Chapter 1, Problem 41RE

(a)

To determine

To find: The value of the function $\left(f\circ g\right)\left(-1\right)$ .

Answer to Problem 41RE

The value of the function $\left(f\circ g\right)\left(-1\right)$ is $1$ .

Explanation of Solution

Given information: The given functions are $f\left(x\right)=\frac{1}{x}$ and $g\left(x\right)=\frac{1}{\sqrt{x+2}}$ .

Calculation:

The composite function $\left(f\circ g\right)\left(-1\right)$ can be written as:

  $\left(f\circ g\right)\left(-1\right)=f\left(g\left(-1\right)\right)$

Substitute $-1$ for $x$ in the function $g\left(x\right)=\frac{1}{\sqrt{x+2}}$ .

  $\begin{array}{c}g\left(-1\right)=\frac{1}{\sqrt{-1+2}}\\ =1\end{array}$

So, $\left(f\circ g\right)\left(-1\right)=f\left(1\right)$ .

Now, substitute 1 for $x$ in the function $f\left(x\right)=\frac{1}{x}$ .

  $\begin{array}{c}f\left(1\right)=\frac{1}{1}\\ =1\end{array}$

Therefore, the value of the function $\left(f\circ g\right)\left(-1\right)$ is 1.

(b)

To determine

To find: The value of the function $\left(g\circ f\right)\left(2\right)$ .

Answer to Problem 41RE

The value of the function $\left(g\circ f\right)\left(2\right)$ is $\sqrt{\frac{2}{5}}$ .

Explanation of Solution

Given information: The given functions are $f\left(x\right)=\frac{1}{x}$ and $g\left(x\right)=\frac{1}{\sqrt{x+2}}$ .

Calculation:

The composite function $\left(g\circ f\right)\left(2\right)$ can be written as:

  $\left(g\circ f\right)\left(2\right)=g\left(f\left(2\right)\right)$

Substitute 2 for $x$ in the function $f\left(x\right)=\frac{1}{x}$ .

  $f\left(2\right)=\frac{1}{2}$

So, $\left(g\circ f\right)\left(2\right)=g\left(\frac{1}{2}\right)$ .

Now, substitute $\frac{1}{2}$ for $x$ in the function $g\left(x\right)=\frac{1}{\sqrt{x+2}}$ .

  $\begin{array}{c}g\left(\frac{1}{2}\right)=\frac{1}{\sqrt{\frac{1}{2}+2}}\\ =\frac{1}{\sqrt{\frac{5}{2}}}\\ =\sqrt{\frac{2}{5}}\end{array}$

Therefore, the value of the function $\left(g\circ f\right)\left(2\right)$ is $\sqrt{\frac{2}{5}}$ .

(c)

To determine

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

Answer to Problem 41RE

The value of the function $\left(f\circ f\right)\left(x\right)$ is $x$ for $x\ne 0$ .

Explanation of Solution

Given information: The given function is $f\left(x\right)=\frac{1}{x}$ .

Calculation:

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

  $\begin{array}{c}\left(f\circ f\right)\left(x\right)=f\left(f\left(x\right)\right)\\ =f\left(\frac{1}{x}\right)\\ =\frac{1}{\frac{1}{x}}\\ =x\end{array}$

Since, $f\left(x\right)=\frac{1}{x}$ has domain $x\ne 0$ .

Therefore, the value of the function $\left(f\circ f\right)\left(x\right)$ is $x$ for $x\ne 0$ .

(d)

To determine

To find: The value of function $\left(g\circ g\right)\left(x\right)$ .

Answer to Problem 41RE

The value of the function $\left(g\circ g\right)\left(x\right)$ is $\frac{\sqrt[4]{x+2}}{\sqrt{1+2\sqrt{x+2}}}$ .

Explanation of Solution

Given information: The given function is $g\left(x\right)=\frac{1}{\sqrt{x+2}}$ .

Calculation:

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

  $\begin{array}{c}\left(g\circ g\right)\left(x\right)=g\left(g\left(x\right)\right)\\ =g\left(\frac{1}{\sqrt{x+2}}\right)\\ =\frac{1}{\sqrt{\frac{1}{\sqrt{x+2}}+2}}\\ =\frac{\sqrt[4]{x+2}}{\sqrt{1+2\sqrt{x+2}}}\end{array}$

Therefore, the value of the function $\left(g\circ g\right)\left(x\right)$ is $\frac{\sqrt[4]{x+2}}{\sqrt{1+2\sqrt{x+2}}}$ .