Inverse Function : Problems and Solutions

1.       Given that $\displaystyle f(x)={{e}^{{x+3}}}$ where $\displaystyle x\in R$, find $\displaystyle {f}^{-1}(x)$ and state the domain of $\displaystyle {f}^{-1}$. Hence solve the equation $\displaystyle {f}^{-1}(x)= \ln \left( {\frac{1}{x}} \right)$.

Show/Hide Solution
$\displaystyle \begin{array}{l}f(x)={{e}^{{x+3}}},\ x\in R\\\\{{f}^{{-1}}}(y)=x\Leftrightarrow f(x)=y\\\\\therefore {{e}^{{x+3}}}=y\\\\\therefore x+3={{\log }_{e}}y\\\\\therefore x+3=\ln y\\\\\therefore x=\ln y-3\\\\\therefore {{f}^{{-1}}}(y)=\ln y-3\\\\\therefore {{f}^{{-1}}}(x)=\ln x-3\\\\\text{Domain of }{{f}^{{-1}}}=\{x|x>0,\ x\in R\}\\\\{{f}^{{-1}}}(x)=\ln \left( {\frac{1}{x}} \right)\\\\\ln x-3=\ln \left( {\frac{1}{x}} \right)\\\\\ln x-\ln \left( {\frac{1}{x}} \right)=3\\\\\ln {{x}^{2}}=3\\\\2\ln x=3\\\\\ln x=\frac{3}{2}\\\\x={{e}^{{\frac{3}{2}}}}\end{array}$

2.        A function f is defined by $\displaystyle f(3x-2) = 5+6x$. Find the value of $\displaystyle {f}^{-1}(29)$.

Show/Hide Solution
$\displaystyle \begin{array}{l}f(3x-2)=5+6x\\\\\therefore {{f}^{{-1}}}(5+6x)=3x-2\\\\\text{Let}\ 5+6x=29,\text{then}\\\\6x=24\Rightarrow x=4\\\\\therefore {{f}^{{-1}}}(29)=3(4)-2=10\end{array}$

3.        A function f is defined by $\displaystyle f(x)=\frac{{x-3}}{{2x-5}}$.

(i) State the value of $\displaystyle x$ for which $\displaystyle f$ is not defined.

(ii) Find the value of $\displaystyle x$ for which $\displaystyle f(x) = 0$.

(iii) Find the inverse function$\displaystyle {f}^{-1}$ and state the domain of $\displaystyle {f}^{-1}$.

Show/Hide Solution
$\displaystyle \ \ \ \ \ \ \ \ f(x)=\frac{{x-3}}{{2x-5}}$

$\displaystyle \text{(i)}\ \ \ \ f\ \text{is not defined when}$

$\displaystyle \ \ \ \ \ \ \ 2x-5=0\Rightarrow x=\frac{5}{2}$

$\displaystyle \text{(ii)}\ \ \ f(x)=0$

$\displaystyle \ \ \ \ \ \ \ \frac{{x-3}}{{2x-5}}=0$

$\displaystyle \ \ \ \ \ \ \ \text{Since }2x-5\ne 0,$

$\displaystyle \ \ \ \ \ \ \ x-3=0\Rightarrow x=3$

$\displaystyle \text{(iii)}\ \ \text{Let }{{f}^{{-1}}}(x)=y,\ \text{then}$

$\displaystyle \ \ \ \ \ \ \ \ f(y)=x$

$\displaystyle \ \ \ \ \ \ \ \frac{{y-3}}{{2y-5}}=x$

$\displaystyle \begin{array}{l}\ \ \ \ \ \ \ y-3=2xy-5x\\\\\ \ \ \ \ \ \ y-2xy=3-5x\\\\\ \ \ \ \ \ \ y(1-2x)=3-5x\end{array}$

$\displaystyle \ \ \ \ \ \ \ y=\frac{{3-5x}}{{1-2x}},\ x\ne \frac{1}{2}$

$\displaystyle \ \ \ \ \ \ \ \text{Domain of }{{f}^{{-1}}}=\{x|x\in R,x\ne \frac{1}{2}\}$

4.       A function $\displaystyle f$ is defined by $\displaystyle f:x\mapsto \frac{a}{x}+1,\ x\ne 0$ where $\displaystyle a$ is a constant. Given that $\displaystyle 6( f \cdot f )(-1) +{f}^{-1}(2) = 0$, find the possible values of $\displaystyle a$.

