.
Key points. Centre (0,0)\to(-1,0); y-intercept (0,1).
Parts (a) and (d) both pass through (-1,0) and (0,1), but they are different curves: compare their centres (0,1) and (-1,0).

\text{Shift 1 unit left; centre }(-1,\,0)

Draw the graph of y=x^{2}+4 and then sketch the graphs of the other functions using translation.

Solution

Basic graph. Table of values for y=x^{2}+4.
\begin{array}{c|ccccc} x & -2 & -1 & 0 & 1 & 2 \\ \hline y & 8 & 5 & 4 & 5 & 8 \end{array}
This is y=x^{2} moved 4 units up: a parabola with vertex (0,4). Domain \mathbb{R}, range [4,\infty).

The basic graph y=x^{2}+4.

(a) y=(x^{2}+4)-3

Rule. f(x)-3 with f(x)=x^{2}+4: the graph moves 3 units down. It simplifies to y=x^{2}+1.
Key point. Vertex (0,4)\to(0,1); range [1,\infty).

\text{Shift 3 units down; vertex }(0,\,1)

(b) y=(x^{2}+4)+3

Rule. f(x)+3: the graph moves 3 units up. It simplifies to y=x^{2}+7.
Key point. Vertex (0,4)\to(0,7); range [7,\infty).

\text{Shift 3 units up; vertex }(0,\,7)

Rule. f(x-3)=(x-3)^{2}+4: x is replaced by x-3, so the graph moves 3 units right.
Key point. Vertex (0,4)\to(3,4); axis of symmetry x=3. The range stays [4,\infty).

\text{Shift 3 units right; vertex }(3,\,4)

(d) y=(x+3)^{2}+4

Rule. f(x+3)=(x+3)^{2}+4: the graph moves 3 units left.
Key point. Vertex (0,4)\to(-3,4); axis of symmetry x=-3.

\text{Shift 3 units left; vertex }(-3,\,4)

Draw the graph of y=x^{2} and then sketch the graphs of the other functions.

Solution

Basic graph. Table of values for y=x^{2}.
\begin{array}{c|ccccccc} x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline y & 9 & 4 & 1 & 0 & 1 & 4 & 9 \end{array}
Parabola with vertex (0,0), opening upwards. Scaling y keeps the vertex at the origin and changes how quickly the arms rise.

The basic parabola y=x^{2}.

(a) y=3x^{2}

Rule. 3x^{2}=3f(x): every y-coordinate is multiplied by 3 while x stays the same. This is a vertical stretch by factor 3.
Points. (1,1)\to(1,3) and (2,4)\to(2,12). The vertex stays at (0,0) and the parabola becomes narrower.

\text{Vertical stretch by factor }3\ (\text{narrower})

(b) y=-3x^{2}

Rule. -3x^{2}=-3f(x): multiply every y-coordinate by -3. The 3 stretches the graph and the minus sign reflects it in the x-axis.
Points. (1,1)\to(1,-3) and (2,4)\to(2,-12). The parabola opens downwards with vertex (0,0); range (-\infty,0].

\text{Stretch by }3\text{ and reflect in the }x\text{-axis}

Rule. \dfrac{x^{2}}{3}=\dfrac{f(x)}{3}: every y-coordinate is divided by 3. This is a vertical compression by factor 3, so the parabola becomes wider.
Points. (3,9)\to(3,3) and (1,1)\to\left(1,\tfrac13\right). The vertex stays at (0,0).

\text{Vertical compression by factor }3\ (\text{wider})

(d) y=-\dfrac{x^{2}}{3}

Rule. -\dfrac{x^{2}}{3}=\dfrac{f(x)}{-3}: divide every y-coordinate by 3 and change its sign. The graph is compressed (wider) and reflected in the x-axis.
Points. (3,9)\to(3,-3); the parabola opens downwards with vertex (0,0).

\text{Compress by }3\text{ and reflect in the }x\text{-axis}

Draw the graph of y=x^{2} and then sketch the graphs of the other functions.

Solution

Basic graph. Table of values for y=x^{2}.
\begin{array}{c|ccccccc} x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline y & 9 & 4 & 1 & 0 & 1 & 4 & 9 \end{array}
Here the number multiplies x inside the bracket, so each change is a horizontal scaling (Table 1.3): f(cx) divides every x-coordinate by c.

The basic parabola y=x^{2}.

(a) y=(3x)^{2}

Rule. (3x)^{2}=f(3x): every x-coordinate is divided by 3 while y stays the same. This is a horizontal compression by factor 3.
Points. (3,9)\to(1,9) and (-3,9)\to(-1,9). The parabola becomes narrower. Algebraically (3x)^{2}=9x^{2}.

\text{Horizontal compression by factor }3;\ y=9x^{2}

(b) y=(-3x)^{2}

Rule. (-3x)^{2}=f(-3x): compress horizontally by 3 and reflect in the y-axis.
Result. The parabola is symmetric about the y-axis, so the reflection changes nothing. Indeed (-3x)^{2}=9x^{2}=(3x)^{2}: the graph is the same as in part (a).

\text{Same graph as (a)}:\ y=9x^{2}

Rule. \left(\dfrac{x}{3}\right)^{2}=f\left(\dfrac{x}{3}\right): every x-coordinate is multiplied by 3. This is a horizontal stretch by factor 3.
Points. (1,1)\to(3,1) and (2,4)\to(6,4). The parabola becomes wider. Algebraically \left(\dfrac{x}{3}\right)^{2}=\dfrac{x^{2}}{9}.

\text{Horizontal stretch by factor }3;\ y=\tfrac{x^{2}}{9}

(d) y=\left(-\dfrac{x}{3}\right)^{2}

Rule. f\left(-\dfrac{x}{3}\right): stretch horizontally by 3 and reflect in the y-axis.
Result. Again the reflection has no effect on this symmetric parabola: \left(-\dfrac{x}{3}\right)^{2}=\dfrac{x^{2}}{9}, the same graph as part (c).

\text{Same graph as (c)}:\ y=\tfrac{x^{2}}{9}

Draw the graph of y=\sqrt{x} and then sketch the graphs of the other functions.

