Open Math›Year 11›Mathematics Advanced›Working with functions›Introduction to functions

Introduction to functions

Determine whether a graph represents a function using the vertical line test, state the domain and range, evaluate function rules, find zeroes, and classify functions as even, odd or neither.

Worked examples

Finding the domain and range of a function

Straightforward

Problem

For the function

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

, state the domain and range in interval notation and explain why it passes the vertical line test.

Find the domain by determining which

x

-values make the expression defined. The square root requires a non-negative radicand.

x - 1 \geq 0 \implies x \geq 1

Write the domain in interval notation.

\text{Domain} = [1, \infty)

Find the range. Since

\sqrt{x - 1} \geq 0

for all

x

in the domain, and the square root can grow without bound, determine the smallest and largest possible output values.

\text{When } x = 1: f(1) = \sqrt{0} = 0 \qquad \text{As } x \to \infty: f(x) \to \infty

Write the range in interval notation.

\text{Range} = [0, \infty)

Explain the vertical line test. For each

x \geq 1

, the rule

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

produces exactly one output, so every vertical line meets the graph at most once.

\text{Each } x \text{ in } [1, \infty) \text{ maps to exactly one value } \sqrt{x-1} \geq 0.

Answer

Domain

= [1, \infty)

, Range

= [0, \infty)

. The graph passes the vertical line test because each

x

-value produces exactly one output.

Evaluating a function and finding its zeroes

Moderate

Problem

Let

f(x) = x^2 - 5x + 4

. Find

f(2)

f(-2)

and all zeroes of

f

Evaluate

f(2)

by substituting

x = 2

into the rule.

f(2) = (2)^2 - 5(2) + 4 = 4 - 10 + 4 = -2

Evaluate

f(-2)

by substituting

x = -2

into the rule.

f(-2) = (-2)^2 - 5(-2) + 4 = 4 + 10 + 4 = 18

Find the zeroes by setting

f(x) = 0

and solving the equation.

x^2 - 5x + 4 = 0

Factorise the quadratic by finding two numbers that multiply to

4

and add to

-5

(x - 1)(x - 4) = 0

Apply the null factor law to find each zero.

x - 1 = 0 \implies x = 1 \qquad \text{or} \qquad x - 4 = 0 \implies x = 4

Answer

f(2) = -2

f(-2) = 18

. The zeroes of

f

are

x = 1

and

x = 4

Classifying a function as even, odd or neither

Challenging

Problem

Determine whether

f(x) = x^4 - 3x^2 + 2

is even, odd or neither. Describe the symmetry of its graph.

Compute

f(-x)

by replacing every

x

with

-x

in the rule.

f(-x) = (-x)^4 - 3(-x)^2 + 2

Simplify each term using the fact that even powers of

-x

equal the corresponding power of

x

(-x)^4 = x^4, \quad (-x)^2 = x^2 \implies f(-x) = x^4 - 3x^2 + 2

Compare

f(-x)

with

f(x)

f(-x) = x^4 - 3x^2 + 2 = f(x) \quad \text{for all } x \in \mathbb{R}

Since

f(-x) = f(x)

, the function is even. State the geometric interpretation.

f(-x) = f(x) \implies \text{the graph is symmetric about the } y\text{-axis.}

Answer

f

is even because

f(-x) = f(x)

for all

x \in \mathbb{R}

. The graph of

y = x^4 - 3x^2 + 2

is symmetric about the

y

-axis.

Practise

Q1·Straightforward

The vertical line test is used to decide whether a graph represents a function. Which statement correctly describes when a graph IS a function?

Every horizontal line crosses the graph at most once.Every vertical line crosses the graph at most once.Every vertical line crosses the graph at least once.The graph passes through the origin.

Explanation

A graph represents a function if and only if every vertical line meets the graph at most once. This guarantees each

x

-value maps to no more than one

y

-value. Option A describes the horizontal line test, which is used to check if a function is one-to-one, not whether it is a function at all.

Q2·Straightforward

The graph of

x^2 + y^2 = 25

is a circle centred at the origin with radius

5

. Does this graph represent a function of

x

Yes, because it is a smooth, closed curve.No, because a vertical line such as $x = 3$ crosses the circle at two points.Yes, because every $x$-value in $[-5, 5]$ appears on the graph.No, because the graph does not pass the horizontal line test.

Explanation

x = 3

, substituting into

x^2 + y^2 = 25

gives

9 + y^2 = 25

, so

y = \pm 4

. The vertical line

x = 3

meets the circle at

(3, 4)

and

(3, -4)

— two points. Because one

x

-value maps to two

y

-values, the circle does not represent a function.

