Chapter 4.1, Problem 32E

To determine

To find : extreme values of the function and where they occur.

Answer to Problem 32E

Minimum $\left(2,-4\right)$ and maximum $\left(-4,10\right),\left(4,-2\right)$

Explanation of Solution

Given information :

  $\left|2x-4\right|-\left|x+2\right|,-4\le x\le 4$

Formula needed :

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

The graph of the function in the domain is

  

Differentiate expression $\left|2x-4\right|-\left|x+2\right|$ with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}\left(\left|2x-4\right|-\left|x+2\right|\right)=\frac{d}{dx}\left(\left|2x-4\right|\right)-\frac{d}{dx}\left(\left|x+2\right|\right)\\ =\frac{\left|2x-4\right|}{2\left(x-2\right)}-\frac{\left|x+2\right|}{x+2}\end{array}$

Thus, the derivative of $\left|2x-4\right|-\left|x+2\right|$ never zero on $-4\le x\le 4$ .

Therefore, the derivative does not exist at $x=-2$ and, $x=2$ .

The critical points are $x=-2$ and, $x=2$ .

Examine the value of expression at critical points and end points.

  $\begin{array}{cc}x& \left|2x-4\right|-\left|x+2\right|\\ -4& 10\\ -2& 8\\ 2& -4\\ 4& -2\end{array}$

Hence, minimum occur at $x=2$ and maximum occur at end points.