Inverse functions and restricted domains | Year 12

01 / LEARN

Understand the idea.

Undo input and output

An inverse function reverses the input-output mapping of a one-to-one function. Its domain is the original range and its range is the original domain. Corresponding graph points swap coordinates, reflecting the graph in $y = x$ .

Write $y = f (x)$ and solve for the original input $x$ , then rename the variables. The notation $f^{- 1}$ denotes this reversed mapping, not the reciprocal $1 / f$ . Verify compositions on the correct domains if a check is needed.

Select the right branch

A quadratic needs a domain restriction before it has an inverse function. Solving its square relation offers two algebraic branches, but the original input restriction selects only one of them.

Carry endpoint conditions through the reversal. An included minimum becomes an included inverse-domain endpoint. For a rational inverse, establish which output value is impossible before dividing by an expression involving that output.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

The increasing parabola branch

Find the inverse of $f (x) = {(x - 3)}^{2} + 2$ , restricted to $x \geq 3$ . Give its domain and range.

Worked solution

1
Find the original range.
$x \geq 3 \Rightarrow {(x - 3)}^{2} \geq 0 \Rightarrow y \geq 2.$
Every value at least two is attained on the increasing branch.
2
Solve for the original input with its restriction.
$\begin{aligned} y - 2 & = {(x - 3)}^{2}, \\ x - 3 & = \sqrt{y - 2}, \\ x & = 3 + \sqrt{y - 2} . \end{aligned}$
The condition $x - 3 \geq 0$ selects the positive square-root branch.
3
Rename the variables and exchange the sets.
$\begin{aligned} f^{- 1} (x) & = 3 + \sqrt{x - 2}, \\ dom (f^{- 1}) & = [2, \infty), \\ range (f^{- 1}) & = [3, \infty) . \end{aligned}$
The inverse accepts exactly the original outputs and returns exactly the original allowed inputs.
4
Check the branch by composition.
$f^{- 1} (f (x)) = 3 + \sqrt{{(x - 3)}^{2}} = 3 + ∣ x - 3 ∣ = x (x \geq 3) .$
The domain condition is what allows the absolute value to become $x - 3$ .

Answer

$f^{- 1} (x) = 3 + \sqrt{x - 2}$ ; domain $[2, \infty)$ , range $[3, \infty)$ .

EXAMPLE 02

A self-inverse rational rule

Find the inverse of $f (x) = x / (x - 1)$ , $x \neq 1$ .

Worked solution

1
Rearrange the input-output equation.
$\begin{aligned} y & = \frac{x}{x - 1}, x \neq 1, \\ y (x - 1) & = x, \\ x (y - 1) & = y . \end{aligned}$
Multiplication by $x - 1$ is valid on the original domain.
2
Identify the excluded output and solve.
$y = 1 would give x - 1 = x; x = \frac{y}{y - 1} (y \neq 1) .$
The value one is impossible as an output, and every other output has the displayed unique preimage.
3
State the inverse with its sets.
$\begin{aligned} f^{- 1} (x) & = \frac{x}{x - 1}, \\ dom (f^{- 1}) & = ℝ ∖ {1}, \\ range (f^{- 1}) & = ℝ ∖ {1} . \end{aligned}$
The original domain and range coincide, so the inverse has the same rule and restriction.
4
Verify that the rule undoes itself.
$\begin{aligned} f (f (x)) & = \frac{\frac{x}{x - 1}}{\frac{x}{x - 1} - 1} \\ = \frac{\frac{x}{x - 1}}{\frac{1}{x - 1}} \\ = x (x \neq 1) . \end{aligned}$
Both compositions agree because the inverse is the original function itself.

Answer

$f^{- 1} (x) = x / (x - 1)$ , $x \neq 1$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

If f has domain $[3, \infty)$ and range $[2, \infty)$ , what is the domain of its inverse?

Show answer & reasoning

$[2, \infty)$ .

An inverse takes original outputs as its inputs. Since the original range is $[2, \infty)$ , that interval is the inverse domain, including the endpoint two.

Keep practising

6 worksheet questions · 20 marks

Open worksheet Worked solutions

Read the questions online

Find the inverse of $f (x) = \frac{2 x + 1}{3}$ .
3 marks
Find the inverse of $f (x) = {(x - 3)}^{2} + 2$ , restricted to $x \geq 3$ . Give its domain and range.
4 marks
Find the inverse of $f (x) = {(x - 3)}^{2} + 2$ when $x \leq 3$ .
3 marks
Find the inverse of $f (x) = \frac{x}{x - 1}$ , $x \neq 1$ .
4 marks
Find the inverse of $f (x) = 3 e^{x} - 4$ and its domain.
3 marks
Find a domain restriction for $f (x) = ∣ x + 1 ∣$ that gives inverse $x - 1$ , and state the inverse domain.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the inverse of $f (x) = \frac{2 x + 1}{3}$ .
3 marks
Write $y = \frac{2 x + 1}{3}$ .
Solve for $x$ : $x = \frac{3 y - 1}{2}$ , then swap the variable names.
$f^{- 1} (x) = \frac{3 x - 1}{2}$ .
Find the inverse of $f (x) = {(x - 3)}^{2} + 2$ , restricted to $x \geq 3$ . Give its domain and range.
4 marks
$y - 2 = {(x - 3)}^{2}$ .
The chosen branch has $x - 3$ nonnegative, so take the positive square root.
Swap the original domain and range.
$f^{- 1} (x) = 3 + \sqrt{x - 2}$ ; domain $[2, \infty)$ , range $[3, \infty)$ .
Find the inverse of $f (x) = {(x - 3)}^{2} + 2$ when $x \leq 3$ .
3 marks
The original branch has $x - 3$ nonpositive.
Choose the negative square-root branch before swapping variables.
$f^{- 1} (x) = 3 - \sqrt{x - 2}$ , $x \geq 2$ .
Find the inverse of $f (x) = \frac{x}{x - 1}$ , $x \neq 1$ .
4 marks
$y (x - 1) = x$ gives $x (y - 1) = y$ .
The function is its own inverse on the stated domain.
$f^{- 1} (x) = \frac{x}{x - 1}$ , $x \neq 1$ .
Find the inverse of $f (x) = 3 e^{x} - 4$ and its domain.
3 marks
Solve $y + 4 = 3 e^{x}$ for $x$ .
The original range is greater than $- 4$ .
$ln (\frac{x + 4}{3})$ , $x > - 4$ .
Find a domain restriction for $f (x) = ∣ x + 1 ∣$ that gives inverse $x - 1$ , and state the inverse domain.
3 marks
On the right branch, $f (x) = x + 1$ .
The branch range is nonnegative.
Restrict $f$ to $x \geq - 1$ ; then $f^{- 1} (x) = x - 1$ for $x \geq 0$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist functions and sketching graphs. Calculus-based analysis assumes concurrent Methods differentiation; real-polynomial division detail is taught in the separate Real Polynomials chapter.

Official curriculum source ↗

In this lesson

Understand the idea.

Undo input and output

Select the right branch

See the method in action.

The increasing parabola branch

Find the original range.

Solve for the original input with its restriction.

Rename the variables and exchange the sets.

Check the branch by composition.

A self-inverse rational rule

Rearrange the input-output equation.

Identify the excluded output and solve.

State the inverse with its sets.

Verify that the rule undoes itself.

Try it yourself.

Check your understanding

Keep practising