Chapter 5.1, Problem 40AYU

In Problems 23-38, for the given functions $f$ and $g$, find:

a. $f\circ g$

b. $g\circ f$

c. $f\circ f$

d. $g\circ g$

State the domain of each composite function.

$f(x)=\sqrt{x-2}$; $g(x)=1-2x$

To determine

To find:

a. $f\circ g$ for the given functions $f$ and $g$, State the domain of each composite function.

$f(x)=\sqrt{x\text{ - 2}}$; $g(x)=\text{ 1 - }2x$

Answer to Problem 40AYU

Solution:

a. The domain of $\left(f\circ g\right)\left(x\right)$ is $x\le \frac{1}{2}$.

Explanation of Solution

Given:

$f(x)=\sqrt{x\text{ - 2}}$; $g(x)=\text{ 1 - }2x$

Calculation:

For $\left(f\circ g\right)\left(x\right)$:

Step 1: Find the domain of the inside function $g\left(x\right)$. i.e., $x\le \frac{1}{2}$.

Step 2: The composite function $\left(f\circ g\right)\left(x\right)=\sqrt{-2x-\text{ 1}}$. Its domain also is $x\ge -\frac{1}{2}$.

This function puts no additional restrictions on the domain, so the composite domain is $x\le \frac{1}{2}$.

This function has a more restrictive domain than $f\left(x\right)$, so the composite domain is $x\le \frac{1}{2}$.

We know the composite function of $f$ and $g$ is defined as

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

$\left(f\circ g\right)\left(x\right)=f\left(1-\text{ 2}x\right)$

$\left(f\circ g\right)\left(x\right)=\sqrt{1-2x-\text{ 2}}$

$\left(f\circ g\right)\left(x\right)=\sqrt{-2x-\text{ 1}}$

To determine

To find:

b. $g\circ f$ for the given functions $f$ and $g$, State the domain of each composite function.

$f(x)=\sqrt{x\text{ - 2}}$; $g(x)=\text{ 1 - }2x$

Answer to Problem 40AYU

Solution:

b. The domain of $\left(g\circ f\right)\left(x\right)$ is $x\ge \text{ 2}$.

Explanation of Solution

Given:

$f(x)=\sqrt{x\text{ - 2}}$; $g(x)=\text{ 1 - }2x$

Calculation:

For $\left(g\circ f\right)\left(x\right)$:

Step 1: Find the domain of the inside function $f\left(x\right)$. i.e., $x\ge \text{ 2}$.

Step 2: The composite function $\left(g\circ f\right)\left(x\right)=\text{ 1 }-\text{ 2}\sqrt{x-\text{ 2}}$. Its domain also is $x\le \frac{9}{4}$.

This function puts no additional restrictions on the domain, so the composite domain is $x\ge \text{ 2}$.

This function has a more restrictive domain than $f\left(x\right)$, so the composite domain is $x\ge \text{ 2}$.

We know the composite function of $f$ and $g$ is defined as

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

Also $\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)$

$\left(g\circ f\right)\left(x\right)=g\left(\sqrt{x-\text{ 2}}\right)$

$\left(g\circ f\right)\left(x\right)=\text{ 1 }-\text{ 2}\sqrt{x-\text{ 2}}$

To determine

To find:

c. $f\circ f$ for the given functions $f$ and $g$, State the domain of each composite function.

$f(x)=\sqrt{x\text{ - 2}}$; $g(x)=\text{ 1 - }2x$

Answer to Problem 40AYU

Solution:

c. The domain of $\left(f\circ f\right)\left(x\right)$ is $x\ge \text{ 2}$.

Explanation of Solution

Given:

$f(x)=\sqrt{x\text{ - 2}}$; $g(x)=\text{ 1 - }2x$

Calculation:

We know the composite function of $f$ and $g$ is defined as

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

Also $\left(f\circ f\right)\left(x\right)=f\left(f\left(x\right)\right)$

$\left(f\circ f\right)\left(x\right)=f\left(\sqrt{x-\text{ 2}}\right)$

$\left(f\circ f\right)\left(x\right)\text{ = }\sqrt{\sqrt{x-\text{ 2}}-\text{ 2}}$

To determine

To find:

d. $g\circ g$ for the given functions $f$ and $g$, State the domain of each composite function.

$f(x)=\sqrt{x\text{ - 2}}$; $g(x)=\text{ 1 - }2x$

Answer to Problem 40AYU

Solution:

d. The domain of $\left(g\circ g\right)\left(x\right)$ is $x\ge \frac{1}{4}$.

Explanation of Solution

Given:

$f(x)=\sqrt{x\text{ - 2}}$; $g(x)=\text{ 1 - }2x$

Calculation:

For $\left(g\circ g\right)\left(x\right)$:

Step 1: Find the domain of the inside function $g\left(x\right)$. i.e., $x\le \frac{1}{2}$.

Step 2: The composite function $\left(g\circ g\right)\left(x\right)=\text{ 4}x-\text{ 1}$. Its domain also is $x\ge \frac{1}{4}$.

This function puts no additional restrictions on the domain, so the composite domain is $x\ge \frac{1}{4}$.

This function has a more restrictive domain than $f\left(x\right)$, so the composite domain is $x\ge \frac{1}{4}$.

We know the composite function of $f$ and $g$ is defined as

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

Also $\left(g\circ g\right)\left(x\right)=g\left(g\left(x\right)\right)$

$\left(g\circ g\right)\left(x\right)=\text{ 1 }-\text{ 2}\left(1-\text{ 2}x\right)$

$\left(g\circ g\right)\left(x\right)=\text{ 1 }-\text{ 2 + 4}x$

$\left(g\circ g\right)\left(x\right)=\text{ 4}x\text{ - 1}$