Graphs - Inverse Functions
Question 1
Find $f(x)$
a) $f(x)=2x$ when $x=3$
a
$f(3)=6$
Question ID: 10060010010
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Answer by: ada
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When $x=3$:
$\begin{align*}f(3)&=2\times3\\&=6\end{align*}$
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b) $f(x)=\frac x5$ when $x=20$
a
$f(20)=4$
Question ID: 10060010020
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Answer by: ada
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When $x=20$:
$\begin{align*}f(20)&=\frac{20}{5}\\&=4\end{align*}$
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c) $f(x)=x+16$ when $x=12$
a
$f(12)=28$
Question ID: 10060010030
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Answer by: ada
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When $x=12$:
$\begin{align*}f(12)&=12+16\\&=28\end{align*}$
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d) $f(x)=x-3$ when $x=10$
a
$f(10)=7$
Question ID: 10060010040
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x = 10: f(10)=10-3=7
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e) $f(x)=3x+11$ when $x=4$
a
$f(4)=23$
Question ID: 10060010050
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When x=4: f(4)= 3*4+11 = 12+11 = 23
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f) $f(x)=\frac x6-2$ when $x=18$
a
$f(18)=1$
Question ID: 10060010060
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Answer by: ada
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When $x=18$:
$\begin{align*}f(18)&=\frac{18}{6}-2\\&=3-2\\&=1\end{align*}$
Question ID: 10060010060
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g) $f(x)=x^2+5$ when $x=2$
a
$f(2)=9$
Question ID: 10060010070
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h) $f(x)=3-\sqrt x$ when $x=36$
a
$f(36)=-3$
Question ID: 10060010080
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Question 2
Find the inverse function for the following functions
a) $f(x)=x+2$
a
$f^{-1}(x)=x-2$
Question ID: 10060020010
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Answer by: ada
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$f(x)=x+2$
The opposite operation to adding $2$ is to minus $2$
Therefore $f^{-1}(x)=x-2$
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b) $g(x)=x-5$
a
$g^{-1}(x)=x+5$
Question ID: 10060020020
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Answer by: ada
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$g(x)=x-5$
The opposite operation to subtracting $5$ is to add $5$
Therefore $g^{-1}(x)=x+5$
Question ID: 10060020020
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c) $h(x)=3x$
a
$h^{-1}(x)=\frac x3$
Question ID: 10060020030
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$h(x)=3x$
The opposite operation to multiplying by $3$ is to divide by $3$
Therefore $h^{-1}(x)=\frac x3$
Question ID: 10060020030
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d) $p(x)=x\div6$
a
$p^{-1}(x)=6x$
Question ID: 10060020040
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Answer by: ada
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$p(x)=x\div6$
The opposite operation to dividing by $6$ is to multiply by $6$
Therefore $p^{-1}(x)=6x$
Question ID: 10060020040
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e) $q(x)=\frac x9$
a
$q^{-1}(x)=9x$
Question ID: 10060020050
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f) $f(y)=y+t$
a
$f^{-1}(y)=y-t$
Question ID: 10060020060
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g) $g(a)=\frac ab$
a
$g^{-1}(a)=ab$
Question ID: 10060020070
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h) $h(z)=\sqrt z$
a
$h^{-1}(z)=z^2$
Question ID: 10060020080
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i) $p(x)=x^5$
a
$p^{-1}(x)=\sqrt[5]x$
Question ID: 10060020090
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Question ID: 10060020090
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j) $m(r)=-r$
a
$m^{-1}(r)=-r$
Question ID: 10060020100
