Chapter 5.1, Problem 35AYU

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.

$f(x)=\frac{3}{x-1}$; $g(x)=\frac{2}{x}$

To determine

To find:

a. $f\circ g$ for the given functions $f$ and $g$, find

State the domain of each composite function.

$f(x)=\frac{3}{x-1};g(x)=\frac{2}{x}$

Answer to Problem 35AYU

Solution:

a. The domain of $f\circ g$ is $\left\{x\text{|}x\ne 1,x\ne 2\right\}$.

i.e., the domain of the composition is all real numbers with the exclusion of 1 and 2.

Explanation of Solution

Given:

$f(x)=\frac{3}{x-1};g(x)=\frac{2}{x}$

Calculation:

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

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

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

$\left(f\circ g\right)\left(x\right)=\frac{3}{\left(\frac{2}{x}\right)-1}$

$\left(f\circ g\right)\left(x\right)=\frac{3}{\frac{2}{x}-1}$

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

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

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 $g(x)$ 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 $g(x)$ cannot contain 0.

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

Also exclude 2 from the domain of $f\circ g$.

The domain of $f\circ g$ is $\left\{x\text{|}x\ne 1,x\ne 2\right\}$.

To determine

To find:

b. $g\circ f$ for the given functions $f$ and $g$ , find

State the domain of each composite function.

$f(x)=\frac{3}{x-1};g(x)=\frac{2}{x}$

Answer to Problem 35AYU

Solution:

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

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

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

Explanation of Solution

Given:

$f(x)=\frac{3}{x-1};g(x)=\frac{2}{x}$

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

$\left(g\circ f\right)\left(x\right)=g\left(\frac{3}{x-1}\right)$

$\left(g\circ f\right)\left(x\right)=\frac{2}{\frac{3}{x-1}}$

$\left(g\circ f\right)\left(x\right)=\frac{2}{3}\left(x-1\right)$

The domain of $f$ and the domain of $g$ are the set of all real numbers.

Because the domains of both $f$ and $g$ are the set of all real numbers, the domain of $g\circ f$ is the set of all real numbers.

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 $g(x)$ 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 $g(x)$ cannot contain 0.

For this composition $\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)$ . shows us that the domain of $g\circ f$ 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 $g\circ f$ is $\left\{x\text{|}x\ne 1\right\}$.

To determine

To find:

c. $f\circ f$ for the given functions $f$ and $g$ , find

State the domain of each composite function.

$f(x)=\frac{3}{x-1};g(x)=\frac{2}{x}$

Answer to Problem 35AYU

Solution:

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

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

Explanation of Solution

Given:

$f(x)=\frac{3}{x-1};g(x)=\frac{2}{x}$

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

$\left(f\circ f\right)\left(x\right)=f\left(\frac{3}{x-1}\right)$

$\left(f\circ f\right)\left(x\right)=\frac{3}{\left(\frac{3}{x-1}\right)-1}$

$\left(f\circ f\right)\left(x\right)=\frac{3}{\frac{3}{x-1}-1}$

$\left(f\circ f\right)\left(x\right)=\frac{3}{\frac{3-\left(x-1\right)}{x-1}}$

$\left(f\circ f\right)\left(x\right)=\frac{-3\left(x-1\right)}{x-4}$

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 $g(x)$ 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 $g(x)$ cannot contain 0

For this composition $\left(f\circ f\right)\left(x\right)=f\left(f\left(x\right)\right)$. shows us that $x=4$ would not be an acceptable domain element, since it creates a zero denominator, setting $x–4=0$. i.e., $x=4$.

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

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

To determine

To find:

d. $g\circ g$ for the given functions $f$ and $g$ , find

State the domain of each composite function.

$f(x)=\frac{3}{x-1};g(x)=\frac{2}{x}$

Answer to Problem 35AYU

Solution:

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:

$f(x)=\frac{3}{x-1};g(x)=\frac{2}{x}$

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

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

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

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

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

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 $g(x)$ 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 $g(x)$ 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 ∘ 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 $\left\{x\text{|}x\ne 0\right\}$.