EXERCISE SOLUTION-0.4, Pag#48, Calculus Book

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Q.01
Determine whether $f$ and $g$ are inverse functions.
Solution:
(a) $$\begin{aligned} & f(x)=4 x, \quad g(x)=\frac{1}{4} x \\ \\ & f(g(x))=4 g(x)=4 \cdot \frac{1}{4} x=x \\ \\ & g(f(x))=\frac{1}{4} f(x)=\frac{1}{4} \cdot 4 x=x \\ \\ & \Rightarrow \quad f(g(x))=x, g(f(x))=x \end{aligned}$$ $\Rightarrow \quad f$ and $g$ are inverse functions.
(b) $$\begin{aligned} f(x)=3 x+1, \quad g(x) & =3 x-1 \\ \\ f(g(x))=3 g(x)+1 & =3(3 x-1)+1 \\ \\ & =9 x-3+1 \\ \\ & =9 x-2 \neq x \end{aligned}$$ $\Rightarrow f$ and $g$ are not inverse function
(c) $$\begin{aligned} f(x) & =\sqrt[3]{x-2}, \quad g(x)=x^3+2 \\ \\ f(g(x)) & =\sqrt[3]{g(x)-2}=\sqrt[3]{x^3+2-2}=\left(x^3\right)^{1/3}=x \\ \\ g(f(x)) & =[f(x)]^3+2 \\ \\ & =[ \sqrt[3]{x-2}]^3+2=\left[(x-2)^{1 / 3}\right]^3+2 \\ \\ & =x-2+2=x \end{aligned}$$ $\Rightarrow f$ and $g$ are inverse functions.
Q.09
Find a formula for $f^{-1}(x)$.
Solution:
$$\begin{aligned} f(x) & =7 x-6  \\ \\ \quad \text {let}  \quad y=f(x) \Rightarrow x=f^{-1}(y) \\ \\ \therefore \quad  y & =7 x-6 \\ \\ y & +6=7 x \\ \\ 7 x & =y+6 \\ \\ x & =\frac{1}{7}(y+6) \Rightarrow f^{-1}(y)=\frac{1}{7}(y+6) \end{aligned}$$ Interchanging $x$ and $y$ $$f^{-1}(x)=\frac{1}{7}(x+6)$$
Q.10
Find a formula for $f^{-1}(x)$.
Solution:
$$\begin{aligned} f(x) & =\frac{x+1}{x-1} \\ \\ \quad \text {let} \quad y=f(x) \Rightarrow x=f^{-1}(y)  \\ \\  \therefore \quad y & =\frac{x+1}{x-1} \end{aligned}$$ solving for $x$ $$\begin{aligned} & y(x-1)=x+1 \\ \\ & y x-y=x+1 \\ \\ & y x-x=y+1 \\ \\ & x(y-1)=y+1 \\ \\ & x=\frac{y+1}{y-1} \\ \\ & f(y)=\frac{y+1}{y-1} \end{aligned}$$ Interchanging $y$ and $x$ $$f^{-1}(x)=\frac{x+1}{x-1}$$
Q.11
Find a formula for $f^{-1}(x)$.
Solution:
$$f(x)=3 x^3-5$$  $$ \quad \text {let} \quad y=f(x) \Rightarrow x=f^{-1}(y) $$ $$ \therefore \quad  y=3 x^3-5  $$ Solving for $x$ $$\begin{aligned} & 3 x^3=y+5 \\ \\ & x^3=\frac{1}{3}(y+5) \\ \\ & x=\left[\frac{y+5}{3}\right]^{1 / 3} \\ \\ & x=\sqrt[3]{\frac{y+5}{3}} \\ \\ & f^{-1}(y)=\sqrt[3]{\frac{y+5}{3}} \end{aligned}$$ Interchanging $x$ and $y$ $$f^{-1}(x)=\sqrt[3]{\frac{x+5}{3}}$$
Q.12
Find a formula for $f^{-1}(x)$.