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Question ID: 10060020100
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k) $f(x)=\frac1x$
a
$f^{-1}(x)=\frac1x$
Question ID: 10060020110
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Question 3
Show that the following are the inverse when $x=4$
a) $f(x)=x^3$ and $f^{-1}(x)=\sqrt[3] x$
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 $x=4$
Question ID: 10060030010
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Answer by: ada, edited by Steven
s
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 $x=4$
Question ID: 10060030010
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b) $f(x)=2+3x$ and $f^{-1}(x)=\frac{x-2}{3}$
a
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 $x=4$
Question ID: 10060030020
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Answer by: ada
s
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 $x=4$
Question ID: 10060030020
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c) $f(x)=\frac{2+x}{4}$ and $f^{-1}(x)=4x-2$
a
no answer
Question ID: 10060030030
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d) $g(x)=\sqrt{x+12}-3$ and $g^{-1}(x)={(x+3)}^2-12$
a
no answer
Question ID: 10060030040
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e) $h(x)=\sqrt{x+12}-3$ and $h^{-1}(x)={(x+3)}^2-12$
a
no answer
Question ID: 10060030050
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f) $q(x)=\frac1x$ and $q^{-1}(x)=\frac1x$
a
no answer
Question ID: 10060030060
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g) $p(x)=\frac14x-7$ and $p^{-1}(x)=4(x+7)$
a
no answer
Question ID: 10060030070
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h) $f(x)=\frac{(2x+3)^2}{11}$ and $f^{-1}(x)=\frac{\sqrt{11x}-3}{2}$
a
no answer
Question ID: 10060030080
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Question 4
Find the inverse function for the following functions
a) $f(x)=2x+3$
a
$f^{-1}(x)=\frac{x-3}{2}$
Question ID: 10060040010
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$f(x)=2x+3$
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 $y=x$:
$f^{-1}(x)=\frac{x-3}{2}$
Question ID: 10060040010
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b) $f(x)=-5x+6$
a
$f^{-1}(x)=\frac{x-6}{-5}$ or $f^{-1}(x)=\frac{6-x}{5}$
Question ID: 10060040020
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$f(x)=-5x+6$
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 $y=x$:
$f^{-1}(x)=\frac{x-6}{-5}\,\,\,\text{or}\,\,\,f^{-1}(x)=\frac{6-x}{5}$
Question ID: 10060040020
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c) $f(x)=\frac12x-3$
a
$f^{-1}(x)=2(x+3)$
Question ID: 10060040030
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d) $g(x)=\frac15x+\frac23$
a
$g^{-1}(x)=5(x-\frac23)$
Question ID: 10060040040
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e) $f(x)=\frac{x+1}{4}$
a
$f^{-1}(x)=4x-1$
Question ID: 10060040050
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Answer by: ada edited by Will Tudehope
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f) $n(x)=\frac{x-2}{-5}$
a
$f^{-1}(x)=-5x+2$
Question ID: 10060040060
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Question ID: 10060040060
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g) $f(a)=\sqrt a+15$
a
$f^{-1}(a)={(a-15)}^2$
Question ID: 10060040070
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$f(a)=\sqrt a+15$
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 $y=a$:
$f^{-1}(a)={(a-15)}^2$
Question ID: 10060040070
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h) $g(y)={(y+3)}^3$
a
$g^{-1}(y)=\sqrt[3]{y}-3$
Question ID: 10060040080
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Question 5
By showing that $f(f^{-1}(x))=x$ prove that the following inverse functions are given:
a) $f(x)=2x$ and $f^{-1}(x)=\frac x2$
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
Question ID: 10060050010
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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
Question ID: 10060050010
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b) $f(x)=\frac14x$ and $f^{-1}(x)=4x$
