Inverse Functions | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

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. “Make y the subject” means rearrange until y is alone on the left:

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}

Check by undoing the original rule: $\mathrm{f}^{-1}(\mathrm{f}(x))=\dfrac{(3x-1)+1}{3}=x$ . A correct inverse returns the starting input wherever the composition is defined.

It only exists when f is one-one

An inverse exists only when f is one-one. For example, $x^2$ sends both $-3$ and $3$ to $9$ , so reversing it would give two outputs. Restricting the domain to one side of the vertex removes that ambiguity.

“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: $y=x^2-6x=(x-3)^2-9$ .
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}

The original range begins at the vertex output $-9$ , so it becomes the inverse domain:

Answer

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

The original domain becomes the inverse range:

Answer

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

Two inverse checks

Common mistake

A restricted quadratic inverse keeps one square-root branch. Use the original domain to choose

+\sqrt{\;}

-\sqrt{\;}

; do not leave

\pm

Common mistake

State the inverse domain when requested: it is the original range and often earns a separate mark. (9709/12 Oct/Nov 2021 Q3)

Now you try

Your turn9709/12 Oct/Nov 2021 Q3(b)

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)

Show worked answer

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

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

Clear the fraction by multiplying both sides by $y-1$ , then 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}

The same rule returns, so f is self-inverse. Its original range is $\mathrm{f}(x)>1$ , so the inverse domain is $x>1$ .

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. “Make y the subject” means rearrange until y is alone on the left:

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}

Check by undoing the original rule: $\mathrm{f}^{-1}(\mathrm{f}(x))=\dfrac{(3x-1)+1}{3}=x$ . A correct inverse returns the starting input wherever the composition is defined.

It only exists when f 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: $y=x^2-6x=(x-3)^2-9$ .
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}

The original range begins at the vertex output $-9$ , so it becomes the inverse domain:

Answer

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

The original domain becomes the inverse range:

Answer

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

Two inverse checks

Common mistake

A restricted quadratic inverse keeps one square-root branch. Use the original domain to choose

+\sqrt{\;}

-\sqrt{\;}

; do not leave

\pm

Common mistake

State the inverse domain when requested: it is the original range and often earns a separate mark. (9709/12 Oct/Nov 2021 Q3)

Now you try

Your turn9709/12 Oct/Nov 2021 Q3(b)

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)

Show worked answer

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

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

Clear the fraction by multiplying both sides by $y-1$ , then 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}

The same rule returns, so f is self-inverse. Its original range is $\mathrm{f}(x)>1$ , so the inverse domain is $x>1$ .