Solution

Basic graph. Table of values for y=\sqrt{x}.
\begin{array}{c|cccc} x & 0 & 1 & 4 & 9 \\ \hline y & 0 & 1 & 2 & 3 \end{array}
Starts at (0,0) and rises more and more slowly. Domain [0,\infty), range [0,\infty).

The basic graph y=\sqrt{x}.

(a) y=\sqrt{2x}

Rule. \sqrt{2x}=f(2x): every x-coordinate is divided by 2 (horizontal compression by factor 2).
Points. (4,2)\to(2,2), (1,1)\to\left(\tfrac12,1\right) and (16,4)\to(8,4). The start (0,0) does not move; domain [0,\infty).

\text{Horizontal compression by factor }2

(b) y=2\sqrt{x}

Rule. 2\sqrt{x}=2f(x): every y-coordinate is doubled (vertical stretch by factor 2).
Points. (1,1)\to(1,2) and (4,2)\to(4,4). Start (0,0); range [0,\infty).

\text{Vertical stretch by factor }2

Rule. Two changes: first stretch vertically by 2 (part b), then move 3 units up.
Points. (0,0)\to(0,0)\to(0,3) and (4,2)\to(4,4)\to(4,7). The curve starts at (0,3); range [3,\infty).

\text{Stretch by }2,\text{ then shift 3 up; starts at }(0,\,3)

(d) y=\sqrt{2x+5}

Rewrite. Take out the factor 2 so that the shift can be read off.
\sqrt{2x+5}=\sqrt{2\left(x+\tfrac52\right)}=f\left(2\left(x+\tfrac52\right)\right)
Rule. Compress horizontally by 2 (as in part a), then move \tfrac52 units left.
Starting point. 2x+5\ge0 gives x\ge-\tfrac52, so the curve starts at \left(-\tfrac52,0\right); domain \left[-\tfrac52,\infty\right). Check: at x=2, y=\sqrt9=3.

\begin{gathered}\text{Compress by }2,\text{ then shift }\tfrac52\text{ left}\\ \text{starts at }\left(-\tfrac52,\,0\right)\end{gathered}

Review Exercisep. 40–42

Revision of the whole unit: types of functions, inverses, domain and range, exponential and logarithmic models, trigonometric graphs and transformations.

1 (i)

Which of the following is an example of exponential growth function?

Af(x)=3x+4
Bf(x)=3^{x}\times5Answer
Cf(x)=x^{3}
Df(x)=x^{2}

Why

In an exponential function the variable is in the exponent. f(x)=3^{x}\times5=5(3)^{x} has base 3>1, so it grows exponentially.

Options (a), (c) and (d) are linear, cubic and quadratic functions: in them x is in the base, not in the exponent.

1 (ii)

The exponential decay function is expressed by:

Af(x)=a\cdot b^{x};\ 0 < b < 1Answer
Bf(x)=a\cdot b^{x};\ b>1
Cf(x)=a\cdot b^{x};\ 0<a<1
Df(x)=a\cdot b^{x};\ a>1

Why

In f(x)=a\cdot b^{x} the base b decides growth or decay. When 0 < b < 1 each unit step in x multiplies the value by a number less than 1, so the function decays; b>1 gives growth.

Options (c) and (d) restrict a, the starting value, which does not decide decay.

1 (iii)

The logarithmic function f(x)=\log_b x is defined for:

Aall real numbers
Bx<0
Cx>0Answer
Dx\ge0

Why

\log_b x=y means b^{y}=x. A positive base raised to any power is positive, so x must be greater than 0. The logarithm of 0 does not exist, so (d) is wrong.

1 (iv)

What is the value of \log_5125?

A25
B5
C4
D3Answer

Why

125=5^{3}, so \log_5125=\log_5 5^{3}=3\log_5 5=3(1)=3.

1 (v)

A function f:A\to B is said to be onto if:

AEvery element of the set A has a unique image in the set B.
BEvery element in the set B has a preimage in the set A.Answer
CSome elements of the set B have no preimage in the set A.
Df is both one to one and onto.

Why

Onto means every element of the codomain B is the image of at least one element of A, so \mathrm{Rang}\,f=B.

Option (a) is true of every function, (c) describes an into function and (d) describes a bijective function.

1 (vi)

The function f(x)=x+1, where f:\{1,2,3\}\to\{2,3,4\}, is:

Aone to one but not onto
Bonto but not one to one
Cboth one to one and ontoAnswer
Dneither one to one nor onto

Why

f(1)=2, f(2)=3 and f(3)=4. Different inputs have different images, so f is one to one. The range \{2,3,4\} equals the codomain, so f is onto. Hence f is bijective.

1 (vii)

The function f:\mathbb{R}\to[0,\infty) defined by f(x)=x^{2}+1, is:

Aonto but not one to one
Bone to one but not onto
Cneither one to one nor ontoAnswer
Dboth one to one and onto

Why

f(-1)=f(1)=2, so f is not one to one. Also x^{2}+1\ge1, so the range is [1,\infty): values such as 0 and 0.5 in the codomain [0,\infty) have no pre-image, so f is not onto.

Option (a) is tempting, but it would be true only if the codomain were [1,\infty).

1 (viii)

A function f:A\to B has an inverse if and only if:

Af is one to one
Bf is onto
Cf is both one to one and ontoAnswer
Df is neither one to one and onto

Why

f^{-1} must send every element of B back to exactly one element of A. Every element of B needs a pre-image (onto), and only one pre-image (one to one). So f must be bijective.

1 (ix)

The inverse function of f(x)=x^{3}, is:

Af^{-1}(x)=x^{-3}
Bf^{-1}(x)=\sqrt{x^{-3}}
Cf^{-1}(x)=\sqrt[3]{x^{3}}
Df^{-1}(x)=\sqrt[3]{x}Answer

Why

Put y=x^{3} and solve: x=\sqrt[3]{y}. Interchanging gives f^{-1}(x)=\sqrt[3]{x}.

Option (a) is the reciprocal \dfrac{1}{x^{3}}, not the inverse, and (c) simplifies to x.

