Chapter 5.1, Problem 37AYU

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.

$f(x)=\frac{x}{x-1}$; $g(x)=-\frac{4}{x}$

To determine

To find:

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

$f\left(x\right) = \frac{x}{x - 1},g\left(x\right) = \frac{-4}{x}$

Answer to Problem 37AYU

a. The domain of $f \circ g$ is $\left\{x\text{|}x \ne -4, x \ne 1\right\}$.

i.e., the domain of the composition is all real numbers with the exclusion of $-4$ and 1.

Explanation of Solution

Given:

$f\left(x\right) = \frac{x}{x - 1},g\left(x\right) = \frac{-4}{x}$

Calculation:

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

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

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

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

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

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

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

Also exclude $-4$ from the domain of $f \circ g$.

The domain of $f \circ g$ is $\left\{x\text{|}x \ne -4, x \ne 1\right\}$.

To determine

To find:

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

$f\left(x\right) = \frac{x}{x - 1}, g\left(x\right) = \frac{-4}{x}$

Answer to Problem 37AYU

b. The domain of $g \circ f$ $\left\{x\text{|}x \ne 0, x \ne 1\right\}$.

i.e., the domain of the composition is all real numbers with the exclusion of 0 and 1.

Explanation of Solution

Given:

$f\left(x\right) = \frac{x}{x - 1}, g\left(x\right) = \frac{-4}{x}$

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

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

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

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

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 $f(x)$ cannot contain 1 and the domain of $g\left(x\right)$ 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 $\left\{x\text{|}x \ne 0, x \ne 1\right\}$

To determine

To find:

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

$f\left(x\right) = \frac{x}{x - 1},g\left(x\right) = \frac{-4}{x}$

Answer to Problem 37AYU

c. The domain of $f \circ f$ is $\left\{x\text{|}x \ne 1\right\}$.

i.e., the domain of the composition is all real numbers with the exclusion of 1.

This function puts no additional restrictions on the domain, so the composite domain is $x \ne 1$.

Explanation of Solution

Given:

$f\left(x\right) = \frac{x}{x - 1},g\left(x\right) = \frac{-4}{x}$

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

$\left(f \circ f\right)\left(x\right) = f\left(\frac{x}{x - 1}\right)$

$\left(f \circ f\right)\left(x\right) = \frac{\frac{x}{x - 1}}{\left(\frac{x}{x - 1}\right) - 1}$

$\left(f\circ f\right)\left(x\right)=\frac{\frac{x}{x-1}}{\frac{x-\left(x-1\right)}{x-1}}$

$\left(f \circ f\right)\left(x\right) = \frac{x}{1}$

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

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 $f(x)$ 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 the domain of $f\circ f$ is the set of all real numbers.

This function puts no additional restrictions on the domain, so the composite domain is $x \ne 1$.

Therefore, the domain of $f \circ f$ is $\left\{x\text{|}x \ne 1\right\}$

To determine

To find:

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

$f\left(x\right) = \frac{x}{x - 1},g\left(x\right) = \frac{-4}{x}$

Answer to Problem 37AYU

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:

$f\left(x\right) = \frac{x}{x - 1},g\left(x\right) = \frac{-4}{x}$

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

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

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

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

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

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 $f(x)$ cannot contain 1 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 $\left\{x\text{|}x \ne 0\right\}$.