a
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
Question ID: 10060050020
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Answer by: ada
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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
Question ID: 10060050020
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c) $f(x)=4x-1$ and $f^{-1}(x)=\frac{x+1}{4}$
a
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
Question ID: 10060050030
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Answer by: ada
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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) $f(x)=\sqrt{5x}+4$ and $f^{-1}(x)=\frac{{(x-4)}^2}{5}$
a
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
Question ID: 10060050040
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Answer by: ada
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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
Question ID: 10060050040
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e) $f(x)=\frac{x+4}{2x-5}$ and $f^{-1}(x)=\frac{5x+4}{2x-1}$
a
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
Question ID: 10060050050
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Answer by: ada
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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
Question ID: 10060050050
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Question ID: 10060050050
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Question 1
Find $f(x)$
a) $f(x)=2x$ when $x=3$
b) $f(x)=\frac x5$ when $x=20$
c) $f(x)=x+16$ when $x=12$
d) $f(x)=x-3$ when $x=10$
e) $f(x)=3x+11$ when $x=4$
f) $f(x)=\frac x6-2$ when $x=18$
g) $f(x)=x^2+5$ when $x=2$
h) $f(x)=3-\sqrt x$ when $x=36$
Question 2
Find the inverse function for the following functions
a) $f(x)=x+2$
b) $g(x)=x-5$
c) $h(x)=3x$
d) $p(x)=x\div6$
e) $q(x)=\frac x9$
f) $f(y)=y+t$
g) $g(a)=\frac ab$
h) $h(z)=\sqrt z$
i) $p(x)=x^5$
j) $m(r)=-r$
k) $f(x)=\frac1x$
Question 3
Show that the following are the inverse when $x=4$
a) $f(x)=x^3$ and $f^{-1}(x)=\sqrt[3] x$
b) $f(x)=2+3x$ and $f^{-1}(x)=\frac{x-2}{3}$
c) $f(x)=\frac{2+x}{4}$ and $f^{-1}(x)=4x-2$
d) $g(x)=\sqrt{x+12}-3$ and $g^{-1}(x)={(x+3)}^2-12$
e) $h(x)=\sqrt{x+12}-3$ and $h^{-1}(x)={(x+3)}^2-12$
f) $q(x)=\frac1x$ and $q^{-1}(x)=\frac1x$
g) $p(x)=\frac14x-7$ and $p^{-1}(x)=4(x+7)$
h) $f(x)=\frac{(2x+3)^2}{11}$ and $f^{-1}(x)=\frac{\sqrt{11x}-3}{2}$
Question 4
Find the inverse function for the following functions
a) $f(x)=2x+3$
b) $f(x)=-5x+6$
c) $f(x)=\frac12x-3$
d) $g(x)=\frac15x+\frac23$
e) $f(x)=\frac{x+1}{4}$
f) $n(x)=\frac{x-2}{-5}$
g) $f(a)=\sqrt a+15$
h) $g(y)={(y+3)}^3$
Question 5
By showing that $f(f^{-1}(x))=x$ prove that the following inverse functions are given:
a) $f(x)=2x$ and $f^{-1}(x)=\frac x2$
b) $f(x)=\frac14x$ and $f^{-1}(x)=4x$
c) $f(x)=4x-1$ and $f^{-1}(x)=\frac{x+1}{4}$
d) $f(x)=\sqrt{5x}+4$ and $f^{-1}(x)=\frac{{(x-4)}^2}{5}$
e) $f(x)=\frac{x+4}{2x-5}$ and $f^{-1}(x)=\frac{5x+4}{2x-1}$
Answers
Question 1
a) $f(3)=6$
b) $f(20)=4$
c) $f(12)=28$
d) $f(10)=7$
e) $f(4)=23$
f) $f(18)=1$
g) $f(2)=9$
h) $f(36)=-3$
Question 2
a) $f^{-1}(x)=x-2$
b) $g^{-1}(x)=x+5$
c) $h^{-1}(x)=\frac x3$
d) $p^{-1}(x)=6x$
e) $q^{-1}(x)=9x$
f) $f^{-1}(y)=y-t$
g) $g^{-1}(a)=ab$
h) $h^{-1}(z)=z^2$
i) $p^{-1}(x)=\sqrt[5]x$
j) $m^{-1}(r)=-r$
k) $f^{-1}(x)=\frac1x$
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 $x=4$
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 $x=4$
c)
d)
e)
f)
g)
h)
Question 4
a) $f^{-1}(x)=\frac{x-3}{2}$
b) $f^{-1}(x)=\frac{x-6}{-5}$ or $f^{-1}(x)=\frac{6-x}{5}$
c) $f^{-1}(x)=2(x+3)$
d) $g^{-1}(x)=5(x-\frac23)$
e) $f^{-1}(x)=4x-1$
f) $f^{-1}(x)=-5x+2$
g) $f^{-1}(a)={(a-15)}^2$
h) $g^{-1}(y)=\sqrt[3]{y}-3$
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
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
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
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
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
