Chapter 1.3, Problem 23E

a.

To determine

To find: The point of discontinuity.

Answer to Problem 23E

The point of discontinuity occurs at $x=0,x=1$ .

Explanation of Solution

Given the graph:

  

Calculation:

Find the point of discontinuity:

  

Based on the graph, the function is continuous at all points in the domain except at

  $x=0,1$ since these points are not in the domain of the function.

Note that, the graph with empty circles indicates a break in the graph.

Thus, the point of discontinuity occurs at $x=0,x=1$ .

b.

To determine

To find: Whether the function is removal or non-removal discontinuity.

Answer to Problem 23E

The function is removable at $x=0$ and non-removable at $x=1$ .

Explanation of Solution

Given the graph:

  

Calculation:

The function has removable discontinuity at $x=0$ , and it can be continuous if:

  $f\left(0\right)=0$

The left- and right-hand limits are not equal at $x=1$ .

At $x=1$ , there is a jump discontinuity, and it is therefore non-removable.