Chapter 5.3, Problem 49E

a.

To determine

To estimate : Where $f$ is increasing.

Answer to Problem 49E

The function is increasing on the interval $\left[0,1\right]\cup \left[3,\text{ 4}\right]\cup \left[5.5,\text{ 6}\right]$ .

Explanation of Solution

Given information :

The graph of $f\text{'}\left(x\right)$ is given below:

  

Calculation :

The original function $f\left(x\right)$ is increasing only when $f\text{'}\left(x\right)$ is positive that is above the x axis.

Since, $f\text{'}\left(x\right)$ is above the x axis on the interval $\left[0,1\right]$ , $\left[3,\text{ 4}\right]$ and $\left[5.5,\text{ 6}\right]$ .

Note: Although there is a hole at $x=4$ but the point is included because increasing takes two points.

Hence,

The function is increasing on the interval $\left[0,1\right]\cup \left[3,\text{ 4}\right]\cup \left[5.5,\text{ 6}\right]$ .

b.

To determine

To estimate : Where $f$ is decreasing.

Answer to Problem 49E

The function is decreasing on the interval $\left[1,\text{ 3}\right]\cup \left[4,5.5\right]$ .

Explanation of Solution

Given information :

The graph of $f\text{'}\left(x\right)$ is given below:

  

Calculation :

The original function $f\left(x\right)$ is decreasing only when $f\text{'}\left(x\right)$ is negative that is below the x axis.

Since, $f\text{'}\left(x\right)$ is below the x axis on the interval $\left[1,\text{ 3}\right]$ and $\left[4,5.5\right]$ .

Hence,

The function is decreasing on the interval $\left[1,\text{ 3}\right]\cup \left[4,5.5\right]$ .

c.

To determine

To estimate : x coordinate of all local extreme values.

Answer to Problem 49E

The original function has maximum at $x=1$ , $x=4$ and $x=6$ and minimum at $x=3$ , $x=5.5$ and $x=0$ .

Explanation of Solution

Given information :

The graph of $f\text{'}\left(x\right)$ is given below,

  

Calculation :

The original function $f\left(x\right)$ has a maximum when $f\text{'}\left(x\right)$ goes from positive to negative. And the function has minimum where $f\text{'}\left(x\right)$ goes from negative to positive.

Since, $f\text{'}\left(x\right)$ goes from positive to negative at $x=1$ and $x=4$ . And $f\text{'}\left(x\right)$ goes from negative to positive at $x=3$ and $x=5.5$ .

Also, the endpoint at $x=0$ and $x=6$ . Since, f has a positive slope to the right of $x=0$ so it must be local minimum and since the f has appositive slope to the left of $x=6$ so it must be a local maximum.

Hence,

The original function has maximum at $x=1$ , $x=4$ and $x=6$ and minimum at $x=3$ , $x=5.5$ and $x=0$ .