One-one and inverse functions

AS · P1 · 12 min

An inverse function undoes a function: if $f$ turns $3$ into $7$ , then $f^{-1}$ turns $7$ back into $3$ . Only one-one functions have inverses, which is why so many Cambridge questions first ask you to restrict a domain, then find $f^{-1}(x)$ , then sketch $f$ and $f^{-1}$ on the same axes. Expect at least one inverse function part on every Paper 1, often the hardest part of the functions question.

Undoing a function

If $f(x) = 2x + 3$ , then $f$ means "double, then add $3$ ". To undo it you reverse the steps in the opposite order: "subtract $3$ , then halve". So

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

Check: $f(5) = 13$ and $f^{-1}(13) = 5$ .

Definition

The inverse of a one-one function $f$ is the function $f^{-1}$ that reverses $f$ : if $f(a) = b$ then $f^{-1}(b) = a$ . It satisfies

f^{-1}f(x) = x \quad\text{for } x \text{ in the domain of } f, \qquad ff^{-1}(x) = x \quad\text{for } x \text{ in the domain of } f^{-1}

Why it must be one-one

If $f$ is many-one, some output comes from two inputs. For $f(x) = x^2$ on $\mathbb{R}$ , $f(3) = f(-3) = 9$ . Undoing $9$ would have to give both $3$ and $-3$ , but a function can only give one output. So a function has an inverse if and only if it is one-one.

The fix is to restrict the domain so that the function is one-one. $x^2$ for $x \ge 0$ is one-one, and its inverse is $\sqrt{x}$ .

Domains and ranges swap

The inverse takes outputs of $f$ back to inputs, so the roles of domain and range swap.

Key result

\text{domain of } f^{-1} = \text{range of } f \qquad\qquad \text{range of } f^{-1} = \text{domain of } f

The graph of $y = f^{-1}(x)$ is the reflection of the graph of $y = f(x)$ in the line $y = x$ .

Finding an inverse algebraically

Finding f inverse

Find the range of $f$ first (you will need it for the domain of $f^{-1}$ ).
Write $y = f(x)$ .
Rearrange to make $x$ the subject.
If you take a square root, use the domain of $f$ to decide whether it is $+\sqrt{\phantom{x}}$ or $-\sqrt{\phantom{x}}$ .
Swap $x$ and $y$ (or just write the answer as $f^{-1}(x) = \dots$ in terms of $x$ ).
State the domain of $f^{-1}$ , which is the range of $f$ .

A linear function

The function $f$ is defined by $f(x) = 5x - 2$ for $x \in \mathbb{R}$ . Find $f^{-1}(x)$ .

Solution

y = 5x - 2 \quad\Rightarrow\quad y + 2 = 5x \quad\Rightarrow\quad x = \frac{y + 2}{5}

So $f^{-1}(x) = \dfrac{x + 2}{5}$ for $x \in \mathbb{R}$ .

A reciprocal function

The function $f$ is defined by $f(x) = \dfrac{4}{x - 1} + 2$ for $x > 1$ . Find $f^{-1}(x)$ and state its domain.

Solution

Range of $f$ : for $x > 1$ , $\dfrac{4}{x - 1} > 0$ (and takes every positive value), so $f(x) > 2$ .

y = \frac{4}{x - 1} + 2 \quad\Rightarrow\quad y - 2 = \frac{4}{x - 1} \quad\Rightarrow\quad x - 1 = \frac{4}{y - 2} \quad\Rightarrow\quad x = 1 + \frac{4}{y - 2}

So $f^{-1}(x) = 1 + \dfrac{4}{x - 2}$ , with domain $x > 2$ (the range of $f$ ).

A completed square: choosing the sign

The function $h$ is defined by $h(x) = (2x - 3)^2 - 4$ for $x > \tfrac{3}{2}$ . Find an expression for $h^{-1}(x)$ and state its domain.

