Chapter 2.3, Problem 5QR

To determine

To find:The functions $g$ and $f\circ g$ for $f\left(x\right)=x^2$ and $\left(g\circ f\right)\left(x\right)=\sin x^2$ such that $g\in \left[0,\infty \right)$ .

Answer to Problem 5QR

The function $g$ is $\sin x$ and $f\circ g$ is ${\left(\sin x\right)}^2$ for the given functions.

Explanation of Solution

Given information:The functions are $f\left(x\right)=x^2$ and $\left(g\circ f\right)\left(x\right)=\sin x^2$ .

Calculation:

Simplify the composite function $g\circ f$ .

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

Substitute $\sin x^2$ for $\left(g\circ f\right)\left(x\right)$ in above equation.

  $\begin{array}{c}\left(g\circ f\right)\left(x\right)=g\left(x^2\right)\\ \sin \left(x^2\right)=g\left(x^2\right)\end{array}$

Compare both the sides of equation to get $g\left(x\right)=\sin x$ .So, the value of function $g$ is $\sin x$ .

Simplify the composite function $f\circ g$ .

  $\begin{array}{c}\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)\\ =f\left(\sin x\right)\end{array}$

Substitute $\sin x$ for $x$ in the function $f\left(x\right)=x^2$ .

  $\left(f\circ g\right)\left(x\right)={\left(\sin x\right)}^2$

So, the value of $f\circ g$ is ${\left(\sin x\right)}^2$ .

Therefore, the function $g$ is $\sin x$ and $f\circ g$ is ${\left(\sin x\right)}^2$ for the given functions.