Chapter 10.2, Problem 40E

a.

To determine

To determine on what interval $f$ is negative and positive.

Answer to Problem 40E

  $x>0$ $f$ is positive

  $x<0$ $f$ is negative

Explanation of Solution

Given information:

Graph of function $f(x)=\sqrt[3]{x}$

  

  $x>0$ $f$ is positive because here the average of graph is increasing.

  $x<0$ $f$ is negative because here the average of graph is decreasing.

b.

To determine

To determine on what interval $f$ is increasing and decreasing.

Answer to Problem 40E

  $x>0$ $f$ is increasing.

  $x<0$ $f$ is decreasing.

Explanation of Solution

Given information:

Graph of function $f(x)=\sqrt[3]{x}$

  

  $x>0$ $f$ is increasing because here the average of graph is increasing.

  $x<0$ $f$ is decreasing because here the average of graph is decreasing.

c.

To determine

To determine if $f$ has minimum or maximum value.

Answer to Problem 40E

No, $f$ has no minimum or maximum value.

Explanation of Solution

Given information:

Graph of function $f(x)=\sqrt[3]{x}$

  

No, $f$ has no minimum or maximum value because this is an infinite graph.

d.

To determine

To determine the average rate of change of function $f$ .

Answer to Problem 40E

The average rate of change between $x=-1$ to $x=1$ is 1.

Explanation of Solution

Given information:

Graph of function $f(x)=\sqrt[3]{x}$

  

Interval: $x=-1$ to $x=1$

Formula used:

Formula for average rate of change of a function,

  $=\frac{f(b)-f(a)}{b-a}$

Calculation:

Formula for average rate of change of $f$ ,

  $=\frac{f(b)-f(a)}{b-a}$

On substituting values,

  $=\frac{f(1)-f(-1)}{1-(-1)}$

So,

  $=\frac{\sqrt[3]{1}-\sqrt[3]{-1}}{1+1}$

So,

  $=\frac{1-(-1)}{1+1}$

On solving,

  $=\frac{1+1}{1+1}$

Therefore ,

  $=\frac{2}{2}$

Therefore , the average rate of change between $x=-1$ to $x=1$ is 1.