Solution

For $x > \tfrac{3}{2}$ , $2x - 3 > 0$ , so $(2x - 3)^2 > 0$ and $h(x) > -4$ . The range of $h$ is $h(x) > -4$ .

y = (2x - 3)^2 - 4 \quad\Rightarrow\quad (2x - 3)^2 = y + 4 \quad\Rightarrow\quad 2x - 3 = \sqrt{y + 4}

Only the positive root is used, because $2x - 3 > 0$ on the domain of $h$ . Then

x = \frac{3 + \sqrt{y + 4}}{2}

So $h^{-1}(x) = \dfrac{3 + \sqrt{x + 4}}{2}$ for $x > -4$ .

Restricting the domain of a quadratic

A quadratic is one-one on any domain that lies entirely on one side of its vertex. Completing the square shows where the vertex is.

Key result

For $f(x) = a(x - p)^2 + q$ :

on $x \ge p$ (or any domain to the right of $p$ ), $f$ is one-one;
on $x \le p$ (or any domain to the left of $p$ ), $f$ is one-one;
on a domain containing values on both sides of $p$ , $f$ is many-one.

The least value of k

The function $f$ is defined by $f(x) = x^2 - 6x + 5$ for $x \ge k$ .

(a) Find the least value of $k$ for which $f$ has an inverse.

(b) For this value of $k$ , find $f^{-1}(x)$ and state its domain.

Solution

(a) $f(x) = (x - 3)^2 - 4$ . The vertex is at $x = 3$ . The function is one-one on $x \ge k$ provided the domain does not extend to the left of the vertex, so the least value is $k = 3$ .

(b) With $x \ge 3$ , the range of $f$ is $f(x) \ge -4$ .

y = (x - 3)^2 - 4 \quad\Rightarrow\quad x - 3 = +\sqrt{y + 4} \quad\Rightarrow\quad x = 3 + \sqrt{y + 4}

(positive root, since $x - 3 \ge 0$ ). So $f^{-1}(x) = 3 + \sqrt{x + 4}$ for $x \ge -4$ .

The left branch: a negative root

The function $f$ is defined by $f(x) = 2(x + 1)^2 - 3$ for $x \le -1$ . Find $f^{-1}(x)$ and state the domain and range of $f^{-1}$ .

Solution

On $x \le -1$ , $f$ is one-one with range $f(x) \ge -3$ .

y = 2(x + 1)^2 - 3 \quad\Rightarrow\quad (x + 1)^2 = \frac{y + 3}{2}

Here $x + 1 \le 0$ , so we need the negative root:

x + 1 = -\sqrt{\frac{y + 3}{2}} \quad\Rightarrow\quad x = -1 - \sqrt{\frac{y + 3}{2}}

So $f^{-1}(x) = -1 - \sqrt{\dfrac{x + 3}{2}}$ . Domain of $f^{-1}$ : $x \ge -3$ . Range of $f^{-1}$ : $f^{-1}(x) \le -1$ .

Check: $f(-3) = 2(4) - 3 = 5$ , and $f^{-1}(5) = -1 - \sqrt{4} = -3$ . Correct.

The graphical relationship

Swapping $x$ and $y$ reflects every point $(a, b)$ to $(b, a)$ , which is reflection in the line $y = x$ . So the graph of $f^{-1}$ is the mirror image of the graph of $f$ in $y = x$ .

y = x^2 + 1 + 0 sqrt(x) y = sqrt(x - 1) y = x (0, 1) (1, 0)

The graph shows $f(x) = x^2 + 1$ for $x \ge 0$ and its inverse $f^{-1}(x) = \sqrt{x - 1}$ for $x \ge 1$ . The point $(0, 1)$ on $f$ reflects to $(1, 0)$ on $f^{-1}$ .

Sketching f and its inverse on one diagram

Draw the line $y = x$ (the syllabus requires the mirror line to be shown).
Sketch $y = f(x)$ over its domain only, labelling end points and intercepts.
Reflect each key point $(a, b)$ to $(b, a)$ and sketch $y = f^{-1}(x)$ through them.
Use equal scales on both axes if you can, so that the symmetry is visible.