1 (x)

The function f(x)=\sin x, where f:\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\to[-1,1], is:

Aone to one but not onto
Bonto but not one to one
Cboth one to one and ontoAnswer
Dneither one to one nor onto

Why

On \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right] the sine curve rises steadily from -1 to 1. No output repeats (one to one) and every value in [-1,1] is reached (onto). This is why this interval is used to define \sin^{-1}x.

1 (xi)

The inverse function of f(x)=\dfrac{1}{x};\ x\neq0, is:

Af^{-1}(x)=1
Bf^{-1}(x)=-x
Cf^{-1}(x)=x
Df^{-1}(x)=\dfrac{1}{x}Answer

Why

Put y=\dfrac1x and solve for x: x=\dfrac1y. Interchanging gives f^{-1}(x)=\dfrac1x. The function is its own inverse; its graph is symmetric about the line y=x.

1 (xii)

Scaling refers to:

Aincreasing the size of an object.
Bdecreasing the size of an object.
Cmaintaining the properties while resizing an object.Answer
Dchanging the shape of an object.

Why

Scaling multiplies or divides the coordinates by a constant, so it resizes a figure. It can enlarge or reduce it, so (a) and (b) each describe only one case.

The figure keeps its basic properties: for example, y=3x^{2} is still a parabola with its vertex at the origin, only narrower. So (c) is the best description; scaling does not turn one kind of curve into a different kind, as (d) suggests.

Note. The book calls scaling a non-rigid transformation because a graph can become narrower or wider (Section 1.13.2). Of the four options, (c) is the most complete description of scaling.

1 (xiii)

Which of the following statements is true for the uniform scaling?

Aboth width and height change proportionally.Answer
Bonly the width changes.
Conly the height changes.
Dwidth and height remain unchanged.

Why

In uniform scaling the same scale factor is used in every direction, so width and height change in the same ratio and the figure keeps its proportions. Options (b) and (c) describe non-uniform scaling, and (d) is no scaling at all.

1 (xiv)

What is the effect on the graph of f(x) when it is replaced by f(x+2)?

AIt shifts 2 units to the right.
BIt shifts 2 units to the left.Answer
CIt shifts 2 units up.
DIt shifts 2 unit down.

Why

Adding a positive constant to x inside the function moves the graph that many units to the left (Table 1.1). For example, y=(x+2)^{2} has its vertex at (-2,0). A shift to the right would need f(x-2).

1 (xv)

The domain of y=\sin^{-1}(x), is:

A[0,\infty)
B(-\infty,\infty)
C[-1,1]Answer
D[0,1]

Why

\sin^{-1}x is the angle whose sine is x. Since -1\le\sin\theta\le1 for every angle, only x from -1 to 1 can be used. Its principal values lie in \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right].

Find the domain of the given functions.

Solution

(a) f(x)=4+\sqrt{x+2}

The constant 4 is defined for every x. The square root needs a non-negative input.
x+2\ge0\;\Longrightarrow\;x\ge-2
So every x from -2 onwards is allowed. The graph starts at (-2,4).

The graph exists only for x\ge-2.

\text{Domain}=[-2,\,\infty)

(b) f(x)=x\sqrt{2x-3}

The factor x is defined for every real x. The square root needs
2x-3\ge0\;\Longrightarrow\;x\ge\dfrac32
Both factors must be defined, so only x\ge\tfrac32 is allowed. The graph starts at \left(\tfrac32,0\right); for example f(2)=2\sqrt1=2.

No graph to the left of x=\tfrac32.

\text{Domain}=\left[\tfrac32,\,\infty\right)

A fraction is not defined when its denominator is zero.
x-2=0\;\Longrightarrow\;x=2
Every other real number is allowed.
\text{Domain}=\mathbb{R}\setminus\{2\}=(-\infty,2)\cup(2,\infty)
On the graph, x=2 is a vertical asymptote. Since \dfrac{x}{x-2}=1+\dfrac{2}{x-2}, the line y=1 is a horizontal asymptote.

The graph breaks at the excluded value x=2.

\text{Domain}=\mathbb{R}\setminus\{2\}

(d) f(x)=\sqrt{x^{2}-5x+4}

The expression under the root must be \ge0. Factorise it.
x^{2}-5x+4=(x-1)(x-4)\ge0
Sign table. The product is positive when both factors have the same sign, and zero at x=1 and x=4.

Interval	x-1	x-4	(x-1)(x-4)
x<1	-	-	+
1<x<4	+	-	-
x>4	+	+	+

Interval	x<1	1<x<4	x>4
x-1	-	+	+
x-4	-	-	+
(x-1)(x-4)	+	-	+

Keep the intervals where the product is positive, together with the zeros x=1 and x=4. Reject 1<x<4, where it is negative.
\text{Domain}=(-\infty,1]\cup[4,\infty)

No part of the graph lies over the interval 1<x<4.

\text{Domain}=(-\infty,\,1]\cup[4,\,\infty)

Find the domain and range of the given functions.

Solution

(a) f(x)=1+x^{2}

Domain. A polynomial is defined for every real x, so the domain is \mathbb{R}.
Range. A square is never negative, and x^{2}=0 only at x=0.
x^{2}\ge0\;\Longrightarrow\;1+x^{2}\ge1
The least value is f(0)=1, and x^{2} can be as large as we like. Every y\ge1 is reached (take x=\sqrt{y-1}).

The lowest point is (0,1), so the range starts at 1.

\text{Domain}=\mathbb{R},\qquad \text{Range}=[1,\,\infty)

(b) f(x)=(2x+1)^{2}

Domain. A polynomial: every real x is allowed.
Range. A square is never negative, and it is zero when 2x+1=0, that is at x=-\tfrac12.
(2x+1)^{2}\ge0
Every y\ge0 is reached: x=\dfrac{-1+\sqrt{y}}{2} gives f(x)=y.

A parabola with vertex \left(-\tfrac12,0\right) on the x-axis.

