Chapter 4.1, Problem 31E

To determine

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

Answer to Problem 31E

Minimum $\left(1,4\right)$ and maximum $\left(-5,14\right),\left(5,16\right)$

Explanation of Solution

Given information :

  $\left|x+3\right|+\left|2x-2\right|,-5\le x\le 5$

Formula needed :

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

Expand the absolute

  $\left|x+3\right|+\left|2x-2\right|=\left\{\begin{array}{l}-\left(x+3\right)-\left(2x-2\right),-5\le x<-3\\ \left(x+3\right)-\left(2x-2\right),-3\le x<1\\ \left(x+3\right)+\left(2x-2\right),1\le x\le 5\end{array}\right.$

Simply the expression

  $\left|x+3\right|+\left|2x-2\right|=\left\{\begin{array}{l}-3x-1,-5\le x<-3\\ -x+5,-3\le x<1\\ 3x+1,1\le x\le 5\end{array}\right.$

The graph of the function in the domain is

  

Differentiate expression $\left|x+3\right|+\left|2x-2\right|$ with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}\left(\left|x+3\right|+\left|2x-2\right|\right)=\frac{d}{dx}\left(\left|x+3\right|\right)+\frac{d}{dx}\left(\left|2x-2\right|\right)\\ =\frac{\left|x+3\right|}{x+3}+\frac{\left|2x-2\right|}{2\left(x-1\right)}\end{array}$

Thus, the derivative of $\left|x+3\right|+\left|2x-2\right|$ never zero on $-5\le x\le 5$ .

Therefore, the derivative does not exist at $x=-3$ and, $x=1$ .

The critical points are $x=-3$ and, $x=1$ .

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

  $\begin{array}{cc}x& \left|x+3\right|+\left|2x-2\right|\\ -5& 14\\ -3& 8\\ 1& 4\\ 5& 16\end{array}$

Hence, minimum occur at $x=1$ and maximum occur at end points.