Chapter 5.1, Problem 39AYU

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.

$f(x)=\sqrt{x}$; $g(x)=2x+3$

To determine

To find:

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

$f(x)=\sqrt{x}$; $g(x)=2x+3$

Answer to Problem 39AYU

Solution:

a. The domain of $\left(f\circ g\right)\left(x\right)$ is $x\le \frac{3}{2}$.

Explanation of Solution

Given:

$f(x)=\sqrt{x}$; $g(x)=2x+3$

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., $x\ge -\frac{3}{2}$.

Step 2: The composite function $\left(f\circ g\right)\left(x\right)=\sqrt{2x+3}$. Its domain also is $x\ge -\frac{3}{2}$.

This function puts no additional restrictions on the domain, so the composite domain is $x\le \frac{3}{2}$.

This function has a more restrictive domain than $f\left(x\right)$, so the composite domain is $x\le \frac{3}{2}$.

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

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

$\left(f\circ g\right)\left(x\right)=f\left(2x+3\right)$

$\left(f\circ g\right)\left(x\right)=\sqrt{2x+3}$

To determine

To find:

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

$f(x)=\sqrt{x}$; $g(x)=2x+3$

Answer to Problem 39AYU

Solution:

b. The domain of $\left(g\circ f\right)\left(x\right)$ is $x\ge 0$.

Explanation of Solution

Given:

$f(x)=\sqrt{x}$; $g(x)=2x+3$

Calculation:

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

Step 1: Find the domain of the inside function $g\left(x\right)$. i.e., $x\ge 0$.

Step 2: The composite function $\left(g\circ f\right)\left(x\right)=\text{ 2}\sqrt{x}\text{ + 3}$. Its domain also is $x\ge \frac{9}{4}$.

This function puts no additional restrictions on the domain, so the composite domain is $x\ge 0$.

This function has a more restrictive domain than $f\left(x\right)$, so the composite domain is $x\ge 0$.

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

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

$\left(g\circ f\right)\left(x\right)=\text{ 2}\sqrt{x}\text{ + 3}$

To determine

To find:

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

$f(x)=\sqrt{x}$; $g(x)=2x+3$

Answer to Problem 39AYU

Solution:

c. The domain of $\left(f\circ f\right)\left(x\right)$ is $x\ge 0$.

Explanation of Solution

Given:

$f(x)=\sqrt{x}$; $g(x)=2x+3$

Calculation:

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

Step 1: Find the domain of the inside function $g\left(x\right)$. i.e., $x\ge 0$.

Step 2: The composite function $\left(f\circ f\right)\left(x\right)=\sqrt{\sqrt{x}}$. Its domain also is $x\ge 0$.

This function puts no additional restrictions on the domain, so the composite domain is $x\ge 0$.

This function has a more restrictive domain than $f\left(x\right)$, so the composite domain is $x\ge 0$. (This is the intersection of the two domains.)

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

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

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

To determine

To find:

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

$f(x)=\sqrt{x}$; $g(x)=2x+3$

Answer to Problem 39AYU

Solution:

d. The domain of $\left(g\circ g\right)\left(x\right)$ is $x\ge -\frac{3}{2}$.

Explanation of Solution

Given:

$f(x)=\sqrt{x}$; $g(x)=2x+3$

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., $x\ge -\frac{3}{2}$.

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

This function puts no additional restrictions on the domain, so the composite domain is $x\ge -\frac{3}{2}$.

This function has a more restrictive domain than $f\left(x\right)$, so the composite domain is $x\ge -\frac{3}{2}$.

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

$\left(g\circ g\right)\left(x\right)=g\left(2x\text{ + 3}\right)$

$\left(g\circ g\right)\left(x\right)=\text{ 2}\left(2x\text{ + 3}\right)\text{ + 3)}$

$\left(g\circ g\right)\left(x\right)=\text{ 4}x\text{ + 6 + 3}$

$\left(g\circ g\right)\left(x\right)=\text{ 4}x\text{ + 9}$