Chapter 2, Problem 43RE

(a)

To determine

To find:The intervals in which $f$ increases or decreases.

Answer to Problem 43RE

The function $f$ is increasing in intervals $\left(-2,0\right)$ and $\left(2,\infty \right)$ , and decreasing in intervals $\left(-\infty ,-2\right)$ and $\left(0,2\right)$ .

Explanation of Solution

Given information:

The given graph is shown in figure ().

  

Figure (1)

Calculation:

If $f\text{'}\left(x\right)>0$ on an interval, then the $f$ is increasing on that interval.

If $f\text{'}\left(x\right)<0$ on an interval, then the $f$ is increasing on that interval.

From the above graph, the function $f$ is increasing in intervals $\left(-2,0\right)$ and $\left(2,\infty \right)$ , and decreasing in intervals $\left(-\infty ,-2\right)$ and $\left(0,2\right)$ .

  $f$ is increases when $f\text{'}\left(x\right)$ is positive and decreases when $f\text{'}\left(x\right)$ is negative.

Therefore, the function $f$ is increasing in intervals $\left(-2,0\right)$ and $\left(2,\infty \right)$ , and decreasing in intervals $\left(-\infty ,-2\right)$ and $\left(0,2\right)$ .

(b)

To determine

To find:The values of $x$ does $f$ have a local maximum and minimum.

Answer to Problem 43RE

The $f$ have a local value maximum is $x=0$ and minimum value is $x=-2,x=2$ .

Explanation of Solution

Given information:

The given graph is shown in figure ().

  

Figure (1)

Calculation:

The point at which $f\text{'}\left(x\right)$ changes from negative to positive is the local minimum point at the point at which $f\text{'}\left(x\right)$ changes from positive to negative is the local maximum point.

Here, at $x=-2$ and $x=2$ , $f\text{'}\left(x\right)$ changes from negative to positive and at $x=0$ , $f\text{'}\left(x\right)$ changes from positive to negative.

  $f$ have a local minimum value is $x=-2,x=2$ and the value of $x$ for which $f$ have a local maximum value is $x=0$ .

Therefore, the $f$ have a localvalue maximum is $x=0$ and minimumvalue is $x=-2,x=2$ .

(c)

To determine

To find:Where is $f$ concave upward or downward.

Answer to Problem 43RE

The $f$ is concave upward in $\left(-\infty ,-1\right)$ and $\left(1,\infty \right)$ , concave downward in $\left(-1,1\right)$ .

Explanation of Solution

Given information:

The given graph is shown in figure (1).

  

Figure (1)

Calculation:

The function $f$ is concave upward when $f\text{'}\left(x\right)$ is increasing and $f$ is concave downward when $f\text{'}\left(x\right)$ is decreasing.

Hence, $f$ is concave upward in $\left(-\infty ,-1\right)$ and $\left(1,\infty \right)$ , concave downward in $\left(-1,1\right)$ .

Therefore, the $f$ is concave upward in $\left(-\infty ,-1\right)$ and $\left(1,\infty \right)$ , concave downward in $\left(-1,1\right)$

(d)

To determine

To graph: For the function $f$ at $f\left(0\right)=0$ .

Explanation of Solution

Given information:

The given graph is shown in figure (1).

  

Figure (1)

Graph:

From the given figure that $f\text{'}\left(x\right)$ is negative when $-\infty <x<-2$ and $0<x<2$ . So, the original function $f$ must be decreasing on the interval $\left(-\infty ,-2\right)$ and $\left(0,2\right)$ .

Similarly,

  $f\text{'}\left(x\right)$ is positive when $-2<x<0$ and $2<x<\infty$ . So, the original function $f$ must be increasing on the interval $\left(-2,0\right)$ and $\left(2,\infty \right)$ .

It is known that $f\left(0\right)=0$ , so, we draw the graph of $f$ at a given point $\left(0,0\right)$ .

The graph for the function $f$ is shown in figure (2).

  

Figure (2)

The rough sketch of graph $F$ as a time $t$ is shown in figure (3).

  

Figure (3)

Interpretation: Graph for the function $f$ is shown in figure (2).