Question 1
Find $f(x)$
a) $f(x)=2x$ when $x=3$
b) $f(x)=\frac x5$ when $x=20$
c) $f(x)=x+16$ when $x=12$
d) $f(x)=x-3$ when $x=10$
e) $f(x)=3x+11$ when $x=4$
f) $f(x)=\frac x6-2$ when $x=18$
g) $f(x)=x^2+5$ when $x=2$
h) $f(x)=3-\sqrt x$ when $x=36$
Question 2
Find the inverse function for the following functions
a) $f(x)=x+2$
b) $g(x)=x-5$
c) $h(x)=3x$
d) $p(x)=x\div6$
e) $q(x)=\frac x9$
f) $f(y)=y+t$
g) $g(a)=\frac ab$
h) $h(z)=\sqrt z$
i) $p(x)=x^5$
j) $m(r)=-r$
k) $f(x)=\frac1x$
Question 3
Show that the following are the inverse when $x=4$
a) $f(x)=x^3$ and $f^{-1}(x)=\sqrt[3] x$
b) $f(x)=2+3x$ and $f^{-1}(x)=\frac{x-2}{3}$
c) $f(x)=\frac{2+x}{4}$ and $f^{-1}(x)=4x-2$
d) $g(x)=\sqrt{x+12}-3$ and $g^{-1}(x)={(x+3)}^2-12$
e) $h(x)=\sqrt{x+12}-3$ and $h^{-1}(x)={(x+3)}^2-12$
f) $q(x)=\frac1x$ and $q^{-1}(x)=\frac1x$
g) $p(x)=\frac14x-7$ and $p^{-1}(x)=4(x+7)$
h) $f(x)=\frac{(2x+3)^2}{11}$ and $f^{-1}(x)=\frac{\sqrt{11x}-3}{2}$
Question 4
Find the inverse function for the following functions
a) $f(x)=2x+3$
b) $f(x)=-5x+6$
c) $f(x)=\frac12x-3$
d) $g(x)=\frac15x+\frac23$
e) $f(x)=\frac{x+1}{4}$
f) $n(x)=\frac{x-2}{-5}$
g) $f(a)=\sqrt a+15$
h) $g(y)={(y+3)}^3$
Question 5
By showing that $f(f^{-1}(x))=x$ prove that the following inverse functions are given:
a) $f(x)=2x$ and $f^{-1}(x)=\frac x2$
b) $f(x)=\frac14x$ and $f^{-1}(x)=4x$
c) $f(x)=4x-1$ and $f^{-1}(x)=\frac{x+1}{4}$
d) $f(x)=\sqrt{5x}+4$ and $f^{-1}(x)=\frac{{(x-4)}^2}{5}$
e) $f(x)=\frac{x+4}{2x-5}$ and $f^{-1}(x)=\frac{5x+4}{2x-1}$
Answers
Question 1
a) $f(3)=6$
b) $f(20)=4$
c) $f(12)=28$
d) $f(10)=7$
e) $f(4)=23$
f) $f(18)=1$
g) $f(2)=9$
h) $f(36)=-3$
Question 2
a) $f^{-1}(x)=x-2$
b) $g^{-1}(x)=x+5$
c) $h^{-1}(x)=\frac x3$
d) $p^{-1}(x)=6x$
e) $q^{-1}(x)=9x$
f) $f^{-1}(y)=y-t$
g) $g^{-1}(a)=ab$
h) $h^{-1}(z)=z^2$
i) $p^{-1}(x)=\sqrt[5]x$
j) $m^{-1}(r)=-r$
k) $f^{-1}(x)=\frac1x$
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 $x=4$
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 $x=4$
c)
d)
e)
f)
g)
h)
Question 4
a) $f^{-1}(x)=\frac{x-3}{2}$
b) $f^{-1}(x)=\frac{x-6}{-5}$ or $f^{-1}(x)=\frac{6-x}{5}$
c) $f^{-1}(x)=2(x+3)$
d) $g^{-1}(x)=5(x-\frac23)$
e) $f^{-1}(x)=4x-1$
f) $f^{-1}(x)=-5x+2$
g) $f^{-1}(a)={(a-15)}^2$
h) $g^{-1}(y)=\sqrt[3]{y}-3$
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
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
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
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
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
Solutions
Question 1
a) When $x=3$:
$\begin{align*}f(3)&=2\times3\\&=6\end{align*}$
b) When $x=20$:
$\begin{align*}f(20)&=\frac{20}{5}\\&=4\end{align*}$
c) When $x=12$:
$\begin{align*}f(12)&=12+16\\&=28\end{align*}$
d) x = 10: f(10)=10-3=7
e) When x=4: f(4)= 3*4+11 = 12+11 = 23
f) When $x=18$:
$\begin{align*}f(18)&=\frac{18}{6}-2\\&=3-2\\&=1\end{align*}$
g)
h)
Question 2
a) $f(x)=x+2$
The opposite operation to adding $2$ is to minus $2$
Therefore $f^{-1}(x)=x-2$
b) $g(x)=x-5$
The opposite operation to subtracting $5$ is to add $5$
Therefore $g^{-1}(x)=x+5$
c) $h(x)=3x$
The opposite operation to multiplying by $3$ is to divide by $3$
Therefore $h^{-1}(x)=\frac x3$
d) $p(x)=x\div6$
The opposite operation to dividing by $6$ is to multiply by $6$
Therefore $p^{-1}(x)=6x$
e)
f)
g)
h)
i)
j)
k)
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 $x=4$
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 $x=4$
c)
d)
e)
f)
g)
h)
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 $y=x$:
$f^{-1}(x)=\frac{x-3}{2}$
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 $y=x$:
$f^{-1}(x)=\frac{x-6}{-5}\,\,\,\text{or}\,\,\,f^{-1}(x)=\frac{6-x}{5}$
c)
d)
e)
f)
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 $y=a$:
$f^{-1}(a)={(a-15)}^2$
h)
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
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
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
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
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