Chapter 5.3, Problem 50E

a.

To determine

To estimate : Where $f$ is increasing.

Answer to Problem 50E

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

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 $\left[0,\text{ 3}\right]$ .

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 $\left[0,\text{ 3}\right]$ .

b.

To determine

To estimate : Where $f$ is decreasing.

Answer to Problem 50E

The function has no decreasing interval.

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 nowhere below the x axis

Hence,

The function has no decreasing interval.

c.

To determine

To estimate : x coordinate of all local extreme values.

Answer to Problem 50E

The original function has local maximum at $x=3$ and local minimum at $x=3$ .

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, $f\text{'}\left(x\right)>0$ for all x in the domain. Therefore, f is increasing for all x in $\left[0,\text{ 3}\right]$ .

Now, as f is increasing on its entire domain, the only local extreme values are at the endpoints

Since, $f\left(0\right)<f\left(3\right)$ so $f\left(0\right)$ must be the local minimum and $f\left(3\right)$ must be the local maximum.

Hence,

The original function has local maximum at $x=3$ and local minimum at $x=3$ .