Graphs - Inverse Functions

Solutions

Question 1

a) When $x=3$:

$\begin{align*}f(3)&=2\times3\\&=6\end{align*}$

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b) When $x=20$:

$\begin{align*}f(20)&=\frac{20}{5}\\&=4\end{align*}$

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c) When $x=12$:

$\begin{align*}f(12)&=12+16\\&=28\end{align*}$

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d) x = 10: f(10)=10-3=7

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e) When x=4: f(4)= 3*4+11 = 12+11 = 23

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f) When $x=18$:

$\begin{align*}f(18)&=\frac{18}{6}-2\\&=3-2\\&=1\end{align*}$

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g)

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h)

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Question 2

a) $f(x)=x+2$

The opposite operation to adding $2$ is to minus $2$

Therefore $f^{-1}(x)=x-2$

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b) $g(x)=x-5$

The opposite operation to subtracting $5$ is to add $5$

Therefore $g^{-1}(x)=x+5$

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c) $h(x)=3x$

The opposite operation to multiplying by $3$ is to divide by $3$

Therefore $h^{-1}(x)=\frac x3$

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d) $p(x)=x\div6$

The opposite operation to dividing by $6$ is to multiply by $6$

Therefore $p^{-1}(x)=6x$

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e)

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f)

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g)

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h)

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i)

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j)

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k)

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Question 3

a) As this is a show that question the answer and solution are the same

$\begin{align*}f(x)&=x^3\\\text{when}\,x=4,\,f(4)&=4^3\\&=64\end{align*}$

Substitute $f(x)=64$ into $f^{-1}(x)=\sqrt[3] x$

$\begin{align*}f^{-1}(64)&=\sqrt[3]{64}\\&=4\end{align*}$

As the input into $f(x)$ is $4$ and the output of $f^{-1}(x)$ is also $4$ they are inverse for $x=4$

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b) As this is a show that question the answer and solution are the same

$\begin{align*}f(x)&=2+3x\\\text{when}\,x=4,\,f(4)&=2+3\times4\\&=14\end{align*}$

Substitute $f(x)=14$ into $f^{-1}(x)=\frac{x-2}{3}$

$\begin{align*}f^{-1}(14)&=\frac{14-2}{3}\\&=\frac{12}{3}\\&=4\end{align*}$

As the input into $f(x)$ is $4$ and the output of $f^{-1}(x)$ is also $4$ they are inverse for $x=4$

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c)

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d)

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e)

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f)

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g)

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h)

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Question 4

a) $f(x)=2x+3$

To find the inverse let $f(x)=y$ and rearrange to make $x$ the subject:

$\begin{align*}y&=2x+3\\y-3&=2x\\\frac{y-3}{2}&=x\\\text{so}\,\,\,\,x&=\frac{y-3}{2}\end{align*}$

Now, let $x=f^{-1}(x)$ and $y=x$:

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

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b) $f(x)=-5x+6$

To find the inverse let $f(x)=y$ and rearrange to make $x$ the subject:

$\begin{align*}y&=-5x+6\\y-6&=-5x\\\frac{y-6}{-5}&=x\\\text{so}\,\,\,\,x&=\frac{y-6}{-5}\,\,\,\text{or}\,\,\,x=\frac{6-y}{5}\end{align*}$

Now, let $x=f^{-1}(x)$ and $y=x$:

$f^{-1}(x)=\frac{x-6}{-5}\,\,\,\text{or}\,\,\,f^{-1}(x)=\frac{6-x}{5}$

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c)

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d)

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e)

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f)

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g) $f(a)=\sqrt a+15$

To find the inverse let $f(a)=y$ and rearrange to make $a$ the subject:

$\begin{align*}y&=\sqrt a+15\\y-15&=\sqrt a\\{(y-15)}^2&=a\\\text{so}\,\,\,\,a&={(y-15)}^2\end{align*}$

Now, let $a=f^{-1}(a)$ and $y=a$:

$f^{-1}(a)={(a-15)}^2$

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h)

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Question 5

a) As this is a show that question the answer and solution are the same

$\begin{align*}f(x)&=2x\\f^{-1}(f(x))&=\frac{2x}{2}\\&=x\end{align*}$

Therefore as $f^{-1}(f(x))=x$ these functions must be the inverse of each other

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b) As this is a show that question the answer and solution are the same

$\begin{align*}f(x)&=\frac14x\\f^{-1}(f(x))&=4\times\frac14x\\&=x\end{align*}$

Therefore as $f^{-1}(f(x))=x$ these functions must be the inverse of each other

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c) As this is a show that question the answer and solution are the same

$\begin{align*}f(x)&=4x-1\\f^{-1}(f(x))&=\frac{(4x-1)+1}{4}\\&=\frac{4x}{4}\\&=x\end{align*}$

Therefore as $f^{-1}(f(x))=x$ these functions must be the inverse of each other

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d) As this is a show that question the answer and solution are the same

$\begin{align*}f(x)&=\sqrt{5x}+4\\f^{-1}(f(x))&=\frac{{(\sqrt{5x}+4-4)}^2}{5}\\&=\frac{{(\sqrt{5x})}^2}{5}\\&=\frac{5x}{5}\\&=x\end{align*}$

Therefore as $f^{-1}(f(x))=x$ these functions must be the inverse of each other

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e) As this is a show that question the answer and solution are the same

$\begin{align*}f(x)&=\frac{x+4}{2x-5}\\f^{-1}(f(x))&=\frac{5\left(\frac{x+4}{2x-5}\right)+4}{2\left(\frac{x+4}{2x-5}\right)-1}\\&=\frac{\frac{5x+20}{2x-5}+4}{\frac{2x+8}{2x-5}-1}\\&=\frac{\frac{5x+20}{2x-5}+\frac{8x-20}{2x-5}}{\frac{2x+8}{2x-5}-\frac{2x-5}{2x-5}}\\&=\frac{\frac{5x+20+8x-20}{2x-5}}{\frac{2x+8-2x+5}{2x-5}}\\&=\frac{\frac{13x}{2x-5}}{\frac{13}{2x-5}}\\&=\frac{13x}{13}\\&=x\end{align*}$

Therefore as $f^{-1}(f(x))=x$ these functions must be the inverse of each other

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