Chapter 3.1, Problem 26E

(a)

To determine

To find: the graph of the derivative of the function.

Answer to Problem 26E

The graph of the required function is shown below.

Explanation of Solution

Given:

The given graph of the given function is shown below,

  

Find the slope of the function between $\left(-4,0\right)$ and $\left(0,2\right)$ .

  $\begin{array}{c}{m}_1=\frac{2-0}{0-\left(-4\right)}\\ =\frac{2}{4}\\ =\frac{1}{2}\end{array}$

Find the slope of the function between $\left(0,2\right)$ and $\left(1,-2\right)$ .

  $\begin{array}{c}{m}_2=\frac{-2-2}{1-0}\\ =\frac{-4}{1}\\ =-4\end{array}$

Find the slope of the function between $\left(1,-2\right)$ and $\left(4,-2\right)$ .

  $\begin{array}{c}{m}_3=\frac{4-1}{-2-\left(-2\right)}\\ =\frac{3}{-2+2}\\ =0\end{array}$

Find the slope of the function between $\left(4,-2\right)$ and $\left(6,2\right)$ .

  $\begin{array}{c}{m}_4=\frac{2-\left(-2\right)}{6-4}\\ =\frac{4}{2}\\ =2\end{array}$

So, the function can be written as,

  $f\text{'}\left(x\right)=\{\begin{array}{l}\frac{1}{2}-4<x<0\\ -40<x<1\\ 01<x<4\\ 24<x<6\end{array}$

So, the graph of the above function is shown below.

  

Therefore, the graph of the required function is shown above.

(b)

To determine

To find: the value between $x=-4$ and $x=6$ of the function at which function is not differentiable.

Answer to Problem 26E

The function is not differentiable at the points $x=0,1,4$ .

Explanation of Solution

Given:

The general equation is $y=\left(x-a\right)\left(x+b\right)$ .

Consider the function.

  $f\text{'}\left(x\right)=\{\begin{array}{l}\frac{1}{2}-4<x<0\\ -40<x<1\\ 01<x<4\\ 24<x<6\end{array}$

As it clear from the function that at the points $x=0,1,4$ , the left-hand side derivative are different from the right hand side derivative.

So, the function is not differentiable at the points $x=0,1,4$ .

Therefore, the function is not differentiable at the points $x=0,1,4$ .