Chapter 4.3, Problem 22E

(a)

To determine

To estimate: Where $f^{\prime }$ is 0, positive, and negative.

Answer to Problem 22E

  $f^{\prime }$ is 0 at $x=0$ and $x=\pm 1.25$

  $f^{\prime }$ is positive on $\left(-1.25,0\right)$ and $\left(1.25,\infty \right)$

  $f^{\prime }$ is negative on $\left(-\infty ,-1.25\right)$ and $\left(0,1.25\right)$

Explanation of Solution

Given information:

Concept used: If any function $f\left(x\right)$ is increasing in any interval $I$ , then $f^{\prime }\left(x\right)$ is positive in that interval.

Similarly, if any function $f\left(x\right)$ is decreasing in any interval $I$ , then $f^{\prime }\left(x\right)$ is negative in that interval.

Also, if $f\left(x\right)$ has a relative maximum or relative minimum at the point $x=p$ , then at that point $f^{\prime }\left(x\right)$ is 0.

Calculation:

It is observed from the given graph that $f\left(x\right)$ has a relative maximum and relative minimum at the point $x=0$ , $x=-1.25$ and $x=1.25$ , respectively. So, this implies that $f^{\prime }\left(x\right)$ is 0 at $x=0$ and $x=\pm 1.25$ .

It is also observed from the given graph that $f\left(x\right)$ is increasing in the interval $\left(-1.25,0\right)$ and $\left(1.25,\infty \right)$ . Hence, $f^{\prime }\left(x\right)$ is positive on $\left(-1.25,0\right)$ , and $\left(1.25,\infty \right)$ .

Similarly, it can be observed from the given graph that $f\left(x\right)$ is decreasing in the interval $\left(-\infty ,-1.25\right)$ and $\left(0,1.25\right)$ . Hence, $f^{\prime }\left(x\right)$ is negative on $\left(-\infty ,-1.25\right)$ , and $\left(0,1.25\right)$ .

(b)

To determine

To estimate: Where $f^{″}$ is 0, positive, and negative.

Answer to Problem 22E

  $f^{″}$ is 0 at $x=\pm 0.7$

  $f^{″}$ is positive on $\left(-\infty ,-0.7\right)$ and $\left(0.7,\infty \right)$ .

  $f^{″}$ is negative on $\left(-0.7,0.7\right)$

Explanation of Solution

Given information:

  

Concept used: If any function $f\left(x\right)$ is concave up in any interval $I$ , then $f^{″}\left(x\right)$ is positive in that interval.

Similarly, if any function $f\left(x\right)$ is concave down in any interval $I$ , then $f^{″}\left(x\right)$ is negative in that interval.

Also, if $f\left(x\right)$ has a inflection point at $x=p$ , then at that point $f^{″}\left(x\right)$ is 0.

Calculation:

It is observed from the given graph that $f\left(x\right)$ has an inflection point at $x=0.7$ and $x=-0.7$ . So, this implies that $f^{″}\left(x\right)$ is 0 at $x=\pm 0.7$ .

It is also observed from the given graph that $f\left(x\right)$ is concave up in the interval $\left(-\infty ,-0.7\right)$ and $\left(0.7,\infty \right)$ . Hence, $f^{″}\left(x\right)$ is positive on $\left(-\infty ,-0.7\right)$ , and $\left(0.7,\infty \right)$ .

Similarly, it can be observed from the given graph that $f\left(x\right)$ is concave down in the interval $\left(-0.7,0.7\right)$ . Hence, $f^{″}\left(x\right)$ is negative on $\left(-0.7,0.7\right)$ .