Chapter 5.1, Problem 42AYU

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.

$f\left(x\right)=x^2+4$; $g(x)=\sqrt{x-2}$

To determine

To find:

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

State the domain of each composite function.

$f\left(x\right)=x^2+4;g(x)=\sqrt{x-2}$

Answer to Problem 42AYU

Solution:

a.The domain of $\left(f\circ g\right)\left(x\right)$ is $x\ge -2$.

Explanation of Solution

Given:

$f\left(x\right)=x^2+4;g(x)=\sqrt{x-2}$

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., $x\ge 2$.

Step 2: The composite function $\left(f\circ g\right)\left(x\right)=x+2$.

Its domain also is $x\ge -2$.

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

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

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

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

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

$\left(f\circ g\right)\left(x\right)={\left(\sqrt{x-2}\right)}^2+4$

$\left(f\circ g\right)\left(x\right)=x-2+4$

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

To determine

To find:

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

State the domain of each composite function.

$f\left(x\right)=x^2+4;g(x)=\sqrt{x-2}$

Answer to Problem 42AYU

Solution:

b. The domain of $\left(g\circ f\right)\left(x\right)$ is $x\in R$ or $x\in \left(-\infty ,\infty \right)$.

Explanation of Solution

Given:

$f\left(x\right)=x^2+4;g(x)=\sqrt{x-2}$

Calculation:

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

Step 1: Find the domain of the inside function$f\left(x\right)$ Any real value of $x$.

Here are no domain restrictions, even though there is a variable under the radical. Since $x^2+4$ can never be negative. So, the domain of $f\left(x\right)$ can be any value of $x$. i.e., $x\in R$ or $x\in \left(-\infty ,\infty \right)$

Step 2: The composite function $\left(g\circ f\right)\left(x\right)=\sqrt{x^2+2}$.

Here are no domain restrictions, even though there is a variable under the radical. Since $\sqrt{x^2+2}$ can never be negative. So, the domain of $f\left(x\right)$ can be any value of $x$ i.e., $x\in R$ or $\in \left(-\infty ,\infty \right)$.

Its domain also is $x\in R$ or $x\in \left(-\infty , \infty \right)$.

This function puts no additional restrictions on the domain, so the composite domain is $x\in R$ or $\in \left(-\infty , \infty \right)$.

This function has a more restrictive domain than $f\left(x\right)$, so the composite domain is $x\in R$ or $x\in \left(-\infty , \infty \right)$.

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

$\left(g\circ f\right)\left(x\right)=g\left(x^2+4\right)$

$\left(g\circ f\right)\left(x\right)=\sqrt{\left(x^2+4\right)-2}$

$\left(g\circ f\right)\left(x\right)=\sqrt{x^2+4-2}$

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

To determine

To find:

(c) $f\circ f$ for the given functions $f$ and $g$.

State the domain of each composite function.

$f\left(x\right)=x^2+4;g(x)=\sqrt{x-2}$

Answer to Problem 42AYU

Solution:

c) The domain of $\left(f\circ f\right)\left(x\right)$ is $x\in R$ or $\in \left(-\infty ,\infty \right)$.

Explanation of Solution

Given:

$f\left(x\right)=x^2+4; g(x)=\sqrt{x-2}$

Calculation:

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

Step 1: Find the domain of the inside function $f\left(x\right)$. Any real value of $x$.

Here are no domain restrictions, even though there is a variable under the radical. Since $x^2+4$ can never be negative. So, the domain of $f\left(x\right)$ can be any value of $x$. i.e., $x\in R$ or $x\in \left(-\infty , \infty \right)$.

Step 2: The composite function $\left(f\circ f\right)\left(x\right)=x^4+8x^2+20$.

Here are no domain restrictions, even though there is a variable under the radical. Since $x^4+8x^2+20$ can never be negative. So, the domain of $f\left(x\right)$ can be any value of $x$ i.e., $x\in R$ or $x\in \left(-\infty , \infty \right)$.

This function puts no additional restrictions on the domain, so the composite domain is $x\in R$ or $\in \left(-\infty , \infty \right)$.

This function has a more restrictive domain than $f\left(x\right)$, so the composite domain is $x\in R$ or $\in \left(-\infty , \infty \right)$.

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

$\left(f\circ f\right)\left(x\right)=f(x^2+4)$

$\left(f\circ f\right)\left(x\right)={\left(x^2+4\right)}^2+4$

$\left(f\circ f\right)\left(x\right)=x^4+8x^2+16+4$

$\left(f\circ f\right)\left(x\right)=x^4+8x^2+20$

To determine

To find:

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

State the domain of each composite function.

$f\left(x\right)=x^2+4; g(x)=\sqrt{x-2}$

Answer to Problem 42AYU

Solution:

d. The domain of $\left(g\circ g\right)\left(x\right)$ is $x\ge 2$.

Explanation of Solution

Given:

$f\left(x\right)=x^2+4; g(x)=\sqrt{x-2}$

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.,$x\ge 2$.

Step 2: The composite function $\left(g\circ g\right)\left(x\right)=\sqrt{\left(\sqrt{x-2}\right)-2}$.

Its domain also is $x\ge 6$.

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

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

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

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

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

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