Mathematics 9709 · Pure Mathematics 1

Inverse functions

The inverse $\mathrm{f}^{-1}$ undoes $\mathrm{f}$ . If $\mathrm{f}$ sends $3\to 10$ , then $\mathrm{f}^{-1}$ sends $10\to 3$ .
Finding it is a short method. The marks come from two other things: when it exists, and what its domain is.

Write $y=\dots$ , swap $x$ and $y$ , then make $y$ the subject. Then state the domain of $\mathrm{f}^{-1}$ (= the range of $\mathrm{f}$ ). The swap goes both ways: $\text{range of }\mathrm{f}^{-1}=\text{domain of }\mathrm{f}$ too.

Swap x and y, then make y the subject

An inverse swaps input and output. So the method is to swap $x$ and $y$ , then tidy up:

Write the function as $y=\dots$
Swap every $x$ and $y$ .
Rearrange to make $y$ the subject. That is $\mathrm{f}^{-1}(x)$ .

Worked example

Textbook step-by-step: find the inverse of $\mathrm{f}(x)=3x-1$

Write as $y$ : $y=3x-1$ .
Swap $x$ and $y$ : $x=3y-1$ .
Make $y$ the subject: $x+1=3y\;\Rightarrow\; y=\dfrac{x+1}{3}$ .

Answer

\mathrm{f}^{-1}(x)=\dfrac{x+1}{3}

It only exists when f is one-one

An inverse only works if $\mathrm{f}$ is one-one (each output comes from exactly one input).
A many-one function like $x^2$ has no inverse. Going back, the machine cannot tell which input to go to.

Why a many-one function has no inverse

$\mathrm{f}(x)=x^2$ sends both $-3$ and $3$ to $9$ . Going back from $9$ , the inverse would have to pick $-3$ or $3$ . That is two outputs for one input, so it is not a function at all. So there is no inverse until we make $\mathrm{f}$ one-one.

That is why exam questions make the domain smaller, like “for $x\ge 3$ ”. This removes one half of the parabola and keeps one rising side, which is one-one.

“Find the smallest $k$ so that $\mathrm{f}$ (for $x\ge k$ ) is one-one” has a one-line answer: $k$ = the $x$ -coordinate of the vertex. Cut the parabola at its turning point and only one side is left.

Worked example

Textbook 2.8: $\mathrm{f}(x)=5-(x-2)^2$ for $k\le x\le 6$ , state the smallest $k$ for which $\mathrm{f}$ has an inverse

$\mathrm{f}$ is a parabola, which is one-one only on one side of its turning point. Read the vertex from the completed-square form $5-(x-2)^2$ : it is at $x=2$ . Keep everything from there on and you have one falling side.

Answer

k=2

f(x) = x² − 6x kept only for x ≥ 3: one rising side, so it is one-one and has an inverse. It starts at the vertex (3, −9).

Worked example

$\mathrm{f}(x)=x^2-6x$ for $x\ge 3$ : find $\mathrm{f}^{-1}(x)$ , its domain and its range

Complete the square so the squared term sits alone (the move from the Quadratics chapter, §1.2): $y=x^2-6x=(x-3)^2-9$ .
Swap $x$ and $y$ , then isolate the bracket: $x=(y-3)^2-9\;\Rightarrow\;(y-3)^2=x+9$ .
Square-root. The domain $x\ge 3$ forces $y-3\ge 0$ , so keep the positive root only (not $\pm$ ): $y-3=\sqrt{x+9}$ .

Answer

\mathrm{f}^{-1}(x)=3+\sqrt{x+9}

For the domain of $\mathrm{f}^{-1}$ , take the range of $\mathrm{f}$ . The smallest output is at the start point $x=3$ , giving $-9$ , and the graph only goes up from there:

Answer

\text{domain of }\mathrm{f}^{-1}:\ x\ge -9

The swap goes both ways. So the range of $\mathrm{f}^{-1}$ is the domain of $\mathrm{f}$ , which is $x\ge 3$ :

Answer

\text{range of }\mathrm{f}^{-1}:\ \mathrm{f}^{-1}(x)\ge 3

Where the marks go

Common mistake

\mathrm{f}^{-1}(x)

is not

\dfrac{1}{\mathrm{f}(x)}

. The

-1

means “inverse”, not “one over”. They are completely different things.

Common mistake

When you take the square root, the smaller domain picks one side: keep

+\sqrt{\;}

-\sqrt{\;}

, never

\pm

Common mistake

Always state the domain of

\mathrm{f}^{-1}

(= the range of

\mathrm{f}

). It gets its own mark, so the expression on its own does not score full marks. (9709/12 Oct/Nov 2021 Q3)

Now you try

The function $\mathrm{f}(x)=\dfrac{x+3}{x-1}$ is defined for $x>1$ . Find an expression for $\mathrm{f}^{-1}(x)$ . (9709/12 Oct/Nov 2021 Q3)

Your turn— tap to reveal the worked answer (9709/12 Oct/Nov 2021 Q3(b))

Write as $y$ , then swap $x$ and $y$ :

$x=\dfrac{y+3}{y-1}$

Clear the fraction and gather the $y$ terms:

$x(y-1)=y+3 \;\Rightarrow\; xy-x=y+3$

$xy-y=x+3 \;\Rightarrow\; y(x-1)=x+3$

Answer

\mathrm{f}^{-1}(x)=\dfrac{x+3}{x-1}

It comes back to the same rule. So $\mathrm{f}$ is its own inverse (self-inverse).

