Chapter 5.3, Problem 22E

a.

To determine

To estimate : Where $f\text{'}$ is zero, positive and negative.

Answer to Problem 22E

  $f\text{'}$ is zero at $x=\pm 1.25$ and $x=0$ , $f\text{'}$ is positive for $\left(-1.25,0\right)$ and $\left(1.25,\infty \right)$ and $f\text{'}$ is negative for $\left(-\infty ,-1.25\right)$ and $\left(0,1.25\right)$ .

Explanation of Solution

Given information :

The graph of the function is provided in the question.

Calculation:

  $f\text{'}$ will be zero where the slope of the tangents to the graph is zero. That is when tangent is horizontal to x axis.

Since, the tangents are horizontal at $x=\pm 1.25$ and $x=0$ . Therefore,

  $f\text{'}$ is zero at $x=\pm 1.25$ and $x=0$ .

  $f\text{'}$ will be positive where the slope of the tangents to the graph is positive.

Since, the slope is positive at $-1.25<x<0$ and $1.25<x<\infty$ . Therefore,

  $f\text{'}$ is positive for $\left(-1.25,0\right)$ and $\left(1.25,\infty \right)$ .

Also, $f\text{'}$ will be negative where the slope of the tangents to the graph is negative.

Since, the slope is negative for $-\infty <x<-1.25$ and $0<x<1.25$ . Therefore,

  $f\text{'}$ is negative for $\left(-\infty ,-1.25\right)$ and $\left(0,1.25\right)$ .

Hence,

  $f\text{'}$ is zero at $x=\pm 1.25$ and $x=0$ , $f\text{'}$ is positive for $\left(-1.25,0\right)$ and $\left(1.25,\infty \right)$ and $f\text{'}$ is negative for $\left(-\infty ,-1.25\right)$ and $\left(0,1.25\right)$ .

b.

To determine

To estimate : Where $f\text{'}\text{'}$ is zero, positive and negative.

Answer to Problem 22E

  $f\text{'}\text{'}$ is zero at $x=0$ , $f\text{'}\text{'}$ is positive for $\left(1,\infty \right)$ and $f\text{'}\text{'}$ is negative for $\left(-\infty ,\text{ 0}\right)$ .

Explanation of Solution

Given information :

The graph of the function is provided in the question.

Calculation:

  $f\text{'}\text{'}$ will be zero where the concavity changes from down to up.

Since, the concavity changes at $x=0$ and $x=\pm 1.25$ . Therefore,

  $f\text{'}\text{'}$ is zero at $x=0$ and $x=\pm 1.25$ .

  $f\text{'}\text{'}$ will be positive where the concavity is up.

Since, the concavity is up for $-\infty <x<0$ and $0<x<\infty$ . Therefore,

  $f\text{'}\text{'}$ is positive for $\left(-\infty ,\text{ 0}\right)$ and $\left(\text{0, }\infty \right)$ .

  $f\text{'}\text{'}$ will be negative where the concavity is down.

Since, the concavity is down for $-1.25<x<1.25$ . Therefore,

  $f\text{'}\text{'}$ is negative for $\left(-1.25,\text{ 1}\text{.25}\right)$ .

Hence,

  $f\text{'}\text{'}$ is zero at $x=0$ and $x=\pm 1.25$ , $f\text{'}\text{'}$ is positive for $\left(-\infty ,\text{ 0}\right)$ and $\left(\text{0, }\infty \right)$ and $f\text{'}\text{'}$ is negative for $\left(-1.25,\text{ 1}\text{.25}\right)$ .