Chapter 1.2, Problem 28E

Solution

To identify: local minimum, a local maximum, or neither and function is decreasing and intervals on which it is increasing.

The function $\left(-1,1\right)$ and $\left(3,1\right)$ are local maximum, $\left(1,6\right)$ is a local maximum, and $\left(5,4\right)$ is a local minimum.

The function increases on $\left[-1,1\right]$ , decreases on $(-\infty ,-1]$ , increases on $\left[3,5\right]$ and decreases on $[5,\infty )$ .

Given information: 

The graph

  

Calculation:

Here,

• $\left(-1,1\right)$ and $\left(3,1\right)$ are local minima, because the graph descends until that point and rises after those points.
• $\left(1,6\right)$ is a local maximum, because there is an interval around $\left(1,6\right)$ where all $y-$ values are less than $6.$ Note that the discontinuity doesn't change this.
• $\left(5,4\right)$ is a local maximum, because the graph rises until that point and then starts falling.

We see that the function

• decreases in the interval $(-\infty ,-1]$;
• increases in the interval $\left[-1,1\right]$;
• decreases in the interval $(1,3]$;
• increases in the interval $\left[3,5\right]$;
• decreases in the interval $[5,\infty )$ .