PreviousComposite functions NextThe graph of a function and its inverse

Inverse functions

The inverse $\mathrm{f}^{-1}$ undoes $\mathrm{f}$ . If $\mathrm{f}$ sends $3\to 10$ , then $\mathrm{f}^{-1}$ sends $10\to 3$ .
Finding it is a short method. The marks come from two other things: when it exists, and what its domain is.

Swap x and y, then make y the subject

An inverse swaps input and output. So the method is to swap $x$ and $y$ , then tidy up:

Write the function as $y=\dots$
Swap every $x$ and $y$ .
Rearrange to make $y$ the subject. That is $\mathrm{f}^{-1}(x)$ .

Worked example

Textbook step-by-step: find the inverse of $\mathrm{f}(x)=3x-1$

Write as $y$ : $y=3x-1$ .
Swap $x$ and $y$ : $x=3y-1$ .
Make $y$ the subject: $x+1=3y\;\Rightarrow\; y=\dfrac{x+1}{3}$ .

Answer

\mathrm{f}^{-1}(x)=\dfrac{x+1}{3}

It only exists when f is one-one

An inverse only works if $\mathrm{f}$ is one-one (each output comes from exactly one input).
A many-one function like $x^2$ has no inverse. Going back, the machine cannot tell which input to go to.

Why a many-one function has no inverse

That is why exam questions make the domain smaller, like “for $x\ge 3$ ”. This removes one half of the parabola and keeps one rising side, which is one-one.

Worked example

Textbook 2.8: $\mathrm{f}(x)=5-(x-2)^2$ for $k\le x\le 6$ , state the smallest $k$ for which $\mathrm{f}$ has an inverse

Answer

k=2

f(x) = x² − 6x kept only for x ≥ 3: one rising side, so it is one-one and has an inverse. It starts at the vertex (3, −9).

Worked example

$\mathrm{f}(x)=x^2-6x$ for $x\ge 3$ : find $\mathrm{f}^{-1}(x)$ , its domain and its range

Complete the square so the squared term sits alone (the move from the Quadratics chapter, §1.2): $y=x^2-6x=(x-3)^2-9$ .
Swap $x$ and $y$ , then isolate the bracket: $x=(y-3)^2-9\;\Rightarrow\;(y-3)^2=x+9$ .
Square-root. The domain $x\ge 3$ forces $y-3\ge 0$ , so keep the positive root only (not $\pm$ ): $y-3=\sqrt{x+9}$ .

Answer

\mathrm{f}^{-1}(x)=3+\sqrt{x+9}

For the domain of $\mathrm{f}^{-1}$ , take the range of $\mathrm{f}$ . The smallest output is at the start point $x=3$ , giving $-9$ , and the graph only goes up from there:

Answer

\text{domain of }\mathrm{f}^{-1}:\ x\ge -9

The swap goes both ways. So the range of $\mathrm{f}^{-1}$ is the domain of $\mathrm{f}$ , which is $x\ge 3$ :

Answer

\text{range of }\mathrm{f}^{-1}:\ \mathrm{f}^{-1}(x)\ge 3

Where the marks go

Common mistake

\mathrm{f}^{-1}(x)

is not

\dfrac{1}{\mathrm{f}(x)}

. The

-1

means “inverse”, not “one over”. They are completely different things.

Common mistake

When you take the square root, the smaller domain picks one side: keep

+\sqrt{\;}

-\sqrt{\;}

, never

\pm

Common mistake

Always state the domain of

\mathrm{f}^{-1}

(= the range of

\mathrm{f}

). It gets its own mark, so the expression on its own does not score full marks. (9709/12 Oct/Nov 2021 Q3)

Now you try

The function $\mathrm{f}(x)=\dfrac{x+3}{x-1}$ is defined for $x>1$ . Find an expression for $\mathrm{f}^{-1}(x)$ . (9709/12 Oct/Nov 2021 Q3)

Your turn— tap to reveal the worked answer (9709/12 Oct/Nov 2021 Q3(b))

Write as $y$ , then swap $x$ and $y$ :

$x=\dfrac{y+3}{y-1}$

Clear the fraction and gather the $y$ terms:

$x(y-1)=y+3 \;\Rightarrow\; xy-x=y+3$

$xy-y=x+3 \;\Rightarrow\; y(x-1)=x+3$

Answer

\mathrm{f}^{-1}(x)=\dfrac{x+3}{x-1}

It comes back to the same rule. So $\mathrm{f}$ is its own inverse (self-inverse).