Chapter 1.5, Problem 58E

(a)

To determine

To calculate: To find $f\circ g$ and $g\circ f$ and to find domain of $f\circ g$

Answer to Problem 58E

  $\left(f\circ g\right)\left(x\right)=x^{\frac{1}{4}}$ and domain is $\left[0,\infty \right)$

  $\left(g\circ f\right)\left(x\right)=x^{\frac{1}{4}}$

Explanation of Solution

Given: $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=\sqrt{x}$

Formula Used:

Composition of functions is when one function is inside of another function.

The notation used for the composition of functions looks like this, $\left(f\circ g\right)\left(x\right)$ .

The composition of the function f with g is defined as follows:

  $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$ , notice that in the case the function g is inside of the function

Also, Domain of a function is the set of all possible inputs for the function

Calculation:

Given function as follows:

  $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=\sqrt{x}$

  $f\circ g$ is calculated as follows:

  $\begin{array}{l}\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)\\ \left(f\circ g\right)\left(x\right)=f\left(\sqrt{x}\right)\\ \left(f\circ g\right)\left(x\right)=\sqrt{\sqrt{x}}\\ \left(f\circ g\right)\left(x\right)=x^{\frac{1}{4}}\end{array}$

For $\left(f\circ g\right)\left(x\right)$ to be defined

  $\begin{array}{l}{x}^{\frac{1}{4}}\ge 0\\ x\ge 0\end{array}$

Thus, domain is $\left[0,\infty \right)$

  $g\circ f$ is calculated as follows:

  $\begin{array}{l}\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)\\ \left(g\circ f\right)\left(x\right)=g\left(\sqrt{x}\right)\\ \left(g\circ f\right)\left(x\right)=\sqrt{\sqrt{x}}\\ \left(g\circ f\right)\left(x\right)=x^{\frac{1}{4}}\end{array}$

Conclusion:

Hence, $\left(f\circ g\right)\left(x\right)=x^{\frac{1}{4}}$ and domain is $\left[0,\infty \right)$

  $\left(g\circ f\right)\left(x\right)=x^{\frac{1}{4}}$

(b)

To determine

To calculate: To graph $f\circ g$ , $g\circ f$ and to find $f\circ g=g\circ f$

Answer to Problem 58E

  $f\circ g$ is equal to $g\circ f$

Explanation of Solution

Given: $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=\sqrt{x}$

Calculation:

From part (a).

  $\left(f\circ g\right)\left(x\right)=x^{\frac{1}{4}}$ and $\left(g\circ f\right)\left(x\right)=x^{\frac{1}{4}}$

Plotting the graph of $f\circ g$ and $g\circ f$ using graphing utility:

  

From the above graph, $f\circ g$ is a curve and $g\circ f$ is line

Thus, $f\circ g=g\circ f$

Conclusion:

Hence, $f\circ g=g\circ f$