Solution:
$$\begin{aligned} f(x) & =\sqrt[5]{4 x+2}  \\ \\ \quad \text {let} \quad y=f(x) \Rightarrow x=f^{-1}(y) \\ \\ \therefore \quad y & =\sqrt[5]{4 x+2} \end{aligned}$$ Solving for $x$ $$(4 x+2)^{1 / 5}=y$$ $$4 x+2=y^5$$ $$\begin{aligned} & 4 x=y^5-2 \\ \\ & x=\frac{1}{4}\left(y^5-2\right) \\ \\ & f^{-1}(y)=\frac{1}{4}\left(y^5-2\right) \end{aligned}$$ Interchanging $x$ and $y$ $$f^{-1}(x)=\frac{1}{4}\left(x^5-2\right) \text {. }$$
Q.13
Find a formula for $f^{-1}(x)$.
Solution:
$$\begin{aligned} & f(x)=3 / x^2, \quad x<0  \\ \\ \quad \text {let} \quad y=f(x) \Rightarrow x=f^{-1}(y) \\ \\ &\therefore \quad  y=3 / x^2 \end{aligned}$$ Solving for $x$ $$\begin{aligned} & x^2=3 / y \\ \\ & x=-\sqrt{3 / y} \\ \\ & f^{-1}(y)=-\sqrt{3 / y} \end{aligned}$$ negative sign because $x<0$. Interchanging $x$ by $y$ $$f(x)=-\sqrt{3} / x$$
Q.14
Solution:
$$\begin{aligned} f(x) & =5 /\left(x^2+1\right), \quad x \geqslant 0 \\ \\ \quad \text {let} \quad y=f(x) \Rightarrow x=f^{-1}(y) \\ \\ \therefore \quad y & =\frac{5}{x^2+1} \end{aligned}$$ Solving for  $x$ we get $$\begin{aligned} x^2+1 & =\frac{5}{y} \\ \\ x^2 & =\frac{5}{y}-1 \\ \\ x^2 & =\frac{5-y}{y} \\ \\ x & =\sqrt{\frac{5-y}{y}} \\ \\ f^{-1}(y) & =\sqrt{\frac{5-y}{y}} \end{aligned}$$ Interchanging $x$ and $y$ we get $$f^{-1}(x)=\sqrt{\frac{5-x}{x}}$$
Q.15
Solution:
$f(x)= \begin{cases}\frac{5}{2}-x, & x<2 \\ \\ \frac{1}{x}, & x \geqslant 2\end{cases}$ 
It will be solved for the domains $x<2$ and $x \geqslant 2$ separately.
  $$ \quad \text {let} \quad y=f(x) \Rightarrow x=f^{-1}(y) $$ $$ \therefore \quad y=\frac{5}{2}-x, \quad x<2 .$$ $$\begin{aligned}  x=\frac{5}{2}-y,  x & <2 \\ \\ \text {Then Range }: y > \frac{1}{2} \\ \\ \text {solving for} x \\ \\ x=\frac{5}{2}-y, \quad y> \frac{1}{2} —–(1)\end{aligned}$$
and 
$$ y=\frac{1}{x}, \quad x \geqslant 2 $$
Range: $0<y \geqslant \frac{1}{2} $
solving for $ x $
$$x=\frac{1}{y}, \quad 0<y \geqslant \frac{1}{2}—-(2) \quad $$ because of Range of $f$= domain of $f^{-1} $
From (1) and (2) we have 
$x= \begin{cases}\frac{5}{2}-y, & y> \frac{1}{2} \\ \\ \frac{1}{y}, & 0<y \leq \frac{1}{ 2} \end{cases}$
$f^{-1}(y)= \begin{cases}\frac{5}{2}-y, & y> \frac{1}{2} \\ \\ \frac{1}{y}, & 0<y \leq \frac{1}{ 2} \end{cases}$
Interchanging $x$ and $y$ 
$f^{-1}(x)= \begin{cases}\frac{5}{2}-x, & x> \frac{1}{2} \\ \\ \frac{1}{x}, & 0<x \leq \frac{1}{ 2} \end{cases}$