Chapter 5, Problem 11RE

a.

To determine

To find the intervals on which the function is increasing by using analytical method.

Answer to Problem 11RE

The function $y=\ln \left|x\right|$ is increasing when $\left(0,2\right]$

Explanation of Solution

Given:

The function is $y=\ln \left|x\right|,-2\le x\le 2,x\ne 0$ .

Calculation:

Since, $\begin{array}{l}\left|x\right|=\{\begin{array}{l}x,x\ge 0\\ -x,x<0\end{array}\\ y=\ln \left|x\right|=\{\begin{array}{l}\ln x,x>0\\ \ln \left(-x\right),x<0\end{array}\\ y^{\prime }=\{\begin{array}{l}\frac{1}{x},x>0\\ \frac{1}{-x}\times -1=\frac{1}{x},x<0\end{array}\end{array}$

At $x=0$ the value of $y^{\prime }$ is undefined therefore $x=0$ is one of the critical point also it is given that $-2\le x\le 2$ so there exist two intervals that is $\left[-2,0\right)$ and $\left(0,2\right]$

The function is increasing when $f^{\prime }\left(x\right)>0$ .

Now , put $x=-1$ to check the whether the function is increasing or decreasing in interval $\left[-2,0\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{1}{x}\\ f^{\prime }\left(-1\right)=\frac{1}{-1}=-1<0\end{array}$

Now , put $x=1$ to check the whether the function is increasing or decreasing in interval $\left(0,2\right]$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{1}{x}\\ f^{\prime }\left(1\right)=\frac{1}{1}=1>0\end{array}$

Therefore, the function $y=\ln \left|x\right|$ is increasing when $\left(0,2\right]$

Below is the graph of the function $y=\ln \left|x\right|$

  

From graph it is clear that the function $y=\ln \left|x\right|$ is increasing when $x\in \left(0,2\right]$ .

b.

To determine

To find the intervals on which the function is decreasing by using analytical method.

Answer to Problem 11RE

The function $y=\ln \left|x\right|$ is decreasing when $\left[-2,0\right)$

Explanation of Solution

Given:

The function is $y=\ln \left|x\right|,-2\le x\le 2,x\ne 0$ .

Calculation:

Since, $\begin{array}{l}\left|x\right|=\{\begin{array}{l}x,x\ge 0\\ -x,x<0\end{array}\\ y=\ln \left|x\right|=\{\begin{array}{l}\ln x,x>0\\ \ln \left(-x\right),x<0\end{array}\\ y^{\prime }=\{\begin{array}{l}\frac{1}{x},x>0\\ \frac{1}{-x}\times -1=\frac{1}{x},x<0\end{array}\end{array}$

At $x=0$ the value of $y^{\prime }$ is undefined therefore $x=0$ is one of the critical point also it is given that $-2\le x\le 2$ so there exist two intervals that is $\left[-2,0\right)$ and $\left(0,2\right]$

The function is increasing when $f^{\prime }\left(x\right)>0$ .

Now , put $x=-1$ to check the whether the function is increasing or decreasing in interval $\left[-2,0\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{1}{x}\\ f^{\prime }\left(-1\right)=\frac{1}{-1}=-1<0\end{array}$

Now , put $x=1$ to check the whether the function is increasing or decreasing in interval $\left(0,2\right]$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{1}{x}\\ f^{\prime }\left(1\right)=\frac{1}{1}=1>0\end{array}$

Therefore, the function $y=\ln \left|x\right|$ is decreasing when $\left[-2,0\right)$

Below is the graph of the function $y=\ln \left|x\right|$

  

From graph it is clear that the function $y=\ln \left|x\right|$ is decreasing when $x\in \left[-2,0\right)$ .

c.

To determine

To find the intervals on which the function is concave up by using analytical method.

Answer to Problem 11RE

The Function $y=\ln \left|x\right|$ is never concave up because there does not exist any interval for which $f^{″}\left(x\right)>0$

Explanation of Solution

Given:

The function is $y=\ln \left|x\right|,-2\le x\le 2,x\ne 0$ .

Calculation:

The graph of a twice differentiable function $y=f\left(x\right)$ is

Concave up on any interval where $f^{″}\left(x\right)>0$ and concave down on any interval where $f^{″}\left(x\right)<0$

Since, $y=\ln \left|x\right|$

  $\begin{array}{l}\left|x\right|=\{\begin{array}{l}x,x\ge 0\\ -x,x<0\end{array}\\ y=\ln \left|x\right|=\{\begin{array}{l}\ln x,x>0\\ \ln \left(-x\right),x<0\end{array}\\ y^{\prime }=\{\begin{array}{l}\frac{1}{x},x>0\\ \frac{1}{-x}\times -1=\frac{1}{x},x<0\end{array}\end{array}$

Second derivative: $y^{″}=-\frac{1}{x^2}$

Now, put $f^{″}\left(x\right)=0$ to find critical points

  $\begin{array}{l}{y}^{″}=-\frac{1}{x^2}\\ -\frac{1}{x^2}=0\end{array}$

At $x=0$ the value of $y^{″}$ is undefined therefore $x=0$ is one of the critical point also it is given that $-2\le x\le 2$ so there exist two intervals that is $\left[-2,0\right)$ and $\left(0,2\right]$

