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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