Where f and its inverse meet

If $f$ is increasing, the graphs of $f$ and $f^{-1}$ can only meet on the mirror line $y = x$ . So $f(x) = f^{-1}(x)$ can be solved by the much easier equation $f(x) = x$ .

Intersection of f and its inverse

The function $f$ is defined by $f(x) = x^2 - 4x + 6$ for $x \ge 2$ . Find the coordinates of the points where the graphs of $y = f(x)$ and $y = f^{-1}(x)$ meet.

Solution

$f(x) = (x - 2)^2 + 2$ is increasing for $x \ge 2$ , so the graphs meet on $y = x$ :

x^2 - 4x + 6 = x \quad\Rightarrow\quad x^2 - 5x + 6 = 0 \quad\Rightarrow\quad (x - 2)(x - 3) = 0

Both $x = 2$ and $x = 3$ are in the domain. The points are $(2, 2)$ and $(3, 3)$ .

y = x^2 - 4x + 6 + 0 sqrt(x - 2) y = 2 + sqrt(x - 2) y = x (2, 2) (3, 3)

Watch out

$f^{-1}(x)$ is not $\dfrac{1}{f(x)}$ . The $-1$ is notation for "inverse", not a power.

Choosing the wrong sign of the square root. Use the domain of $f$ : if $x \ge p$ take $+\sqrt{\phantom{x}}$ , if $x \le p$ take $-\sqrt{\phantom{x}}$ . Writing $\pm$ in the final answer scores no accuracy mark.

Forgetting the domain of $f^{-1}$ . It is the range of $f$ , and many mark schemes award a mark for it.

Giving the answer in terms of $y$ . Your final answer must be $f^{-1}(x) = \dots$ in terms of $x$ .

Reflecting in the wrong line. The mirror is $y = x$ , not the $x$ -axis or $y$ -axis.

Exam tip

"Find an expression for $f^{-1}(x)$ " usually earns 3 marks: one for the first rearrangement, one for the correct root or step, one for the final form in terms of $x$ .
"State the domain of $f^{-1}$ " is a separate 1-mark part. Work out the range of $f$ before you start inverting.
"Explain why $f$ has an inverse" or "show that $f$ is one-one": say it is increasing (or decreasing) throughout its domain, or that the domain lies on one side of the vertex.
On a sketch of $f$ and $f^{-1}$ , examiners look for: the line $y = x$ drawn, the two curves as reflections, and end points or intercepts swapped correctly.

Summary

Only one-one functions have inverses. Restrict the domain of a quadratic to one side of its vertex.
To find $f^{-1}$ : write $y = f(x)$ , make $x$ the subject, then write in terms of $x$ .
Use the domain of $f$ to choose the sign of any square root.
Domain of $f^{-1}$ = range of $f$ ; range of $f^{-1}$ = domain of $f$ .
$ff^{-1}(x) = f^{-1}f(x) = x$ .
The graph of $f^{-1}$ is the reflection of the graph of $f$ in $y = x$ ; draw the line $y = x$ on sketches.
For increasing $f$ , solve $f(x) = f^{-1}(x)$ by solving $f(x) = x$ .

