Chapter 2.6, Problem 35E

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 35E

The functions are $\left(f\circ g\right)\left(x\right)=x^2+2x+1$ with domain $\left(-\infty ,+\infty \right)$ , $\left(g\circ f\right)\left(x\right)=x^2+1$ with domain $\left(-\infty ,+\infty \right)$ , $\left(f\circ f\right)\left(x\right)=x^2$ with domain $\left(-\infty ,+\infty \right)$ and $\left(g\circ g\right)\left(x\right)=x+2$ with domain $\left(-\infty ,+\infty \right)$ .

Explanation of Solution

Given information:

The given functions are $f\left(x\right)=x^2$ and $g\left(x\right)=x+1$ .

Calculation:

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

The composite function $$ can be calculated as:

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

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

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

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

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

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

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

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:

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

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

Therefore, the functions are $\left(f\circ g\right)\left(x\right)=x^2+2x+1$ with domain $\left(-\infty ,+\infty \right)$ , $\left(g\circ f\right)\left(x\right)=x^2+1$ with domain $\left(-\infty ,+\infty \right)$ , $\left(f\circ f\right)\left(x\right)=x^2$ with domain $\left(-\infty ,+\infty \right)$ and $\left(g\circ g\right)\left(x\right)=x+2$ with domain $\left(-\infty ,+\infty \right)$ .