\text{Domain}=\mathbb{R},\qquad \text{Range}=[0,\,\infty)

Domain. \sqrt{x} needs x\ge0, so the domain is [0,\infty).
Range. \sqrt{x}\ge0, so 9-\sqrt{x}\le9. The greatest value is f(0)=9.
\sqrt{x}\ge0\;\Longrightarrow\;9-\sqrt{x}\le9
As x increases, \sqrt{x} grows without limit, so f(x) decreases without limit: for example f(81)=0 and f(144)=-3. Every value y\le9 is reached.

The graph starts at its highest point (0,9) and falls for ever.

\text{Domain}=[0,\,\infty),\qquad \text{Range}=(-\infty,\,9]

(d) f(x)=3+\sqrt{4-x^{2}}

Domain. The root needs a non-negative input.
4-x^{2}\ge0\;\Longrightarrow\;x^{2}\le4\;\Longrightarrow\;-2\le x\le2
Range. For -2\le x\le2, x^{2} runs from 0 to 4, so 4-x^{2} runs from 0 to 4 and its square root from 0 to 2.
0\le\sqrt{4-x^{2}}\le2\;\Longrightarrow\;3\le f(x)\le5
The least value 3 occurs at x=\pm2 and the greatest value 5 at x=0.
Graph. y-3=\sqrt{4-x^{2}} gives x^{2}+(y-3)^{2}=4 with y\ge3: the upper half of the circle with centre (0,3) and radius 2.

A semicircle: x runs from -2 to 2 and y from 3 to 5.

\text{Domain}=[-2,\,2],\qquad \text{Range}=[3,\,5]

Draw the graph of f(x)=\sqrt{x}, then sketch the graphs of the following functions.

Solution

Basic graph. Table of values for f(x)=\sqrt{x}.
\begin{array}{c|cccc} x & 0 & 1 & 4 & 9 \\ \hline f(x) & 0 & 1 & 2 & 3 \end{array}
The graph starts at (0,0) and rises more and more slowly. Domain [0,\infty), range [0,\infty). Each new graph below is obtained from it by a shift, a scaling or a reflection.

The basic graph f(x)=\sqrt{x}.

(a) f(x)=\sqrt{x-2}

Rule. \sqrt{x-2} is f(x-2): the graph moves 2 units right.
Key points. Start (0,0)\to(2,0) and (4,2)\to(6,2). Domain [2,\infty), range [0,\infty).

\text{Shift 2 units right; starts at }(2,\,0)

(b) f(x)=\sqrt{x}+4

Rule. \sqrt{x}+4 is f(x)+4: the graph moves 4 units up.
Key points. Start (0,0)\to(0,4) and (4,2)\to(4,6). Domain [0,\infty), range [4,\infty).

\text{Shift 4 units up; starts at }(0,\,4)

Rule. -\sqrt{x} is -f(x): every y-coordinate changes sign, so the graph is reflected in the x-axis.
Key points. (4,2)\to(4,-2) and (9,3)\to(9,-3). Domain [0,\infty), range (-\infty,0].

\text{Reflection in the }x\text{-axis}

(d) f(x)=1+\sqrt{x-2}

Rule. Two shifts: x-2 inside the root moves the graph 2 units right, and +1 outside moves it 1 unit up.
Key points. Start (0,0)\to(2,1) and (4,2)\to(6,3). Domain [2,\infty), range [1,\infty).

\text{Shift 2 right and 1 up; starts at }(2,\,1)

(e) f(x)=4\sqrt{x}

Rule. 4\sqrt{x} is 4f(x): every y-coordinate is multiplied by 4 (vertical stretch by factor 4).
Key points. (1,1)\to(1,4) and (4,2)\to(4,8). The start (0,0) stays; range [0,\infty).

\text{Vertical stretch by factor }4

(f) f(x)=-\dfrac{1}{3}\sqrt{x}

Rule. -\tfrac13\sqrt{x} is -\tfrac13f(x): every y-coordinate is divided by 3 (vertical compression by factor 3) and changes sign (reflection in the x-axis).
Key points. (9,3)\to(9,-1) and (4,2)\to\left(4,-\tfrac23\right). Range (-\infty,0].

\text{Compress by }3\text{ and reflect in the }x\text{-axis}

Graph the given functions.

Solution

(a) y=2+2\sin x

Build from the sine curve. Multiplying \sin x by 2 gives amplitude 2 (values from -2 to 2). Adding 2 moves the wave 2 units up (values from 0 to 4).
Table of values over one period.
\begin{array}{c|ccccc} x & 0 & \tfrac{\pi}{2} & \pi & \tfrac{3\pi}{2} & 2\pi \\ \hline \sin x & 0 & 1 & 0 & -1 & 0 \\ \hline y & 2 & 4 & 2 & 0 & 2 \end{array}
Features. Period 2\pi and midline y=2. Maximum 4 at x=\tfrac{\pi}{2}; minimum 0 at x=\tfrac{3\pi}{2}, where the curve just touches the x-axis. Domain \mathbb{R}, range [0,4].

A sine wave of amplitude 2 about the midline y=2.

\text{Period }2\pi,\qquad \text{Range}=[0,\,4]

(b) y=-\dfrac{1}{2}\tan x

Build from the tangent curve. Compared with \tan x, the factor \tfrac12 makes the curve less steep, and the minus sign reflects it in the x-axis, so each branch now falls from left to right.
Asymptotes. As for \tan x, where \cos x=0: x=(2n+1)\tfrac{\pi}{2}.
Table of values between two asymptotes.
\begin{array}{c|ccccc} x & -\tfrac{\pi}{3} & -\tfrac{\pi}{4} & 0 & \tfrac{\pi}{4} & \tfrac{\pi}{3} \\ \hline y & 0.87 & 0.5 & 0 & -0.5 & -0.87 \end{array}
Features. Period \pi; passes through the origin. Domain: x\neq(2n+1)\tfrac{\pi}{2}; range \mathbb{R}.

Dashed lines are the asymptotes x=\pm\tfrac{\pi}{2},\ \pm\tfrac{3\pi}{2}. Each branch falls from +\infty to -\infty.

