Week 1 · Day 5 Lesson 1.11 About 10 minutes Free

Inverse functions

By the end: Find the inverse of a function and state its domain, evaluate $f^{-1}(k)$ without finding a formula, and use the fact that a graph and its inverse mirror each other in the line $y=x$ .

Understand it

A function sends each input $x$ to one output $y$ . Its inverse function $f^{-1}$ goes back: it sends that $y$ to the original $x$ . So

$f(a)=b\iff f^{-1}(b)=a.$

That single fact answers most exam questions. To find $f^{-1}(7)$ you do not need a formula: solve $f(x)=7$ , and the solution is the answer.

When an inverse exists. The function must be one-to-one: no two different inputs may give the same output. A function that is always increasing, or always decreasing, on its domain is one-to-one. The function $f(x)=x^2$ on $\mathbb{R}$ is not (since $f(2)=f(-2)$ ), but on the restricted domain $x\ge0$ it is, and its inverse is $\sqrt{x}$ .

Domain and range swap. The inputs of $f^{-1}$ are the outputs of $f$ :

domain of $f^{-1}$ = range of $f$ ,
range of $f^{-1}$ = domain of $f$ .

So always state the domain of $f^{-1}$ , and find it from the range of $f$ , not from the formula alone.

Finding the formula, in four steps.

Write $y=f(x)$ .
Solve for $x$ in terms of $y$ .
Swap the names: write $x$ instead of $y$ and $f^{-1}(x)$ for the result.
State the domain, which is the range of $f$ .

The graph. If the point $(a,b)$ is on the graph of $f$ , then $(b,a)$ is on the graph of $f^{-1}$ . So the two graphs are mirror images in the line $y=x$ . Careful: the inverse is not the reciprocal. $f^{-1}(x)$ is not $\dfrac1{f(x)}$ .

Check your answer by testing one number: $f\bigl(f^{-1}(x)\bigr)$ must give back $x$ .

Essential formulas

f(a)=b\iff f^{-1}(b)=a

Use it to evaluate an inverse:

f^{-1}(b)

is the input

a

that

f

turns into

b

. Solve

f(x)=b

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

The inverse reverses the roles of inputs and outputs.

(a,b)\text{ on }f\iff(b,a)\text{ on }f^{-1}

The graphs of

f

and

f^{-1}

are reflections of each other in the line

y=x

See it

The point (a, b) on f becomes (b, a) on the inverse: (1, 0) goes to (0, 1), and (5, 2) to (2, 5). The graphs are mirror images in the line y = x.

Worked examples

Worked example 1

Find the inverse of $f(x)=3x-5$ .

Write $y=3x-5$ and solve for $x$ : $y+5=3x$ , so $x=\dfrac{y+5}{3}$ .
Swap the names: $f^{-1}(x)=\dfrac{x+5}{3}$ .
The range of $f$ is all real numbers, so the domain of $f^{-1}$ is $\mathbb{R}$ .
Check with $x=1$ : $f(1)=-2$ , and $f^{-1}(-2)=\dfrac{-2+5}{3}=1$ brings it back.

Answer: $f^{-1}(x)=\dfrac{x+5}{3}$ , with domain $\mathbb{R}$ .

Worked example 2

Find the inverse of $f(x)=\sqrt{x-1}$ , and state its domain.

The domain of $f$ is $x\ge1$ . Write $y=\sqrt{x-1}$ . A square root is never negative, so the range of $f$ is $y\ge0$ .
Square both sides: $y^2=x-1$ , so $x=y^2+1$ .
Swap the names: $f^{-1}(x)=x^2+1$ .
The domain of $f^{-1}$ is the range of $f$ , which is $x\ge0$ . Without this restriction $x^2+1$ would not be the inverse (it is not one-to-one on $\mathbb{R}$ ).

Answer: $f^{-1}(x)=x^2+1$ for $x\ge0$ .

Worked example 3

Let $f(x)=x^3+x$ . Find $f^{-1}(10)$ .

Use the shortcut: $f^{-1}(10)$ is the number $a$ with $f(a)=10$ .
Solve $x^3+x=10$ . Try small integers: $f(2)=8+2=10$ .
$f$ is increasing, so it is one-to-one and $x=2$ is the only solution.

Answer: $f^{-1}(10)=2$

Try it yourself

Work it on paper first. Open a hint only when you are stuck, then compare with the solution.

Find the inverse of $f(x)=\dfrac{2x+1}{x-1}$ and state its domain.

Hint 1 of 2

Write $y=\dfrac{2x+1}{x-1}$ , multiply out the denominator, and collect the terms with $x$ on one side.

Hint 2 of 2

The range of $f$ misses the value $2$ (the ratio of the coefficients of $x$ ).

Show the step-by-step solution

$y(x-1)=2x+1$ , so $xy-y=2x+1$ .
Collect $x$ : $xy-2x=y+1$ , so $x(y-2)=y+1$ and $x=\dfrac{y+1}{y-2}$ .
Swap the names: $f^{-1}(x)=\dfrac{x+1}{x-2}$ .
The range of $f$ is $y\ne2$ (because $y=2+\dfrac{3}{x-1}$ ), so the domain of $f^{-1}$ is $x\ne2$ .

Answer: $f^{-1}(x)=\dfrac{x+1}{x-2}$ , $x\ne2$ .

Common mistakes

Using the reciprocal: $f^{-1}(x)\ne\dfrac1{f(x)}$ . The inverse undoes $f$ ; it does not flip it.
Forgetting to state the domain of $f^{-1}$ . It is the range of $f$ , and it can change the answer.
Looking for an inverse of a function that is not one-to-one without restricting the domain, as with $x^2$ on $\mathbb{R}$ .
Writing the answer in terms of $y$ . After solving for $x$ , swap the names so the inverse is in terms of $x$ .
Mixing up $f(a)=b$ and $f^{-1}(a)=b$ . The inverse reverses the roles: $f^{-1}(b)=a$ .

Exam tips

If the question asks for one value such as $f^{-1}(5)$ , solve $f(x)=5$ . No formula is needed.
If the graph of $f$ passes through $(1,3)$ , then $f^{-1}$ passes through $(3,1)$ .
Test your formula: $f(f^{-1}(x))$ should simplify to $x$ .
For a fraction such as $\dfrac{ax+b}{cx+d}$ , the inverse is again a fraction, and the missing value of $f$ becomes the excluded input of $f^{-1}$ .

In short

$f(a)=b\iff f^{-1}(b)=a$ .
Domain of $f^{-1}$ is the range of $f$ ; range of $f^{-1}$ is the domain of $f$ .
Steps: $y=f(x)$ , solve for $x$ , swap names, state the domain.
The graphs mirror in $y=x$ , and the inverse is not the reciprocal.

Knowledge check

4 questions on this lesson, marked as you go. Take them in ACE CSCA with a free account: passing them completes the lesson, and your progress is saved on every device.

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