Chapter 1, Problem 43RE

(a)

To determine

To find: The formula for the functions $\left(f\circ g\right)\left(x\right)$ and $\left(g\circ f\right)\left(x\right)$ .

Answer to Problem 43RE

The formula for the function $\left(f\circ g\right)\left(x\right)$ is $-x$ , $x\ge -2$ and the formula for the function $\left(g\circ f\right)\left(x\right)$ is $\sqrt{4-x^2}$ .

Explanation of Solution

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

Calculation:

The formula for the function $\left(f\circ g\right)\left(x\right)$ can be calculated as:

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

Substitute $\sqrt{x+2}$ for $x$ in the function $f\left(x\right)=2-x^2$ .

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

Since, the domain of the function $g\left(x\right)=\sqrt{x+2}$ . So, the formula for the function $\left(f\circ g\right)\left(x\right)$ is $-x$ , such that $x\ge -2$ .

The formula for the function $\left(g\circ f\right)\left(x\right)$ can be calculated as:

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

Substitute $2-x^2$ for $x$ in the function $g\left(x\right)=\sqrt{x+2}$ .

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

Therefore, the formula for the function $\left(f\circ g\right)\left(x\right)$ is $-x$ , $x\ge -2$ and the formula for the function $\left(g\circ f\right)\left(x\right)$ is $\sqrt{4-x^2}$ .

(b)

To determine

To find: The domain of the function of $\left(f\circ g\right)\left(x\right)$ and $\left(g\circ f\right)\left(x\right)$ .

Answer to Problem 43RE

The domain of the function $\left(f\circ g\right)\left(x\right)$ is $\left[-2,\infty \right)$ and the domain of the function $\left(g\circ f\right)\left(x\right)$ is $\left[-2,2\right]$ .

Explanation of Solution

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

Calculation:

From part (a), the formula of the function $\left(f\circ g\right)\left(x\right)$ is $-x$ , $x\ge -2$ and the formula of the function $\left(g\circ f\right)\left(x\right)$ is $\sqrt{4-x^2}$ .

The domain of the function $\left(f\circ g\right)\left(x\right)=-x$ is $\left[-2,\infty \right)$ .

The domain of the function $\left(g\circ f\right)\left(x\right)$ is:

  $\begin{array}{c}4-x^2\ge 0\\ -x^2\ge -4\\ x\ge -2\text{or}x\le 2\end{array}$

Therefore, the domain of the function $\left(f\circ g\right)\left(x\right)$ is $\left[-2,\infty \right)$ and the domain of the function $\left(g\circ f\right)\left(x\right)$ is $\left[-2,2\right]$ .

(c)

To determine

To find: The range of the function of $\left(f\circ g\right)\left(x\right)$ and $\left(g\circ f\right)\left(x\right)$ .

Answer to Problem 43RE

The range of the function $\left(f\circ g\right)\left(x\right)$ is $\left(-\infty ,2\right]$ and the range of the function $\left(g\circ f\right)\left(x\right)$ is $\left[0,2\right]$ .

Explanation of Solution

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

Calculation:

From part (a), the formula of the function $\left(f\circ g\right)\left(x\right)$ is $-x$ , $x\ge -2$ and the formula of the function $\left(g\circ f\right)\left(x\right)$ is $\sqrt{4-x^2}$ .

The range is the set of all output values of the function such that the given function is defined.

The range of the function $\left(f\circ g\right)\left(x\right)=-x$ is $\left(-\infty ,2\right]$ . The range of the function $\left(g\circ f\right)\left(x\right)=\sqrt{4-x^2}$ is $\left[0,2\right]$ .

Therefore, the range of the function $\left(f\circ g\right)\left(x\right)$ is $\left(-\infty ,2\right]$ and the range of the function $\left(g\circ f\right)\left(x\right)$ is $\left[0,2\right]$ .