Chapter 3.2, Problem 124E

(a).

To determine

To graph:the given functions.

Explanation of Solution

Given:

The given functions:

  $\begin{array}{l}f(x)=\ln x\\ g(x)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}\\ n=2\end{array}$

Graph:

As per the given problem

Substitute $n=2$ in $g(x)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}$

  $g(x)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\text{ [Substitute }n=2]$

Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expressions $f(x)=\ln x$ and $g(x)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ , which is required to graph.

Step 4: Press GRAPH button to graph the functions.

  

Interpretation: the graph of $f(x)$ and $g(x)$ do not intersect at any point.

(b).

To determine

To infer:the function that is increasing at a greater rate as x approaches infinity.

Answer to Problem 124E

The function $g(x)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is increasing at a greater rate as x approaches to infinity.

Explanation of Solution

Given:

The given functions:

  $\begin{array}{l}f(x)=\ln x\\ g(x)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}\\ n=2\end{array}$

Consider the graph of the functions obtained in part (a).

As per the given problem,

For $n=2$ ,

  $g(x)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}=g(x)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

From the graph of $f(x)$ and $g(x)$ (for $n=2$ ) (obtained in part (a)), it can be concluded that the function $g(x)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is increasing at a greater rate as x approaches to infinity, as it has a greater value than the value of $f(x)$ at the same value of x .

(c).

To determine

To graph:the given functions for $n=3,4,\text{ and 5}$ and infer which function is increasing at a greater rate as x approaches to infinity for each case.

Explanation of Solution

Given:

The given functions:

  $\begin{array}{l}f(x)=\ln x\\ g(x)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}\\ n=3,\text{ 4, and 5 }\end{array}$

Graph:

As per the given problem

For $n=3$,

  $\begin{array}{l}f(x)=\ln x\\ g(x)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\text{ [Substitute }n=3]\end{array}$

Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expressions $f(x)=\ln x$ and $g(x)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ , which is required to graph.

Step 4: Press GRAPH button to graph the functions.

  

The function $f(x)=\ln x$ is increasing at a greater rate as x approaches to infinity.

For $n=4$,

  $\begin{array}{l}f(x)=\ln x\\ g(x)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\text{ [Substitute }n=4]\end{array}$

Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expressions $f(x)=\ln x$ and $g(x)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$ , which is required to graph.

Step 4: Press GRAPH button to graph the functions.

  

The function $f(x)=\ln x$ is increasing at a greater rate as x approaches to infinity.

For $n=5$,

  $\begin{array}{l}f(x)=\ln x\\ g(x)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}\text{ [Substitute }n=5]\end{array}$

Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expressions $f(x)=\ln x$ and $g(x)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$ , which is required to graph.

Step 4: Press GRAPH button to graph the functions.

  

The function $f(x)=\ln x$ is increasing at a greater rate as x approaches to infinity.

Interpretation: In all three cases, $n=3,4,\text{ and 5}$ , the function $f(x)=\ln x$ is increasing at a greater rate as x approaches to infinity.