Chapter 1.4, Problem 51E

Solution

To verify : The function $f,g\text{and}\text{a}\text{domain}D$ so that $f^{0}g\left(x\right)=x^2$ with the domain $D$ .

  

  

Given information:

The functions are $f\left(x\right)$ and $g\left(x.\right)$ and domain are given

  

  

Concept used:

Domain is the set of independent value..

Let $f$ and $g$ be two functions such that the domain of $f$ intersects the range of $g$ . The

composition $f$ of $g$ , denoted , is defined by the rule

  $f^{0}g\left(x\right)=fg\left(x\right)$ .

Calculations:

Case $1.$

Consider the function $f=e^x,g=\sqrt{2-x}$ .

Then

  $\begin{array}{c}{f}^{o}g\left(x\right)=f(g(x))\\ =e^{\sqrt{2-x}}\end{array}$

domain will be $(-\infty ,2].$

Case $2.$

Consider the function $f={\left(x^2+2\right)}^2,g=\left(x+1\right)$ .

Then

  $\begin{array}{c}{f}^{o}g\left(x\right)=f(g(x))\\ ={\left({\left(x+1\right)}^2+2\right)}^2\end{array}$

So domain will be $\left(-\infty ,\infty \right).$

Case $3.$

Consider the function $f={\left(x^2-2\right)}^2,g=2\ln x$ .

Then

  $\begin{array}{c}{f}^{o}g\left(x\right)=f(g(x))\\ ={\left(4{\left(\ln x\right)}^2-2\right)}^2\end{array}$

domain will be $\left(0,\infty \right).$

Case $4.$

Consider the function $f=\frac{1}{{\left(x-1\right)}^2},g=\frac{1}{\left(x-1\right)}$ .

Then

  $\begin{array}{c}{f}^{o}g\left(x\right)=f(g(x))\\ =\frac{1}{{\left(\frac{1}{x-1}-1\right)}^2}\\ =\frac{1}{{\left(\frac{2-x}{x-1}\right)}^2}\end{array}$

The domain will be $x<1,2<x<1\text{or}\text{x>2}\text{.}$

Case $5.$

Consider the function $f=x^2-2x+1,g=\sqrt{x-2}$ .

Then

  $\begin{array}{c}{f}^{o}g\left(x\right)=f(g(x))\\ ={\left(\sqrt{x-2}\right)}^2-2\left(\sqrt{x-2}\right)+1\\ =\left(x-1\right)-2\left(\sqrt{x-2}\right)\end{array}$

So domain will be $[2,\infty ).$

Case $6.$

Consider the function $f={\left(\frac{x+1}{x}\right)}^2,g=\frac{x+1}{x}$ .

Then

  $\begin{array}{c}{f}^{o}g\left(x\right)=f(g(x))\\ ={\left(\frac{\frac{x+1}{x}+1}{\frac{x+1}{x}}\right)}^2\\ ={\left(\left(\frac{2x+1}{x}\right)\left(\frac{x}{x+1}\right)\right)}^2\\ ={\left(\frac{2x+1}{x+1}\right)}^2\end{array}$

So domain will be $x\ne -1.$