\text{Period }\pi,\qquad \text{Range}=\mathbb{R}

Asymptotes. \operatorname{cosec}x=\dfrac{1}{\sin x} is not defined where \sin x=0, that is at x=n\pi (0,\ \pm\pi,\ \pm2\pi,\dots).
Build from the cosecant curve. The minus sign reflects its branches in the x-axis, and adding 3 moves everything 3 units up.
Table of values.
\begin{array}{c|cccccc} x & \tfrac{\pi}{6} & \tfrac{\pi}{2} & \tfrac{5\pi}{6} & \tfrac{7\pi}{6} & \tfrac{3\pi}{2} & \tfrac{11\pi}{6} \\ \hline \operatorname{cosec}x & 2 & 1 & 2 & -2 & -1 & -2 \\ \hline y & 1 & 2 & 1 & 5 & 4 & 5 \end{array}
Range. Where \operatorname{cosec}x\ge1 (on 0<x<\pi), y\le2: these branches open downwards with highest point 2 at x=\tfrac{\pi}{2}. Where \operatorname{cosec}x\le-1 (on \pi<x<2\pi), y\ge4: these branches open upwards with lowest point 4 at x=\tfrac{3\pi}{2}. No value between 2 and 4 is reached.

Asymptotes at x=n\pi (dashed; x=0 is the y-axis). The band between the dotted lines y=2 and y=4 is never reached.

\text{Domain: }x\neq n\pi,\qquad \text{Range}=(-\infty,\,2]\cup[4,\,\infty)

(d) y=\cos(x+\pi)

Rule. \cos(x+\pi)=f(x+\pi) with f(x)=\cos x, so the cosine graph moves \pi units to the left.
Simplify. The compound angle formula shows that the same graph is the reflection of \cos x in the x-axis.
\cos(x+\pi)=\cos x\cos\pi-\sin x\sin\pi=-\cos x
Table of values.
\begin{array}{c|ccccc} x & 0 & \tfrac{\pi}{2} & \pi & \tfrac{3\pi}{2} & 2\pi \\ \hline x+\pi & \pi & \tfrac{3\pi}{2} & 2\pi & \tfrac{5\pi}{2} & 3\pi \\ \hline y & -1 & 0 & 1 & 0 & -1 \end{array}
Features. Period 2\pi, domain \mathbb{R}, range [-1,1]. Minimum -1 at x=0 and maximum 1 at x=\pi.

\cos(x+\pi) (solid) is \cos x (dashed) moved \pi units left, which is the same as turning it upside down.

y=-\cos x;\qquad \text{Range}=[-1,\,1]

Find the domain and range of the inverse function of f(x)=\log(x^{2}+1).

Solution

Domain. x^{2}+1\ge1>0 for every real x, so the logarithm is always defined.
\mathrm{Dom}\,f=\mathbb{R}
Range. x^{2}+1\ge1 and \log is an increasing function, so \log(x^{2}+1)\ge\log1=0. As x grows, x^{2}+1 and its logarithm grow without limit.
\mathrm{Rang}\,f=[0,\infty)
One to one test. f is not one to one, because f(-x)=f(x): for example f(-3)=f(3)=\log10=1, so the graph fails the horizontal line test. An inverse exists only when the domain is restricted to one side of 0. Take x\ge0; this keeps the whole range [0,\infty).
Find the inverse (for x\ge0). Put y=f(x) and use \log u=y\iff u=10^{y}.
y=\log(x^{2}+1)\;\Longrightarrow\;x^{2}+1=10^{y}\;\Longrightarrow\;x=\sqrt{10^{y}-1}
Interchange x and y.
f^{-1}(x)=\sqrt{10^{x}-1}
Domain and range swap (Section 1.12.4).
\mathrm{Dom}\,f^{-1}=\mathrm{Rang}\,f=[0,\infty),\qquad \mathrm{Rang}\,f^{-1}=\mathrm{Dom}\,f\ (x\ge0)=[0,\infty)
Check. f^{-1}(x)=\sqrt{10^{x}-1} needs 10^{x}\ge1, that is x\ge0, which agrees with the domain found.

The right half of f and its inverse are mirror images in y=x: (3,1) on f matches (1,3) on f^{-1}.

\mathrm{Dom}\,f^{-1}=[0,\infty),\qquad \mathrm{Rang}\,f^{-1}=[0,\infty)

Note. f is not one to one on \mathbb{R}, so its inverse exists only on a restricted domain. The solution uses x\ge0 and explains the other choice.

Show that f\left(g(x)\right)=g\left(f(x)\right)=x, when: f(x)=e^{x} and g(x)=\ln x.

Solution

Facts used. \ln x is the logarithm to base e, so \ln x=y\iff e^{y}=x. Hence e^{\ln x}=x for x>0 (the property a^{\log_a y}=y), and \ln e^{x}=x\ln e=x by the power rule, since \ln e=1.
First composition. Replace x in f by g(x)=\ln x (this needs x>0).
f\left(g(x)\right)=f(\ln x)=e^{\ln x}=x
Second composition. Replace x in g by f(x)=e^{x} (defined for every real x, since e^{x}>0).
g\left(f(x)\right)=g\left(e^{x}\right)=\ln\left(e^{x}\right)=x\ln e=x
Conclusion. Both compositions give back x, so f\left(g(x)\right)=g\left(f(x)\right)=x. Hence f and g are inverses of each other, and their graphs are reflections in the line y=x.

e^{x} and \ln x are mirror images in y=x: (0,1)\leftrightarrow(1,0) and (1,e)\leftrightarrow(e,1).

f\left(g(x)\right)=g\left(f(x)\right)=x

The population of a town grows exponentially according to the formula P(t)=1000\,e^{0.05t} where t is the time in years. After how many years, will the population reach 5000?

Solution

Put P(t)=5000.
5000=1000\,e^{0.05t}
Divide both sides by 1000.
e^{0.05t}=5
Take natural logs; \ln e^{u}=u.
0.05t=\ln5=1.6094
Solve for t.
t=\dfrac{1.6094}{0.05}\approx32.2
So the population reaches 5000 after about 32.2 years.