Show/Hide Solution
$\displaystyle f(x)=\frac{a}{x}+1,\ x\ne 0$

$\displaystyle (f\cdot f)(-1)=f\left( {f(-1)} \right)$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {\frac{a}{{-1}}+1} \right)$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {1-a} \right)$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{a}{{1-a}}+1$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{{1-a}}$

$\displaystyle {{f}^{{-1}}}(2)=k\Leftrightarrow f(k)=2$

$\displaystyle \therefore \frac{a}{k}+1=2$

$\displaystyle \ \ \ \frac{a}{k}=1\Rightarrow k=a$

$\displaystyle \therefore {{f}^{{-1}}}(2)=a$

$\displaystyle 6(f\cdot f)(-1)+{{f}^{{-1}}}(2)=0$

$\displaystyle \therefore \frac{6}{{1-a}}+a=0$

$\displaystyle \ \ \ 6+a-{{a}^{2}}=0$

$\displaystyle \therefore (3-a)(2+a)=0$

$\displaystyle \therefore a=3\ \text{or}\ a=-2$

5.       A function $\displaystyle g$ is defined by $\displaystyle g:x\mapsto \frac{{x+1}}{{x-2}},\ x\ne 2,x\ne 5$ and $\displaystyle h$ is defined by is defined by $\displaystyle h:x\mapsto \frac{{ax+3}}{{x}},\ x\ne 0$. Given that $\displaystyle (h\cdot {g}^{–1})(4) = 6$, calculate the value of $\displaystyle a$.

Show/Hide Solution
$\displaystyle \ \ \ \ \ \ g(x)=\frac{{x+1}}{{x-2}},\ x\ne 2,x\ne 5$

$\displaystyle \ \ \ \ \ \ h(x)=\frac{{ax+3}}{x},\ x\ne 0$

$\displaystyle \begin{array}{l}\ \ \ \ \ \ \left( {h\cdot {{g}^{{-1}}}} \right)(4)=6\\\\\ \ \ \ \ \ {{g}^{{-1}}}(4)=p\Leftrightarrow g(p)=4\end{array}$

$\displaystyle \therefore \ \ \ \ \frac{{p+1}}{{p-2}}=4$

$\displaystyle \begin{array}{l}\ \ \ \ \ \ p+1=4p-8\\\\\ \ \ \ \ \ 3p=9\Rightarrow p=3\\\\\therefore \ \ \ \ {{g}^{{-1}}}(4)=3\\\\\ \ \ \ \ \ \left( {h\cdot {{g}^{{-1}}}} \right)(4)=6\\\\\ \ \ \ \ \ h\left( {{{g}^{{-1}}}(4)} \right)=6\\\\\ \ \ \ \ \ h(3)=6\end{array}$

$\displaystyle \ \ \ \ \ \ \frac{{a(3)+3}}{3}=6$

$\displaystyle \therefore \ \ \ \ a+1=6\Rightarrow a=5$

6.       Let $\displaystyle f:R\to R$ and $\displaystyle g:R\to R$ be defined by $\displaystyle f(x) = 3x - 1$ and $\displaystyle g(x) = x + 7$. Find $\displaystyle ({f}^{-1}\cdot g)(x)$ and what is the value of $\displaystyle a\in R$ for which $\displaystyle ({f}^{-1}\cdot g)(a)=3$.

Show/Hide Solution
$\displaystyle \begin{array}{l}f:R\to R,\ f(x)=3x-1\\\\g:R\to R,\ g(x)=x+7\\\\({{f}^{{-1}}}\cdot g)(x)={{f}^{{-1}}}\left( {g(x)} \right)\\\\\text{Let }{{f}^{{-1}}}\left( {g(x)} \right)=y\ \text{then }g(x)=f(y).\\\\\therefore x+7=3y-1\end{array}$

$\displaystyle \ \ \ y=\frac{{x+8}}{3}$

$\displaystyle \therefore ({{f}^{{-1}}}\cdot g)(x)=\frac{{x+8}}{3}$

$\displaystyle \ \ \ ({{f}^{{-1}}}\cdot g)(a)=3$

$\displaystyle \ \ \ \frac{{a+8}}{3}=3\Rightarrow a=1$

7.       For the function $\displaystyle f(x)=\frac{{2x}}{{3x+1}},\ x\ne -\frac{1}{3}$ find $\displaystyle {f}^{-1}$ and verify that $\displaystyle (f\cdot {f}^{-1})$ and $\displaystyle ({f}^{-1}\cdot f)$ both equal $\displaystyle I$.

