Chapter 1, Problem 19RE

(a)

To determine

To find: The function $f\circ g$ and its domain.

Answer to Problem 19RE

Solution:

The function $\left(f\circ g\right)\left(x\right)=\ln \left(x^2-9\right)$ and its domain is $\left(-\infty ,-3\right)\cup \left(3,\infty \right)$.

Explanation of Solution

Given:

The functions are $f\left(x\right)=\ln x,g\left(x\right)=x^2-9$.

Calculation:

The composite function $\left(f\circ g\right)\left(x\right)$ is defined as follows.

$\begin{array}{c}\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)\\ =f\left(x^2-9\right)\end{array}$

Substitute $x^2-9$ for x in $f\left(x\right)$,

$f\left(x^2-9\right)=\ln \left(x^2-9\right)$

Thus, the composite function $\left(f\circ g\right)\left(x\right)=\ln \left(x^2-9\right)$.

Note that the function $f\left(x\right)$ is defined on all positive real numbers and hence the domain of $f\left(x\right)$ is $\left(0,\infty \right)$.

The function $g\left(x\right)$ is defined on real numbers and hence the domain of the function $g\left(x\right)$ is $\left(-\infty ,\infty \right)$.

To find the domain of $\left(f\circ g\right)\left(x\right)=\ln \left(x^2-9\right)$ set the expression $x^2-9>0$ and solve.

$\begin{array}{l}{x}^2-9>0\\ x^2>9\\ \left|x\right|>3\end{array}$

Thus, the required domain is $\left(-\infty ,-3\right)\cup \left(3,\infty \right)$.

(b)

To determine

To find: The function $g\circ f$ and its domain.

Answer to Problem 19RE

Solution:

The function $\left(g\circ f\right)\left(x\right)={\left(\ln x\right)}^2-9$ and its domain is $\left(0,\infty \right)$.

Explanation of Solution

Given:

The functions are $f\left(x\right)=\ln x,g\left(x\right)=x^2-9$.

Calculation:

The composite function $\left(g\circ f\right)\left(x\right)$ is defined as follows.

$\begin{array}{c}\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left(\ln x\right)\end{array}$

Substitute $\ln x$ for x in $g\left(x\right)$.

$g\left(\ln x\right)={\left(\ln x\right)}^2-9$

Thus, the composite function $\left(g\circ f\right)\left(x\right)={\left(\ln x\right)}^2-9$.

Since the function $\left(g\circ f\right)\left(x\right)$ is defined on positive real numbers, the domain of the composite function $\left(g\circ f\right)\left(x\right)$ is $\left(0,\infty \right)$.

(c)

To determine

To find: The function $f\circ f$ and its domain.

Answer to Problem 19RE

Solution:

The composite function $\left(f\circ f\right)\left(x\right)=\ln \left(\ln x\right)$ and its domain is $\left(\infty ,-\infty \right)$.

Explanation of Solution

Given:

The function is $f\left(x\right)=\ln x$.

The composite function $\left(f\circ f\right)\left(x\right)$ is defined as follows.

$\begin{array}{c}\left(f\circ f\right)\left(x\right)=f\left(f\left(x\right)\right)\\ =f\left(\ln x\right)\end{array}$

Substitute $\ln x$ for x in $f\left(x\right)$.

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

Thus, the composite function $\left(f\circ f\right)\left(x\right)=\ln \left(\ln x\right)$.

Since, the function $f\left(x\right)$ is defined on positive real numbers, the domain of the function $f\left(x\right)$ is $\left(0,\infty \right)$.

Find the domain of the function $\left(f\circ f\right)\left(x\right)=\ln \left(\ln x\right)$ as follows.

Set the expression, $\ln x>0$.

Take exponential on both sides,

$\begin{array}{l}{e}^{\ln x}>e^0\\ x>1\end{array}$

Therefore, the domain of the composite function $\left(f\circ f\right)\left(x\right)$ is $\left(1,\infty \right)$.

(d)

To determine

To find: The function $g\circ g$ and its domain.

Answer to Problem 19RE

Solution:

The composite function $\left(g\circ g\right)\left(x\right)=x^4-18x^2+72$ and its domain is $\left(-\infty ,\infty \right)$.

Explanation of Solution

Given:

The function is $g\left(x\right)=x^2-9$.

Calculation:

The composite function $\left(g\circ g\right)\left(x\right)$ is defined as follows.

$\begin{array}{c}\left(g\circ g\right)\left(x\right)=g\left(g\left(x\right)\right)\\ =g\left(x^2-9\right)\end{array}$

Substitute $x^2-9$ for x in $g\left(x\right)$,

$\begin{array}{c}g\left(x^2-9\right)={\left(x^2-9\right)}^2-9\\ =x^4-2x^2\left(9\right)+81-9\\ =x^4-18x^2+72\end{array}$

Thus, the composite function $\left(g\circ g\right)\left(x\right)=x^4-18x^2+72$.

Since the function $\left(g\circ g\right)\left(x\right)$ is defined on all real numbers, the domain of the composite function $\left(g\circ g\right)\left(x\right)$ is $\left(-\infty ,\infty \right)$.