The growth curve reaches the level P=5000 at t\approx32.2 years.

t\approx32.2\ \text{years}

A company has the following cost and revenue functions: C(x)=5x+10, where x is the number of units produced, and D(x)=15x, where x is the number of units sold. Find the equilibrium point where cost equals revenue.

Solution

Equilibrium (break-even) point. Here D(x) is the revenue. At the equilibrium point cost equals revenue, so put C(x)=D(x).
5x+10=15x
Solve for x.
10=10x\;\Longrightarrow\;x=1
Find the common value.
C(1)=5(1)+10=15,\qquad D(1)=15(1)=15
Graphically. Draw both lines: they intersect at (1,15). For x<1 the cost line is above the revenue line (loss); for x>1 revenue is above cost (profit).

The cost and revenue lines cross at the equilibrium point (1,15).

\text{Equilibrium point }(1,\,15):\ x=1\text{ unit},\ C=D=15

FBISE 2026 questions

Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 11 mark

If f(x)=3x-4, then f^{-1}(5) equals:

A1
B3Answer
C5
D11

Why

f^{-1}(5) is the input that f sends to 5. Solve 3x-4=5: x=3. Equivalently f^{-1}(x)=\dfrac{x+4}{3}, so f^{-1}(5)=\dfrac{9}{3}=3.

Option D (11) is the trap: it is f(5), not f^{-1}(5).

2026 Annual I, Set B, MCQ 81 mark

If f(x)=\sqrt{x+1} for x\ge-1, then the range of f^{-1}(x) is:

A(-\infty,0]
B[0,\infty)
C(-\infty,-1]
D[-1,\infty)Answer

Why

The range of an inverse function is the domain of the original function: \text{Rang}\,f^{-1}=\text{Dom}\,f=[-1,\infty). (Indeed f^{-1}(x)=x^2-1 for x\ge0, whose values start at -1.)

Option B, [0,\infty), is the range of f, which is the domain of f^{-1}.

2026 Annual I, Set B, Q.2 (i)4 marks

If f(x)=\dfrac{2x-3}{x+1}, find f^{-1}(x). Also determine the domain and range of f^{-1}(x).

Solution

Finding the inverse

The denominator of f cannot be zero, so \text{Dom}\,f=\mathbb{R}-\{-1\}.
Put y=f(x) and clear the fraction.
y=\dfrac{2x-3}{x+1}\;\Rightarrow\;y(x+1)=2x-3\;\Rightarrow\;xy+y=2x-3
Collect the x terms on one side and factor out x.
xy-2x=-3-y\;\Rightarrow\;x(y-2)=-(y+3)\;\Rightarrow\;x=\dfrac{y+3}{2-y}
This is f^{-1}(y). Replace y with x.
f^{-1}(x)=\dfrac{x+3}{2-x}

Domain and range of the inverse

Domain. f^{-1}(x) is not defined when its denominator is zero, that is at x=2.
\text{Dom}\,f^{-1}=\mathbb{R}-\{2\}
Range. The range of f^{-1} is the domain of f.
\text{Rang}\,f^{-1}=\text{Dom}\,f=\mathbb{R}-\{-1\}
Confirm: \dfrac{x+3}{2-x}=-1 would give x+3=x-2, that is 3=-2, which is impossible. So -1 is never an output.

Check

Multiply top and bottom by (2-x).
f\big(f^{-1}(x)\big)=\dfrac{2\cdot\frac{x+3}{2-x}-3}{\frac{x+3}{2-x}+1}=\dfrac{2x+6-6+3x}{x+3+2-x}=\dfrac{5x}{5}=x

f has asymptotes x=-1 and y=2; f^{-1} has asymptotes x=2 and y=-1. The curves are reflections in y=x.

f^{-1}(x)=\dfrac{x+3}{2-x};\quad \text{Dom}\,f^{-1}=\mathbb{R}-\{2\},\quad \text{Rang}\,f^{-1}=\mathbb{R}-\{-1\}

2026 Annual I, Set B, Q.2 (xii) OR4 marks

Let f(x)=\sqrt{x-1} for x\ge1. Find f^{-1}(x) and verify that f(f^{-1}(x))=x.

Solution

Finding the inverse

For x\ge1 the outputs \sqrt{x-1} are never negative, so \text{Rang}\,f=[0,\infty). This will be the domain of f^{-1}.
Put y=\sqrt{x-1} with y\ge0 and square both sides.
y^2=x-1\;\Rightarrow\;x=y^2+1
Replace y with x.
f^{-1}(x)=x^2+1,\qquad x\ge0

Verification

Substitute f^{-1}(x) into f.
f\big(f^{-1}(x)\big)=\sqrt{(x^2+1)-1}=\sqrt{x^2}=|x|
Since x\ge0 in the domain of f^{-1}, |x|=x.
f\big(f^{-1}(x)\big)=x
The other order also works: for x\ge1, f^{-1}\big(f(x)\big)=\big(\sqrt{x-1}\big)^2+1=x.

f starts at (1,0) and f^{-1} at (0,1); each is the reflection of the other in y=x.

f^{-1}(x)=x^2+1\ (x\ge0),\qquad f\big(f^{-1}(x)\big)=x

2026 Annual I, Set D, MCQ 21 mark

If f(x)=4x+9, then f^{-1}(13) equals:

A4
B1Answer
C2
D3

Why

f^{-1}(13) is the input that gives the output 13: 4x+9=13\Rightarrow x=1. Equivalently f^{-1}(x)=\dfrac{x-9}{4} and f^{-1}(13)=\dfrac{4}{4}=1.

2026 Annual I, Set D, Q.2 (i)4 marks

If f(x)=\dfrac{5x-2}{x+4}, find f^{-1}(x) then determine the domain and range of f^{-1}(x).