Show/Hide Solution
$\displaystyle \ \ \ \ f(x)=\frac{{2x}}{{3x+1}},\ x\ne -\frac{1}{3}$

$\displaystyle \ \ \ \ {{f}^{{-1}}}(x)=y\Leftrightarrow f(y)=x$ $\displaystyle \ \ \ \ \frac{{2y}}{{3y+1}}=x$

$\displaystyle \begin{array}{l}\ \ \ \ 2y=3xy+x\\\\\ \ \ \ y(2-3x)=x\end{array}$

$\displaystyle \ \ \ \ y=\frac{x}{{2-3x}}$

$\displaystyle \therefore \ \ {{f}^{{-1}}}(x)=\frac{x}{{2-3x}},\ x\ne \frac{2}{3}$

$\displaystyle \therefore \ \ (f\cdot {{f}^{{-1}}})(x)=f\left( {{{f}^{{-1}}}(x)} \right)$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {\frac{x}{{2-3x}}} \right)$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\frac{{2x}}{{2-3x}}}}{{\frac{{3x}}{{2-3x}}+1}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2x}}{{2-3x}}\times \frac{{2-3x}}{{3x+2-3x}}$

$\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =I(x)\\\\\therefore \ \ ({{f}^{{-1}}}\cdot f)(x)={{f}^{{-1}}}\left( {(x)} \right)\end{array}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{f}^{{-1}}}\left( {\frac{{2x}}{{3x+1}}} \right)$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\frac{{2x}}{{3x+1}}}}{{2-\frac{{6x}}{{3x+1}}}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2x}}{{3x+1}}\times \frac{{3x+1}}{{6x+2-6x}}$

$\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =I(x)\\\\\therefore \ \ (f\cdot {{f}^{{-1}}})(x)=({{f}^{{-1}}}\cdot f)(x)=I(x)\end{array}$

8.       Functions $\displaystyle f$ and $\displaystyle g$ are defined, for $\displaystyle x\in R$, by $\displaystyle f : x\mapsto 5x - 2, g:x\mapsto \frac{1}{2x-1},\ x\ne \frac{1}{2}$. Find the value of $\displaystyle x$ for which

Show/Hide Solution
$\displaystyle \text{(i)}\ f(x)={f}^{-1}(x)$.

$\displaystyle \text{(ii)}\ (f\cdot g)(x)+3g(x)=0$.

$\displaystyle \ \ \ \ \ \ \ \ \ \ f(x)=5x-2,$

$\displaystyle \ \ \ \ \ \ \ \ \ \ g(x)=\frac{1}{{2x-1}},\ x\ne \frac{1}{2}$

$\displaystyle \begin{array}{l}\text{(i)}\ \ \ \ \ \ \ f(x)={{f}^{{-1}}}(x)\\\\\ \ \ \ \ \therefore \ \ \ f\left( {f(x)} \right)=x\\\\\ \ \ \ \ \therefore \ \ \ f\left( {5x-2} \right)=x\\\\\ \ \ \ \ \therefore \ \ \ 5(5x-2)-2=x\\\\\ \ \ \ \ \therefore \ \ \ 24x=12\end{array}$

$\displaystyle \ \ \ \ \ \therefore \ \ \ x=\frac{1}{2}$

$\displaystyle \begin{array}{l}\text{(ii)}\ \ \ \ \ \ (f\cdot g)(x)+3g(x)=0\\\\\ \ \ \ \ \ \ \ \ \ f\left( {g(x)} \right)+3g(x)=0\end{array}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ f\left( {\frac{1}{{2x-1}}} \right)+\frac{3}{{2x-1}}=0$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \frac{5}{{2x-1}}-2+\frac{3}{{2x-1}}=0$

$\displaystyle \ \ \ \ \ \therefore \ \ \ \frac{8}{{2x-1}}=2$

$\displaystyle \ \ \ \ \ \therefore \ \ \ 2x-1=4$

$\displaystyle \ \ \ \ \ \therefore \ \ \ x=\frac{5}{2}$

မဂၤလာပါ