EXERCISE SOLUTION-0.2, Pag#24, Calculus Book

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Q.32
Solution: 
$$f(x)=\sqrt{x-3}, g(x)=\sqrt{x^{2}+3}$$
Domain of $f$ consists of all real numbers $x$ such that $x-3 \geq 0 \Rightarrow x \geq 3$ or $[3, \infty)$
Range of $f:[0, \infty)$
Domain of $g$ : All real numbers or $(-\infty, \infty)$
$$(f \circ g)(x)=f(g(x))$$
$$=f\left(\sqrt{x^{2}+3}\right)=\sqrt{\sqrt{x^{2}+3}-3}$$
$$(f \circ g)(x)=\sqrt{\sqrt{x^{2}+3}-3}$$
Domain of $f$ og: for all $x \geq \sqrt{6}$ and $x \leq-\sqrt{6}$, the values of $g(x)$ lies in the domain of $f$. Thus the domain of $f \circ g$ is $|x| \geq \sqrt{6}$
Composition of functions:
$$(g \circ f)(x)=g(f(x))=g(\sqrt{x-3})=$$
$$\sqrt{(\sqrt{x-3})^{2}+3}=\sqrt{x-3+3}=\sqrt{x}$$
Domain of $g o f$ : For all $x \geq 3$, the values of $f$ lies in the domain of $g$. Hence the domain of $g o f$ is $x \geq 3$
Q.33
Solution: 
$$f(x)=\frac{1+x}{1-x}, \quad g(x)=\frac{x}{1-x}$$
Domain of $f$ and $g$ consist of all real number $x$, except $x=1$. At $x=1$ the denominator becomes zero.
Range of $f$ and $g$ consist of all real numbers.
Composition of functions:
$$(f \circ g)(x)=f(g(x))=f\left(\frac{x}{1-x}\right)=\frac{1+\frac{x}{1-x}}{1-\frac{x}{1-x}}$$
$$=\frac{\frac{1-x+x}{1-x}}{\frac{1-x-x}{1-x}}=\frac{1}{1-2 x}$$
Domain of $fog$: $x=1$ is not part of domain of $g$ so it can not be in the domain of fog. Also $x=\frac{1}{2}$ lies in the domain of $g$ and $g\left(\frac{1}{2}\right)=1$ does not belong to domain of $f$. Hence domain of $f \circ g$ consists of all real numbers $x$ except $x=\frac{1}{2}, 1$.
$$(g \circ f)(x)=g(f(x))=g\left(\frac{1+x}{1-x}\right)=\frac{\frac{1+x}{1-x}}{1-\frac{1+x}{1-x}}$$
$$=\frac{\frac{1+x}{1-x}}{\frac{1-x-1-x}{1-x}}=\frac{1+x}{-2 x}=-\frac{1}{2 x}-\frac{1}{2}$$
Domain of $g o f: x=1$ is not part of domain of $f$ so it can not be in the domain of $g \circ f$. Also $x=0$ lies in the domain of $f$ and $f(0)=1$ does not belong to domain of $g$. Hence domain of $g o f$ consists of all real numbers $x$ except $x=0$ and $1$.
Q.34
Solution: $$\quad f(x)=\frac{x}{1+x^{2}}, \quad g(x)=\frac{1}{x}$$
Domain of $f$ : All real numbers $x.$
Range of $f$ : All real numbers $x.$
Domain of $g$ : All real numbers $x$ except $x=0, \quad$ Range of $g$ : All real numbers $x.$
Composition of functions:
$$ (f \circ g)(x)=f(g(x))=f\left(\frac{1}{x}\right)=\frac{\frac{1}{x}}{1+\frac{1}{x^{2}}}=\frac{1 / x}{\frac{x^{2}+1}{x^{2}}}=\frac{x}{x^{2}+1} $$ Domain of $fog$: All real numbers $x$ except $x=0$ or $x \neq 0$ $$ (g \circ f)(x)=g(f(x))=g\left(\frac{x}{1+x^{2}}\right)=\frac{1}{\frac{x}{1+x^{2}}}=\frac{1+x^{2}}{x}=\frac{1}{x}+x $$ Domain of $gof$: for $x=0, f(0)=0$ does not belong to domain of $g$. hence domain of $g$ of is $x \neq$ 0 . 
Q.35-36 Find a formula for $f \circ g \circ h$.
Solution:
(35)
$$f(x)=x^{2}+1, \quad g(x)=\frac{1}{x}, \quad h(x)=x^{3}$$ $$g(h(x))=\frac{1}{h(x)}=\frac{1}{x^{3}}$$ $$f(g(h(x)))=(g(h(x)))^{2}+1$$ $$f(g(h(x)))=\left(\frac{1}{x^{3}}\right)^{2}+1=\frac{1}{x^{6}}+1$$
(36)
$$\quad f(x)=\frac{1}{x+1}, \quad g(x)=\sqrt[3]{x}, \quad h(x)=\frac{1}{x^{3}}$$ $$g(h(x))=\sqrt[3]{h(x)}=\left(\frac{1}{x^{3}}\right)^{\frac{1}{3}}=\frac{1}{x}$$ $$f(g(h(x)))=\frac{1}{g(h(x))+1}=\frac{1}{1 / x+1}=\frac{x}{x+1}$$

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