Chapter 4.1, Problem 52E

(a)

To determine

To find : $f^{\prime }\left(0\right)$ does exist or not.

Answer to Problem 52E

  $f^{\prime }\left(0\right)$ does not exist.

Explanation of Solution

Given information :

  $f\left(x\right)=\left|x^3-9x\right|$

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

Differentiate $f\left(x\right)=\left|x^3-9x\right|$ with respect to $x$ .

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left|x^3-9x\right|\\ =\frac{\left(x^3-9x\right)}{\left|x^3-9x\right|}\cdot \frac{d}{dx}\left(x^3-9x\right)\left[\frac{d}{dx}\left|u\right|=\frac{u}{\left|u\right|}\cdot \frac{d}{dx}\left(u\right)\right]\\ =\frac{\left(x^3-9x\right)}{\left|x^3-9x\right|}\cdot \left(3x^2-9\right)\\ =\frac{\left(x^3-9x\right)\left(3x^2-9\right)}{\left|x^3-9x\right|}\end{array}$

So, the derivative of $f\left(x\right)=\left|x^3-9x\right|$ is $f^{\prime }\left(x\right)=\left\{\begin{array}{l}-\left(3x^2-9\right),x<-3\&0<x<3\\ \left(3x^2-9\right),-3<x<0\&x>3\end{array}\right.$

As $x$ approach to 0 from left hand side, the value of function $f^{\prime }\left(x\right)$ tends to $-9$ .

As $x$ approach to 0 from right hand side, the value of function $f^{\prime }\left(x\right)$ tends to $9$ .

So, the derivative of function $f$ does not exist at $x=0$ .

  

(b)

To determine

To find : $f^{\prime }\left(3\right)$ does exist or not.

Answer to Problem 52E

  $f^{\prime }\left(3\right)$ does not exist.

Explanation of Solution

Given information :

  $f\left(x\right)=\left|x^3-9x\right|$

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

The derivative of $f\left(x\right)=\left|x^3-9x\right|$ is $f^{\prime }\left(x\right)=\left\{\begin{array}{l}-\left(3x^2-9\right),x<-3\&0<x<3\\ \left(3x^2-9\right),-3<x<0\&x>3\end{array}\right.$

As $x$ approach to 3 from left hand side, the value of function $f^{\prime }\left(x\right)$ tends to $-18$ .

As $x$ approach to 3 from right hand side, the value of function $f^{\prime }\left(x\right)$ tends to $18$ .

So, the derivative of function $f$ does not exist at $x=3$ .

  

(c)

To determine

To find : $f^{\prime }\left(-3\right)$ does exist or not.

Answer to Problem 52E

  $f^{\prime }\left(-3\right)$ does not exist.

Explanation of Solution

Given information :

  $f\left(x\right)=\left|x^3-9x\right|$

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

The derivative of $f\left(x\right)=\left|x^3-9x\right|$ is $f^{\prime }\left(x\right)=\left\{\begin{array}{l}-\left(3x^2-9\right),x<-3\&0<x<3\\ \left(3x^2-9\right),-3<x<0\&x>3\end{array}\right.$

As $x$ approach to $-3$ from left hand side, the value of function $f^{\prime }\left(x\right)$ tends to $-18$ .

As $x$ approach to $-3$ from right hand side, the value of function $f^{\prime }\left(x\right)$ tends to $18$ .

So, the derivative of function $f$ does not exist at $x=-3$ .

  

(d)

To determine

To find : all extreme values of $f$ .

Answer to Problem 52E

Maximum occur at $\left(-\sqrt{3},6\sqrt{3}\right)$ and, $\left(\sqrt{3},6\sqrt{3}\right)$ .

Minimum occurs at $\left(-3,0\right),\left(0,0\right)$ and, $\left(3,0\right)$ .

Explanation of Solution

Given information :

  $f\left(x\right)=\left|x^3-9x\right|$

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

The derivative of $f\left(x\right)=\left|x^3-9x\right|$ is $f^{\prime }\left(x\right)=\left\{\begin{array}{l}-\left(3x^2-9\right),x<-3\&0<x<3\\ \left(3x^2-9\right),-3<x<0\&x>3\end{array}\right.$

For critical points, take $f^{\prime }\left(x\right)=0$ .

  $\begin{array}{c}3x^2-9=0\\ 3\left(x^2-3\right)=0\\ x^2-{\left(\sqrt{3}\right)}^2=0\left[3\ne 0\right]\\ \left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\end{array}$

Solving further,

  $x=-\sqrt{3},\sqrt{3}$

As $x$ approach to $-\sqrt{3}$ from both ends, the function $f\left(x\right)$ decreases. So, the maximum occurs at $x=-\sqrt{3}$ .

As $x$ approach to $\sqrt{3}$ from both ends, the function $f\left(x\right)$ decreases. So, the maximum occurs at $x=\sqrt{3}$ .

Therefore, maximum occur at $\left(-\sqrt{3},6\sqrt{3}\right)$ and, $\left(\sqrt{3},6\sqrt{3}\right)$ .

Hence, the minimum occurs at $\left(-3,0\right),\left(0,0\right)$ and, $\left(3,0\right)$ .