Chapter 4, Problem 33RE

(a)

To determine

To find : The value of interval on that the function $y=f(x)$ is increasing.

Answer to Problem 33RE

The answer: $[-3,-2],[1,2]$ .

Explanation of Solution

Given information:

The graph is:

  

Calculation:

The interval where $f\text{'}$ is positive is where the function $f$ is growing. The endpoints are incorporated into the growing interval.

Here, $f^{\prime }>0$ on the interval $(-3,-2)$ and $(1,2)$ .

So, the intervals of increasing are: $[-3,-2],[1,2]$ .

(b)

To determine

To find: The value of interval on that the function $y=f(x)$ is decreasing.

Answer to Problem 33RE

The answer: $[-2,0),(0,1]$

Explanation of Solution

Given information:

The graph is:

  

Calculation:

On the interval where $f\text{'}$ is negative, the function $f$ is decreasing. The interval of diminishing is expanded to cover the endpoints.

Here, $f^{\prime }<0$ on the intervals $(-2,0)$ and $(0,1)$ .

So, the value of interval on that the function $y=f(x)$ is decreasing: $[-2,0),(0,1]$ .

(c)

To determine

To find: The any global extreme value and they occur.

Answer to Problem 33RE

The answer:

Local maximum at $x=-2$ and at $x=2$ .

Local minimum at $x=1$ and at $x=-3$ .

Explanation of Solution

Given information:

The graph is:

  

Calculation:

The function $f$ has a local extreme at the interior point $x_0$ of the domain if $f^{\prime }\left(x_0\right)=0\text{ or }{f}^{\prime }\left(x_0\right)$ is not defined and, $f\text{'}$ changes its sign when passing through $x_0$ .

Here, $f\text{'}$ has two zeros,

  $\begin{array}{c}{x}_1=-2\\ x_2=1\end{array}$

Since it changes its sign from positive to negative when passing through $x_1=-2$ there is local maximum at $-2$ .

  $f\text{'}$ changes its sign from negative to positive when passing through $x_2=1$ so there is a local minimum 1.

At $x=0$ , $f\text{'}$ is not defined, but it also does not changes sign, so there is no a local extreme at

  $x=0$ .

Because the function is increasing on $[-3,-2]$ there is a local minimum at $x=-3$ .

Because the function is increasing on $[1,2]$ there is a local maximum at $x=2$ .