Q3·Straightforward

The domain of

f(x) = \sqrt{x - 4}

can be written as

x \geq k

. What is the value of

k

Explanation

For

f(x) = \sqrt{x - 4}

to be defined, we need

x - 4 \geq 0

, which gives

x \geq 4

. So the domain is

[4, \infty)

and

k = 4

Q4·Straightforward

The range of

f(x) = x^2 + 3

for all real

x

can be written as

y \geq k

. What is the value of

k

Explanation

Since

x^2 \geq 0

for all real

x

, the smallest value of

f(x) = x^2 + 3

0 + 3 = 3

, achieved at

x = 0

. The range is

[3, \infty)

and

k = 3

Q5·Moderate

Let

f(x) = 3x^2 - 2x + 5

. Find

f(3)

Explanation

Substituting

x = 3

f(3) = 3(3)^2 - 2(3) + 5 = 3 \times 9 - 6 + 5 = 27 - 6 + 5 = 26

Q6·Moderate

Let

f(x) = 3x^2 - 2x + 5

. Find

f(-3)

Explanation

Substituting

x = -3

f(-3) = 3(-3)^2 - 2(-3) + 5 = 3 \times 9 + 6 + 5 = 27 + 6 + 5 = 38

Q7·Moderate

Find the zero of

f(x) = 4x - 8

by solving

f(x) = 0

. What is the value of

x

Explanation

Setting

f(x) = 0

4x - 8 = 0 \Rightarrow 4x = 8 \Rightarrow x = 2

. The zero is at

x = 2

, which is also the

x

-intercept of the graph.

Q8·Moderate

Find the positive zero of

f(x) = x^2 - 16

by solving

f(x) = 0

. What is the positive value of

x

Explanation

Setting

f(x) = 0

x^2 - 16 = 0 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4

. The positive zero is

x = 4

Q9·Challenging

Let

f(x) = 2x^2 - 5

. Compute

f(-x)

and classify the function as even, odd or neither.

Even, because $f(-x) = f(x)$ for all $x$.Odd, because $f(-x) = -f(x)$ for all $x$.Neither even nor odd, because $f(-x) \neq f(x)$ and $f(-x) \neq -f(x)$.Both even and odd.

Explanation

Computing

f(-x) = 2(-x)^2 - 5 = 2x^2 - 5 = f(x)

. Since

f(-x) = f(x)

for all real

x

, the function is even. Geometrically, the graph of

y = 2x^2 - 5

is a parabola symmetric about the

y

-axis.

Q10·Challenging

Let

f(x) = 4x^3 - 2x

. Compute

f(-x)

and classify the function as even, odd or neither.

Even, because $f(-x) = f(x)$ for all $x$.Odd, because $f(-x) = -f(x)$ for all $x$.Neither even nor odd.Cannot be determined without a graph.

Explanation

Computing

f(-x) = 4(-x)^3 - 2(-x) = -4x^3 + 2x = -(4x^3 - 2x) = -f(x)

. Since

f(-x) = -f(x)

for all real

x

, the function is odd. Geometrically, the graph has rotational symmetry of

180°

about the origin.

Q11·Challenging

Let

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

. Compute

f(-x)

and classify the function as even, odd or neither.

Even, because $f(-x) = f(x)$.Odd, because $f(-x) = -f(x)$.Neither even nor odd, because $f(-x) \neq f(x)$ and $f(-x) \neq -f(x)$.Both even and odd.

Explanation

Computing

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

. Comparing:

f(x) = x^2 + 2x + 1 \neq x^2 - 2x + 1 = f(-x)

, so

f

is not even. Also

-f(x) = -x^2 - 2x - 1 \neq x^2 - 2x + 1 = f(-x)

, so

f

is not odd. The function is neither even nor odd, which can be seen from the fact that

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

is a parabola shifted left, not centred on the

y

-axis.

Q12·Challenging

An even function satisfies

f(-x) = f(x)

and an odd function satisfies

f(-x) = -f(x)

. Which statement correctly describes the symmetry of each type?

An even function has symmetry about the $x$-axis; an odd function has symmetry about the $y$-axis.An even function has symmetry about the $y$-axis; an odd function has rotational symmetry about the origin.An even function has rotational symmetry about the origin; an odd function has symmetry about the $y$-axis.An even function has symmetry about the $y$-axis; an odd function has symmetry about the $x$-axis.

Explanation

f(-x) = f(x)

, then the point

(-x, f(x))

lies on the graph whenever

(x, f(x))

does. This means the graph is symmetric about the

y

-axis (a mirror reflection). If

f(-x) = -f(x)

, then

(-x, -f(x))

lies on the graph whenever

(x, f(x))

does. Reflecting through the origin maps

(x, y)

(-x, -y)

, so the graph has

180°

rotational symmetry about the origin.