1.3 Inverse Functions Worksheet

Quick Formulae & Methods

Key definitions and steps for solving Inverse Function problems.

Definition

f^{- 1} (y) = x ⟺ f (x) = y

. The inverse undoes the function.

Domain & Range

Dom (f^{- 1}) = Range (f)

and

Range (f^{- 1}) = Dom (f)

Existence

Inverse exists only if the function is One-to-One (Horizontal Line Test).

Reflection

Graphs of

f

and

f^{- 1}

are reflections in the line

y = x

Algebraic Method

1. Set

y = f (x)

. 2. Swap

x

and

y

. 3. Solve for

y

Composition

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

and

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

Past Paper Style Questions

Question 1: Linear Inverse

A function is defined by

f : x \mapsto 2 x + 5

, where

x \in R

(a) Show that

f

is one–one.

(b) Find

f^{- 1} (x)

f^{- 1}

and the range of

f^{- 1}

(d) Solve

2 x + 5 = 17

using the inverse function.

Question 2: Quadratic and Domain Restriction

A function is defined by

f (x) = x^{2} - 6 x + 5

(a) Write

f (x)

in completed-square form and hence state the minimum value of

f (x)

(b) Explain why

f

does not have an inverse if the domain is

R

x \geq 3

and (ii)

x \leq 3

(d) For each restriction in (c), find

f^{- 1} (x)

and state its domain.

Question 3: Rational Function

A function is defined by

f (x) = \frac{3 x - 4}{2 x + 1}

, with

x \neq - \frac{1}{2}

(a) Find

f^{- 1} (x)

algebraically.

(b) State the domain of

f

and the domain of

f^{- 1}

f

(d) Solve

\frac{3 x - 4}{2 x + 1} = 2

and explain why the solution is or isn’t allowed.

Question 4: Square Root Function

The function

f

is given by

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

(a) State the domain and range of

f

(b) Find

f^{- 1} (x)

f^{- 1}

(d) Solve

\sqrt{x - 1} = 5

by applying the inverse function idea.

Question 5: Reciprocal Function

A function is defined by

f (x) = \frac{1}{x - 2}

, where

x \neq 2

(a) Find

f^{- 1} (x)

(b) State the range of

f

f^{- 1} (x)

has domain equal to the range of

f

(d) Evaluate

f (f (5))

and explain why this does not mean

f

is its own inverse.

Question 6: Cubic Function

A function is defined by

f (x) = x^{3} + 1

x \in R

(a) Explain why

f

is one–one.

(b) Find

f^{- 1} (x)

x^{3} + 1 = 28

using the inverse function.

(d) Find

f^{- 1} (9)

Question 7: Two Functions

Functions

f

and

g

are given by

f (x) = 2 x - 3

and

g (x) = \frac{x + 1}{3}

(a) Find

f^{- 1} (x)

and

g^{- 1} (x)

(b) Find

f (g (x))

and

g (f (x))

f \circ g = g \circ f

. Give a reason.

(d) Show that

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

is false, but

g^{- 1} (x)

may equal a simple expression involving

f

Question 8: Graph Points

The graph of

y = f (x)

passes through the points

(- 2, - 1)

(0, 1)

(3, 4)

and is one–one.

(a) Write down three points on the graph of

y = f^{- 1} (x)

(b) State the geometric relationship between the graphs of

y = f (x)

and

y = f^{- 1} (x)

f (5) = 7

, write down

f^{- 1} (7)

and explain why.

Question 9: Self-Inverse Check

A function is defined by

f (x) = \frac{x - 4}{x + 2}

x \neq - 2

(a) Find

f^{- 1} (x)

(b) Show that

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

is not true.

x

such that

f (x) = x

(d) Hence find the intersection points of

y = f (x)

with the line

y = x

Question 10: Range and Verification

A function is defined by

f (x) = \frac{5}{x + 1}

x \neq - 1

(a) State the range of

f

(b) Find

f^{- 1} (x)

and its domain.

f^{- 1}

exists,

f

must be one–one.” Explain why this statement is correct.

(d) Solve

\frac{5}{x + 1} = 2

and check your solution by substitution into

f (x)