Practice questions

Question

Find $f^{-1}(x)$ for $f(x) = 4x + 7$ , $x \in \mathbb{R}$ .
Find $f^{-1}(x)$ for $f(x) = \dfrac{3x - 1}{2}$ , $x \in \mathbb{R}$ .
$f(x) = \dfrac{2}{x + 3}$ for $x > -3$ . Find $f^{-1}(x)$ and state its domain.
$f(x) = (x - 2)^2 + 1$ for $x \ge 2$ . Find $f^{-1}(x)$ and state its domain.
$f(x) = x^2 + 6x$ for $x \ge k$ . Find the least value of $k$ for which $f^{-1}$ exists, and for this value find $f^{-1}(x)$ and its domain.
$f(x) = 5 - (x + 1)^2$ for $x \le -1$ . Find $f^{-1}(x)$ and state its domain and range.
$f(x) = 2x + 1$ for $x \ge 0$ and $g(x) = x^2$ for $x \ge 0$ . Find $gf(x)$ and hence find $(gf)^{-1}(x)$ , stating its domain.
$f(x) = 3x - 4$ for $x \in \mathbb{R}$ . Solve $f(x) = f^{-1}(x)$ .
(a) Express $2x^2 - 12x + 11$ in the form $a(x + b)^2 + c$ . (b) The function $f$ is defined by $f(x) = 2x^2 - 12x + 11$ for $x \le 1$ . Find the range of $f$ and an expression for $f^{-1}(x)$ .
$f(x) = (x - 1)^2 + 1$ for $x \ge 1$ . Sketch $y = f(x)$ and $y = f^{-1}(x)$ on the same diagram, and find the coordinates of the points where they meet.

Answers

$y = 4x + 7$ , $x = \dfrac{y - 7}{4}$ . $f^{-1}(x) = \dfrac{x - 7}{4}$ .
$2y = 3x - 1$ , $x = \dfrac{2y + 1}{3}$ . $f^{-1}(x) = \dfrac{2x + 1}{3}$ .
Range of $f$ : $f(x) > 0$ . $x + 3 = \dfrac{2}{y}$ , so $f^{-1}(x) = \dfrac{2}{x} - 3$ for $x > 0$ .
Range of $f$ : $f(x) \ge 1$ . $x - 2 = +\sqrt{y - 1}$ , so $f^{-1}(x) = 2 + \sqrt{x - 1}$ for $x \ge 1$ .
$f(x) = (x + 3)^2 - 9$ ; least $k = -3$ . Range $f(x) \ge -9$ . $x + 3 = \sqrt{y + 9}$ , so $f^{-1}(x) = -3 + \sqrt{x + 9}$ for $x \ge -9$ .
Range of $f$ : $f(x) \le 5$ . $(x + 1)^2 = 5 - y$ , and $x + 1 \le 0$ , so $x + 1 = -\sqrt{5 - y}$ . $f^{-1}(x) = -1 - \sqrt{5 - x}$ , domain $x \le 5$ , range $f^{-1}(x) \le -1$ .
$gf(x) = (2x + 1)^2$ for $x \ge 0$ , with range $gf(x) \ge 1$ . $2x + 1 = \sqrt{y}$ (positive, since $2x + 1 > 0$ ), so $(gf)^{-1}(x) = \dfrac{\sqrt{x} - 1}{2}$ for $x \ge 1$ .
$f$ is increasing, so solve $f(x) = x$ : $3x - 4 = x$ , $x = 2$ . (Check: $f^{-1}(x) = \dfrac{x + 4}{3}$ , and $f^{-1}(2) = 2 = f(2)$ .)
(a) $2(x - 3)^2 - 7$ . (b) For $x \le 1$ the function is decreasing (left of the vertex $x = 3$ ), with $f(1) = 1$ and $f(x)$ increasing without limit as $x$ decreases. Range $f(x) \ge 1$ . $(x - 3)^2 = \dfrac{y + 7}{2}$ and $x - 3 < 0$ , so $x = 3 - \sqrt{\dfrac{y + 7}{2}}$ . $f^{-1}(x) = 3 - \sqrt{\dfrac{x + 7}{2}}$ for $x \ge 1$ .
$f$ starts at $(1, 1)$ and increases; $f^{-1}(x) = 1 + \sqrt{x - 1}$ starts at $(1, 1)$ , its reflection in $y = x$ . They meet on $y = x$ : $(x - 1)^2 + 1 = x$ , so $x^2 - 3x + 2 = 0$ , $x = 1$ or $2$ . Points $(1, 1)$ and $(2, 2)$ .

Undoing a function

Why it must be one-one

Domains and ranges swap

Finding an inverse algebraically

Restricting the domain of a quadratic

The graphical relationship

Where f and its inverse meet

Practice questions

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