Chapter 5.1, Problem 38AYU

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.

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

To determine

To find:

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

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

Answer to Problem 38AYU

a. The domain of $f \circ g$ is $\left\{x\text{|}x \ne -3, x \ne -\frac{3}{2}\right\}$.

i.e., the domain of the composition is all real numbers with the exclusion of $-3$ and $-\frac{3}{2}$.

Explanation of Solution

Given:

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

Calculation:

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

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

$\left(f \circ g\right)\left(x\right) = f\left(\frac{2}{x}\right)$

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

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

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

When working with rational functions, domain elements must not create division by zero.

1. Any $x$ not in the domain of $g$ must be excluded.

2. Any $x$ for which $g\left(x\right)$ is not in the domain of $f$ must be excluded.

So, we know that the domain of $f(x)$ cannot contain $-3$ and the domain of $g\left(x\right)$ cannot contain 0.

For this composition $\left(f \circ g\right)\left(x\right) = f\left(g\left(x\right)\right)$, shows us that $x = -\frac{3}{2}$ would not be an acceptable domain element, since it creates a zero denominator, setting $3x + 2 = 0$, $x = -\frac{3}{2}$.

Also exclude $-\frac{3}{2}$ from the domain of $f \circ g$. The domain of $f \circ g$ is $\left\{x\text{|}x \ne -3, x \ne -\frac{3}{2}\right\}$.

To determine

To find:

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

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

Answer to Problem 38AYU

b. The domain of $g \circ f$ is $\left\{x\text{|}x \ne -3, x \ne 0\right\}$.

i.e., the domain of the composition is all real numbers with the exclusion of $-3$ and 0.

Explanation of Solution

Given:

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

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

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

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

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

When working with rational functions, domain elements must not create division by zero.

1. Any $x$ not in the domain of $g$ must be excluded.

2. Any $x$ for which $g\left(x\right)$ is not in the domain of $f$ must be excluded.

So, we know that the domain of $f\left(x\right)$ cannot contain $-3$ and the domain of $g(x)$ cannot contain 0.

For this composition $\left(g \circ f\right)\left(x\right) = g\left(f\left(x\right)\right)$, shows us that $x = 0$ would not be an acceptable domain element, since it creates a zero denominator, setting $x = 0$, $x = 0$.

Also exclude 0 from the domain of $g \circ f$. The domain of $g \circ f$ is $\left\{x\text{|}x \ne -3, x \ne 0\right\}$.

To determine

To find:

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

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

Answer to Problem 38AYU

c. The domain of $f \circ f$ is $\left\{x\text{|}x \ne -3, x \ne -\frac{9}{4}\right\}$.

i.e., the domain of the composition is all real numbers with the exclusion of $-3$ and $-\frac{9}{4}$.

Explanation of Solution

Given:

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

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

$\left(f \circ f\right)\left(x\right) = f\left(\frac{x}{x + 3}\right)$

$\left(f \circ f\right)\left(x\right) = \frac{\frac{x}{x + 3}}{\left(\frac{x}{x + 3}\right) + 3}$

$\left(f \circ f\right)\left(x\right) = \frac{\frac{x}{x + 3}}{\frac{x + 3\left(x + 3\right)}{x + 3}}$

$\left(f \circ f\right)\left(x\right) = \frac{x}{4x + 9}$

When working with rational functions, domain elements must not create division by zero.

1. Any $x$ not in the domain of $g$ must be excluded.

2. Any $x$ for which $g\left(x\right)$ is not in the domain of $f$ must be excluded.

So, we know that the domain of $f(x)$ cannot contain 1 and the domain of $g\left(x\right)$ cannot contain 0.

For this composition $\left(f \circ f\right)\left(x\right) = f\left(f\left(x\right)\right)$, shows us that $x = -\frac{9}{4}$ would not be an acceptable domain element, since it creates a zero denominator, setting $4x + 9 = 0$, $x = -\frac{9}{4}$.

Also exclude $-\frac{9}{4}$ from the domain of $f \circ f$. The domain of $f \circ f$ is $\left\{x\text{|}x \ne -3, x \ne -\frac{9}{4}\right\}$.

To determine

To find:

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

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

Answer to Problem 38AYU

d. The domain of $g \circ g$ is $\left\{x\text{|}x \ne 0\right\}$.

i.e., the domain of the composition is all real numbers with the exclusion of 0.

Because, this function puts no additional restrictions on the domain, so the composite domain is $x \ne 0$.

Explanation of Solution

Given:

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

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

$\left(g \circ g\right)\left(x\right) = g\left(\frac{4}{x}\right)$

$\left(g \circ g\right)\left(x\right) = \frac{2}{\left(\frac{2}{x}\right)}$

$\left(g \circ g\right)\left(x\right) = \frac{2x}{2}$

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

When working with rational functions, domain elements must not create division by zero.

1. Any $x$ not in the domain of $g$ must be excluded.

2. Any $x$ for which $g\left(x\right)$ is not in the domain of $f$ must be excluded.

So, we know that the domain of $f\left(x\right)$ cannot contain $-3$ and the domain of $g\left(x\right)$ cannot contain 0.

For this composition $\left(g \circ g\right)\left(x\right) = g\left(g\left(x\right)\right)$. shows us that the domain of $g\circ g$ is the set of all real numbers.

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

Therefore, the domain of $g \circ g$ is $\left\{x\text{|}x \ne 0\right\}$.