Properties of functions

Evaluate composite functions, analyse piecewise-defined functions for continuity and points of discontinuity, and solve and sketch absolute value functions.

Worked examples

Evaluating a composite function

Straightforward

Problem

Let

f(x) = x^2 + 2

and

g(x) = 4x - 1

. Find

f(g(2))

Analysing a piecewise function for continuity

Moderate

Problem

Consider the piecewise function

f(x) = \begin{cases} x^2 & \text{if } x < 2 \\ 3x - 2 & \text{if } x \geq 2 \end{cases}

Evaluate

f(2)

, find the left-hand limit at

x = 2

, and determine whether the function is continuous at

x = 2

Solving an absolute value equation and finding the vertex

Challenging

Problem

Solve

|2x + 3| = 11

and find the vertex of the graph

y = |2x + 3|

Practise

Q1·Straightforward

Let

f(x) = 2x + 3

and

g(x) = x^2

. Find the value of

f(g(3))

Q2·Straightforward

Let

f(x) = x^2 - 1

and

g(x) = 3x

. Find the value of

g(f(2))

Q3·Straightforward

Evaluate

f(4)

for the piecewise function

f(x) = \begin{cases} x + 1 & \text{if } x < 4 \\ 2x - 3 & \text{if } x \geq 4 \end{cases}

Q4·Straightforward

Evaluate

f(-3)

for the piecewise function

f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ x + 5 & \text{if } x \geq 0 \end{cases}

Q5·Moderate

Consider the piecewise function

f(x) = \begin{cases} 2x + 1 & \text{if } x < 3 \\ x^2 - 2 & \text{if } x \geq 3 \end{cases}

Find

f(3)

Q6·Moderate

For the piecewise function

f(x) = \begin{cases} 2x + 1 & \text{if } x < 3 \\ x^2 - 2 & \text{if } x \geq 3 \end{cases}

find the value that

f(x)

approaches as

x \to 3

from the left. (Substitute

x = 3

into the first piece.)

Q7·Moderate

Consider the piecewise function

f(x) = \begin{cases} x + 4 & \text{if } x \leq 2 \\ x^2 + 1 & \text{if } x > 2 \end{cases}

Find

f(2)

Q8·Moderate

For the piecewise function

f(x) = \begin{cases} x + 4 & \text{if } x \leq 2 \\ x^2 + 1 & \text{if } x > 2 \end{cases}

the value at

x = 2

f(2) = 6

, and the value the function approaches from the right is

\lim_{x \to 2^+} f(x) = 2^2 + 1 = 5

. Which statement correctly describes the behaviour at

x = 2

The function is continuous at

x = 2

because

f(2)

exists.There is a jump discontinuity at

x = 2

because the left value (

6

) differs from the right limit (

5

).There is a removable discontinuity at

x = 2

because the function has a hole at that point.The function is undefined at

x = 2

Q9·Challenging

Solve

|2x - 3| = 7

algebraically by considering two cases. What is the larger of the two solutions?

Q10·Challenging

Solve

|x + 4| = 6

algebraically by considering two cases. What is the sum of the two solutions?

Q11·Challenging

The graph of

y = |2x - 6|

is V-shaped. Its vertex is the point where

2x - 6 = 0

. What is the

x

-coordinate of the vertex?

Q12·Challenging

Find the

y

-intercept of the graph of

y = |3x - 9|

Open Math

Properties of functions

Algebra and Functions · MAV-11-01

Name:

Date:

Q1Straightforward

Let

f(x) = 2x + 3

and

g(x) = x^2

. Find the value of

f(g(3))

Q2Straightforward

Let

f(x) = x^2 - 1

and

g(x) = 3x

. Find the value of

g(f(2))

Q3Straightforward

Evaluate

f(4)

for the piecewise function

f(x) = \begin{cases} x + 1 & \text{if } x < 4 \\ 2x - 3 & \text{if } x \geq 4 \end{cases}

Q4Straightforward

Evaluate

f(-3)

for the piecewise function

f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ x + 5 & \text{if } x \geq 0 \end{cases}

Q5Moderate

Consider the piecewise function

f(x) = \begin{cases} 2x + 1 & \text{if } x < 3 \\ x^2 - 2 & \text{if } x \geq 3 \end{cases}

Find

f(3)

Q6Moderate

For the piecewise function

f(x) = \begin{cases} 2x + 1 & \text{if } x < 3 \\ x^2 - 2 & \text{if } x \geq 3 \end{cases}

find the value that

f(x)

approaches as

x \to 3

from the left. (Substitute

x = 3

into the first piece.)

Q7Moderate

Consider the piecewise function

f(x) = \begin{cases} x + 4 & \text{if } x \leq 2 \\ x^2 + 1 & \text{if } x > 2 \end{cases}

Find

f(2)

Q8Moderate

For the piecewise function

f(x) = \begin{cases} x + 4 & \text{if } x \leq 2 \\ x^2 + 1 & \text{if } x > 2 \end{cases}

the value at

x = 2

f(2) = 6

, and the value the function approaches from the right is

\lim_{x \to 2^+} f(x) = 2^2 + 1 = 5

. Which statement correctly describes the behaviour at

x = 2

A.The function is continuous at $x = 2$ because $f(2)$ exists.
B.There is a jump discontinuity at $x = 2$ because the left value ( $6$ ) differs from the right limit ( $5$ ).
C.There is a removable discontinuity at $x = 2$ because the function has a hole at that point.
D.The function is undefined at $x = 2$ .

Q9Challenging

Solve

|2x - 3| = 7

algebraically by considering two cases. What is the larger of the two solutions?

Q10Challenging

Solve

|x + 4| = 6

algebraically by considering two cases. What is the sum of the two solutions?

Q11Challenging

The graph of

y = |2x - 6|

is V-shaped. Its vertex is the point where

2x - 6 = 0

. What is the

x

-coordinate of the vertex?

Q12Challenging

Find the

y

-intercept of the graph of

y = |3x - 9|

Worked solutions and answers at openmath.au/year-11/advanced/working-with-functions/properties-of-functions