Chapter 2.6, Problem 38E

To determine

To find: The functions $f\circ g$ , $g\circ f$ , $f\circ f$ and $g\circ g$ , and their domains.

Answer to Problem 38E

The functions are $\left(f\circ g\right)\left(x\right)=x-3$ with domain $\left[3,+\infty \right)$ , $\left(g\circ f\right)\left(x\right)=\sqrt{x^2-3}$ with domain $\left(-\infty ,-\sqrt{3}\right]\cup \left[\sqrt{3},+\infty \right)$ , $\left(f\circ f\right)\left(x\right)=x^4$ with domain $\left(-\infty ,+\infty \right)$ and $\left(f\circ f\right)\left(x\right)=x^4$ with domain $\left[12,+\infty \right)$ .

Explanation of Solution

Given information:

The given functions are $f\left(x\right)=x^2$ and $g\left(x\right)=\sqrt{x-3}$ .

Calculation:

The domain of a function is the set of all numbers for which the given function is defined.

The composite function $$ can be calculated as:

  $\begin{array}{c}\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)\\ =f\left(\sqrt{x-3}\right)\\ ={\left(\sqrt{x-3}\right)}^2\\ =x-3\end{array}$

The domain of the function $g\left(x\right)$ is $\left[3,+\infty \right)$ . So, the domain of the function $$ is $\left[3,+\infty \right)$ .

The composite function $\left(g\circ f\right)$ can be calculated as:

  $\begin{array}{c}\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left(x^2\right)\\ =\sqrt{x^2-3}\end{array}$

The domain of the function $f\left(x\right)$ is $\left(-\infty ,+\infty \right)$ . But the composite function is not defined at $x=\pm \sqrt{3}$ . So, the domain of the function $\left(g\circ f\right)$ is $\left(-\infty ,-\sqrt{3}\right]\cup \left[\sqrt{3},+\infty \right)$ .

The composite function $\left(f\circ f\right)$ can be calculated as:

  $\begin{array}{c}\left(f\circ f\right)\left(x\right)=f\left(f\left(x\right)\right)\\ =f{\left(x^2\right)}^2\\ =x^4\end{array}$

The domain of the function $f\left(x\right)$ is $\left(-\infty ,+\infty \right)$ . So, the domain of the function $\left(f\circ f\right)$ is $\left(-\infty ,+\infty \right)$ .

The composite function $\left(g\circ g\right)$ can be calculated as:

  $\begin{array}{c}\left(g\circ g\right)\left(x\right)=g\left(g\left(x\right)\right)\\ =g\left(\sqrt{x-3}\right)\\ =\sqrt{\sqrt{x-3}-3}\end{array}$

The domain of the function $g\left(x\right)$ is $\left[3,+\infty \right)$ . The composite function is defined for $x\ge 12$ . So, the domain of the function $\left(g\circ g\right)$ is $\left[12,+\infty \right)$ .

Therefore, the functions are $\left(f\circ g\right)\left(x\right)=x-3$ with domain $\left[3,+\infty \right)$ , $\left(g\circ f\right)\left(x\right)=\sqrt{x^2-3}$ with domain $\left(-\infty ,-\sqrt{3}\right]\cup \left[\sqrt{3},+\infty \right)$ , $\left(f\circ f\right)\left(x\right)=x^4$ with domain $\left(-\infty ,+\infty \right)$ and $\left(f\circ f\right)\left(x\right)=x^4$ with domain $\left[12,+\infty \right)$ .