Solution

Finding the inverse

The denominator of f cannot be zero, so \text{Dom}\,f=\mathbb{R}-\{-4\}.
Put y=f(x) and clear the fraction.
y=\dfrac{5x-2}{x+4}\;\Rightarrow\;y(x+4)=5x-2\;\Rightarrow\;xy+4y=5x-2
Collect the x terms and factor out x.
xy-5x=-2-4y\;\Rightarrow\;x(y-5)=-(4y+2)\;\Rightarrow\;x=\dfrac{4y+2}{5-y}
Replace y with x.
f^{-1}(x)=\dfrac{4x+2}{5-x}

Domain and range of the inverse

Domain. The denominator 5-x must not be zero.
\text{Dom}\,f^{-1}=\mathbb{R}-\{5\}
Range. The range of f^{-1} is the domain of f.
\text{Rang}\,f^{-1}=\text{Dom}\,f=\mathbb{R}-\{-4\}
Confirm: \dfrac{4x+2}{5-x}=-4 would give 4x+2=4x-20, that is 2=-20, impossible. So -4 is never an output.

Check

Multiply top and bottom by (5-x).
f\big(f^{-1}(x)\big)=\dfrac{5\cdot\frac{4x+2}{5-x}-2}{\frac{4x+2}{5-x}+4}=\dfrac{20x+10-10+2x}{4x+2+20-4x}=\dfrac{22x}{22}=x

f: asymptotes x=-4, y=5. f^{-1}: asymptotes x=5, y=-4. Mirror images in y=x.

f^{-1}(x)=\dfrac{4x+2}{5-x};\quad \text{Dom}\,f^{-1}=\mathbb{R}-\{5\},\quad \text{Rang}\,f^{-1}=\mathbb{R}-\{-4\}

2026 Annual I, Set D, Q.2 (xii) OR4 marks

Find domain and range of the following functions: (a) f(x)=e^{x-2}+3, (b) g(x)=\ln(x+5). Write your answers in interval notation.

Solution

(a) f(x)=e^{x-2}+3

Domain. The exponential function accepts every real input, and x-2 is real for every real x.
D_f=(-\infty,\infty)
Range. e^{x-2} takes every positive value but is never 0 or negative: e^{x-2}>0. Adding 3 gives every value greater than 3.
e^{x-2}>0\;\Rightarrow\;f(x)>3\;\Rightarrow\;R_f=(3,\infty)
The graph is y=e^{x} moved 2 units right and 3 units up; its asymptote y=0 moves up to y=3.

D_f=(-\infty,\infty),\qquad R_f=(3,\infty)

(b) g(x)=\ln(x+5)

Domain. A logarithm needs a positive input.
x+5>0\;\Rightarrow\;x>-5\;\Rightarrow\;D_g=(-5,\infty)
Range. As x+5 runs through all positive numbers, \ln(x+5) takes every real value (very negative near x=-5, growing without limit as x grows).
R_g=(-\infty,\infty)
The graph is y=\ln x moved 5 units left; its asymptote x=0 moves to x=-5.

D_g=(-5,\infty),\qquad R_g=(-\infty,\infty)

2026 Annual I, Old syllabus (Curriculum 2000), MCQ 1, old syllabus1 mark

Which of the following functions is odd?

Af(x)=x^2+1
Bf(x)=\cos x
Cf(x)=x^3-xAnswer
Df(x)=e^{x}

Why

f(x)=x^3-x gives f(-x)=-x^3+x=-(x^3-x)=-f(x), so it is odd. x^2+1 and \cos x are even (f(-x)=f(x)), and e^{x} is neither.

2026 Annual I, Old syllabus (Curriculum 2000), MCQ 15, old syllabus1 mark

f(x)=x^2 and g(x)=\sqrt{x}, then (f\circ g)(x) is:

Ax^3
Bx^2
CxAnswer
Dx^2\sqrt{x}

Why

(f\circ g)(x)=f\big(g(x)\big)=\big(\sqrt{x}\big)^2=x for x\ge0. Options A and D come from multiplying the functions instead of composing them.

2026 Annual I, Old syllabus (Curriculum 2000), Q.2 (i), old syllabus4 marks

A function is defined as: x(t)=t^2, y(t)=t^4-2t^2+1, t\in\mathbb{R}. (a) Eliminate the parameter t to express y as a function of x. (b) Determine whether the resulting function y=f(x) is even, odd, or neither. Justify your answer.

Solution

(a) Eliminate the parameter t.

Notice that t^4=(t^2)^2, and t^2=x.
y=t^4-2t^2+1=(t^2)^2-2t^2+1=x^2-2x+1
Factorise. Because x=t^2, only x\ge0 occurs.
y=f(x)=(x-1)^2,\qquad x\ge0

y=x^2-2x+1=(x-1)^2,\quad x\ge0

(b) Even, odd, or neither.

Replace x by -x.
f(-x)=(-x)^2-2(-x)+1=x^2+2x+1=(x+1)^2
Compare with f(x)=(x-1)^2 and -f(x)=-(x-1)^2. A numerical check shows neither is equal.
f(1)=0,\quad f(-1)=4:\quad f(-1)\neq f(1)\ \text{and}\ f(-1)\neq-f(1)
So f is neither even nor odd. Also, x=t^2\ge0 means the domain is [0,\infty), which is not symmetric about 0, so the function cannot be even or odd in any case.

The curve is symmetric about x=1, not about the y-axis or the origin.

\text{Neither even nor odd}

Note. This question is from the old-syllabus (Curriculum 2000) paper. Parametric equations are not part of the new book; even and odd functions appear only in its Key Facts table.

2026 Annual I, Old syllabus (Curriculum 2000), Q.2 (iii) OR, old syllabus4 marks

Prove the identity \operatorname{csch}^2x=\coth^2x-1.

