Chapter 4.1, Problem 34E

To determine

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

Answer to Problem 34E

No extreme values exist.

Explanation of Solution

Given information :

  $k\left(x\right)=\left|x+1\right|+\left|x-3\right|,-\infty <x<\infty$

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 $k\left(x\right)=\left|x+1\right|+\left|x-3\right|$ with respect to $x$ .

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

Thus, the derivative of $k\left(x\right)$ never zero on $ℝ$ .

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

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

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

  $\begin{array}{cc}x& k\left(x\right)\\ -\infty & \infty \\ -1& 4\\ 3& 4\\ \infty & \infty \end{array}$

Hence, no extreme values exist.