Now in interval $\left[-2,0\right)$ test for $x=-1$

  $\begin{array}{l}{f}^{″}\left(x\right)=-\frac{1}{x^2}\\ f^{″}\left(-1\right)=-\frac{1}{{\left(-1\right)}^2}=-1<0\end{array}$

Now in interval $\left(0,2\right]$ test for $x=1$

  $\begin{array}{l}{f}^{″}\left(x\right)=-\frac{1}{x^2}\\ f^{″}\left(1\right)=-\frac{1}{{\left(1\right)}^2}=-1<0\end{array}$

Therefore, the Function $y=\ln \left|x\right|$ is never concave up because there does not exist any interval for which $f^{″}\left(x\right)>0$

Below is the graph of the function

  

From graph it is clear that ,the Function $y=\ln \left|x\right|$ is never concave up because there does not exist any interval for which $f^{″}\left(x\right)>0$ .

d.

To determine

To find the intervals on which the function is concave down by using analytical method.

Answer to Problem 11RE

  $y=\ln \left|x\right|$ is concave down in interval $\left[-2,0\right)$ and $\left(0,2\right]$ .

Explanation of Solution

Given:

The function is $y=\ln \left|x\right|,-2\le x\le 2,x\ne 0$ .

Calculation:

The graph of a twice differentiable function $y=f\left(x\right)$ is

Concave up on any interval where $f^{″}\left(x\right)>0$ and concave down on any interval where $f^{″}\left(x\right)<0$

Since, $y=\ln \left|x\right|$

  $\begin{array}{l}\left|x\right|=\{\begin{array}{l}x,x\ge 0\\ -x,x<0\end{array}\\ y=\ln \left|x\right|=\{\begin{array}{l}\ln x,x>0\\ \ln \left(-x\right),x<0\end{array}\\ y^{\prime }=\{\begin{array}{l}\frac{1}{x},x>0\\ \frac{1}{-x}\times -1=\frac{1}{x},x<0\end{array}\end{array}$

Second derivative: $y^{″}=-\frac{1}{x^2}$

Now, put $f^{″}\left(x\right)=0$ to find critical points

  $\begin{array}{l}{y}^{″}=-\frac{1}{x^2}\\ -\frac{1}{x^2}=0\end{array}$

At $x=0$ the value of $y^{″}$ is undefined therefore $x=0$ is one of the critical point also it is given that $-2\le x\le 2$ so there exist two intervals that is $\left[-2,0\right)$ and $\left(0,2\right]$

Now in interval $\left[-2,0\right)$ test for $x=-1$

  $\begin{array}{l}{f}^{″}\left(x\right)=-\frac{1}{x^2}\\ f^{″}\left(-1\right)=-\frac{1}{{\left(-1\right)}^2}=-1<0\end{array}$

Now in interval $\left(0,2\right]$ test for $x=1$

  $\begin{array}{l}{f}^{″}\left(x\right)=-\frac{1}{x^2}\\ f^{″}\left(1\right)=-\frac{1}{{\left(1\right)}^2}=-1<0\end{array}$

Therefore, the Function $y=\ln \left|x\right|$ is concave down in interval $\left[-2,0\right)$ and $\left(0,2\right]$

Below is the graph of the function

  

From graph it is clear that , the Function $y=\ln \left|x\right|$ is concave down in interval $\left[-2,0\right)$ and $\left(0,2\right]$ .

e.

To determine

To find any local extreme values.

Answer to Problem 11RE

Local extreme values exist at point $\left(-2,\ln 2\right)$ and $\left(2,\ln 2\right)$ .

Explanation of Solution

Given:

The function is $y=\ln \left|x\right|,-2\le x\le 2,x\ne 0$ .

Calculation:

Since, in the interval $\left[-2,0\right)$ and $\left(0,2\right]$ , $f^{″}\left(x\right)<0$ therefore the function $y=\ln \left|x\right|$ has only local maximum value at $x=2$ and $x=-2$ .

Maximum values are

  $\begin{array}{l}f\left(x\right)=\ln \left|x\right|\\ f\left(2\right)=\ln \left|2\right|=\ln 2\\ f\left(-2\right)=\ln \left|-2\right|=\ln 2\end{array}$

Therefore, local extreme values exist at point $\left(-2,\ln 2\right)$ and $\left(2,\ln 2\right)$ .

f.

To determine

To find inflections points.

Answer to Problem 11RE

No inflection point exist because function does not changes its concavity.

Explanation of Solution

Given:

The function is $y=\ln \left|x\right|$ .

Calculation:

Inflection point of any function is a point where the graph of function has a tangent line and where the concavity changes.

Since, $y=\ln \left|x\right|$ always remain concave down in interval $\left[-2,0\right)$ and $\left(0,2\right]$ therefore no inflection point exist because function does not changes its concavity.