Solution

Use the definitions of the hyperbolic functions (here x\neq0, so that \sinh x\neq0):

\begin{aligned}\sinh x&=\dfrac{e^{x}-e^{-x}}{2}\\ \cosh x&=\dfrac{e^{x}+e^{-x}}{2}\\ \coth x&=\dfrac{\cosh x}{\sinh x}\\ \operatorname{csch}x&=\dfrac{1}{\sinh x}\end{aligned}

Start from the right-hand side and write it over a common denominator.
\coth^2x-1=\dfrac{\cosh^2x}{\sinh^2x}-1=\dfrac{\cosh^2x-\sinh^2x}{\sinh^2x}
Work out the numerator from the definitions. Use e^{x}e^{-x}=1 when squaring.
\cosh^2x-\sinh^2x=\dfrac{e^{2x}+2+e^{-2x}}{4}-\dfrac{e^{2x}-2+e^{-2x}}{4}=\dfrac{4}{4}=1
Substitute this back.
\coth^2x-1=\dfrac{1}{\sinh^2x}=\operatorname{csch}^2x

\operatorname{csch}^2x=\coth^2x-1\quad(x\neq0)

Note. Old-syllabus (Curriculum 2000) question. The new book only names the hyperbolic functions (1.10.2); their definitions are summarised in the notes.

2026 Annual I, Old syllabus (Curriculum 2000), Q.2 (viii), old syllabus4 marks

If f(x)=2x-3 and g(x)=x^2+1, then find f(g(x)), and g(f(x)).

Solution

For f(g(x)), replace x in f by g(x)=x^2+1.
f\big(g(x)\big)=2(x^2+1)-3=2x^2+2-3=2x^2-1
For g(f(x)), replace x in g by f(x)=2x-3 and expand.
g\big(f(x)\big)=(2x-3)^2+1=4x^2-12x+9+1=4x^2-12x+10
The two results are different, which shows that composition depends on the order: f\circ g\neq g\circ f.

f(g(x))=2x^2-1,\qquad g(f(x))=4x^2-12x+10

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 1, old syllabus1 mark

The graph of (f\circ g)^{-1}(x) can be obtained by reflecting (f\circ g)(x) along:

Ax-axis
By-axis
Cy=xAnswer
Dy=-x

Why

The graph of the inverse of any one-one function, here f\circ g, is the reflection of its graph in the line y=x, because the point (a,b) becomes (b,a).

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (i), old syllabus4 marks

Sketch the graph of y=|x-2| and find its domain and range.

Solution

The corner of the V is where the expression inside the modulus is zero: x-2=0, so x=2 and y=0.
\text{vertex}=(2,0)
Remove the modulus on each side of the corner.
y=\begin{cases}x-2, & x\ge2\\ 2-x, & x<2\end{cases}
Table of values on both sides of x=2.
\begin{array}{c|ccccccc} x & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline y & 3 & 2 & 1 & 0 & 1 & 2 & 3 \end{array}
Domain. Every real x can be used.
D=\mathbb{R}=(-\infty,\infty)
Range. A modulus is never negative, and the value 0 occurs at x=2.
R=[0,\infty)

The graph of y=|x| shifted 2 units to the right: a V with its corner at (2,0) and y-intercept (0,2).

D=\mathbb{R},\qquad R=[0,\infty)

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (v), old syllabus4 marks

If f(x)=\sqrt{x} and g(x)=x, then find f(g(x)), g(f(x)), f(f(x)) and g(g(x)).

Solution

g is the identity function: it returns its input unchanged. f needs x\ge0.

Replace x in f by g(x)=x.
f\big(g(x)\big)=f(x)=\sqrt{x}
Replace x in g by f(x)=\sqrt{x}.
g\big(f(x)\big)=g(\sqrt{x})=\sqrt{x}
Apply f twice.
f\big(f(x)\big)=f(\sqrt{x})=\sqrt{\sqrt{x}}=x^{1/4}=\sqrt[4]{x}
Apply g twice.
g\big(g(x)\big)=g(x)=x
The first three need x\ge0; g(g(x))=x holds for every real x.

f(g(x))=\sqrt{x},\quad g(f(x))=\sqrt{x},\quad f(f(x))=\sqrt[4]{x},\quad g(g(x))=x

Exam focus

Find f^{-1}(x) of a fractional function such as \dfrac{2x-3}{x+1}, and state the domain and range of f^{-1}. (problem)
Asked in both new 2026 papers as Q.2 (i) (Set B and Set D); the same method as Exercise 1.1 Q9.
Go to answer
Find the inverse of a square root function and verify f(f^{-1}(x))=x. (problem)
2026 Set B Q.2 (xii) OR; remember the restriction x\ge0 on the inverse.
Go to answer
MCQs on inverses: value of f^{-1}(a), and range of f^{-1} = domain of f. (mcq)
Three 2026 MCQs (Set B MCQ 1 and 8, Set D MCQ 2) test exactly this.
Go to answer
Find the domain and range of functions with fractions, square roots, modulus, e^{x} and \ln x. (problem)
Core skill of the unit (Exercise 1.1 Q2); 2026 Set D Q.2 (xii) OR asked it for e^{x-2}+3 and \ln(x+5).
Go to answer
Show that a given function is one-one and onto (bijective). (problem)
Standard proof question (Exercise 1.1 Q7 and Q8) using f(x_1)=f(x_2) and solving y=f(x).
Go to answer
Find the intercepts and vertex of a quadratic and sketch it, directly or using factors. (problem)
Exercise 1.2 Q3 and Q4 types; the axis x=-\frac{b}{2a} and the vertex carry the marks.
Go to answer
Solve a linear and a quadratic function graphically and check algebraically. (problem)
Exercise 1.2 Q7 and Example 16; also daily-life versions (supply and demand, ball and wall).
Go to answer
Sketch a modulus graph such as y=|x-2| and state its domain and range. (problem)
Asked in the 2026 old-syllabus paper; the corner is where the inside of the modulus is zero.
Go to answer
Derive the product, quotient, power and change of base laws of logarithms. (long)
The SLO says derive and apply; each law is a short proof from the laws of indices.
Go to answer
Compound interest and exponential growth: time to reach an amount, doubling time, future population. (problem)
Examples 20 and 21 and the whole of Exercise 1.3 use this method (take \ln of both sides).
Go to answer
Domain, range and period of trigonometric functions from their graphs. (problem)
Exercise 1.4 Q1 type; MCQs on periods such as \tan 3x appear in the papers.
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Describe the shift or scaling that turns y=f(x) into f(x\pm c), f(x)\pm c, cf(x) or f(cx). (short)
Exercise 1.5 and review MCQs test these rules; the sign inside the bracket is the usual trap.
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