Chapter 2.3, Problem 23E

(a)

To determine

To find: The points of discontinuity of $y=f\left(x\right)$ .

Answer to Problem 23E

The points of discontinuity of $y=f\left(x\right)$ are $x=0$ and $x=1$ .

Explanation of Solution

Given information:

The graph of the function $y=f\left(x\right)$ is given below:

  

Calculation:

As observed from the graph the function $y=f\left(x\right)$ is continuous for all the values of $x$ between $-1$ and $0$ .

Now check the continuity for $x=0$ .

As observed from the graph of the function $y=f\left(x\right)$ , the left hand and right hand limit of function at $x=0$ are equal to $0$ . So, the value of $\underset{x\to 0}{\lim }f\left(x\right)$ is equal to $0$ .

As observed from the graph, the value of the function $y=f\left(x\right)$ at $x=0$ is equal to $1$ . So, the value of $f\left(0\right)$ is equal to $1$ .

The limit value of $y=f\left(x\right)$ at $x=0$ is not equal to the value of $f\left(0\right)$ .

So, the function $y=f\left(x\right)$ is discontinuous at $x=0$ .

As observed from the graph, the function $y=f\left(x\right)$ is continuous for all the values of $x$ between $0$ and $1$ .

Now check the continuity of the function at $x=1$ .

As observed from the graph, the value of $f\left(x\right)$ approaches to $1$ as $x$ approaches to $1$ from the left side. So, the left hand limit of $f\left(x\right)$ at $x=1$ is equal to $1$ .

As observed from the graph, the value of $f\left(x\right)$ approaches to $0$ as $x$ approaches to $1$ from the right side. So, the right hand limit of $f\left(x\right)$ at $x=1$ is equal to $0$ .

The left hand limit of $f\left(x\right)$ is not equal to the right hand limit at $x=1$ . So, the value of $\underset{x\to 1}{\lim }f\left(x\right)$ doesn’t exist.

So, the function $y=f\left(x\right)$ is discontinuous at $x=1$ .

As observed from the graph the function $y=f\left(x\right)$ is continuous for all the values of $x$ between $1$ and $2$ .

Therefore, the points of discontinuity of $y=f\left(x\right)$ are $x=0$ and $x=1$ .

(b)

To determine

To check: Whether the discontinuity of $y=f\left(x\right)$ is removable or not, explain.

Answer to Problem 23E

The discontinuity of $y=f\left(x\right)$ at $x=0$ is removable and the discontinuity $x=1$ is non removable.

Explanation of Solution

Given information:

The graph of the function $y=f\left(x\right)$ is given below:

  

Calculation:

As calculated in part(a), the value of $\underset{x\to 0}{\lim }f\left(x\right)$ is equal to $0$ and the value of $f\left(0\right)$ is equal to $1$ .

The function $f\left(x\right)$ can be made continuous be setting the value of $f\left(0\right)$ equal to $1$ . So, the point of discontinuity $x=0$ is non removable.

As calculated in part(a), the left hand limit of $f\left(x\right)$ is not equal to the right hand limit at $x=1$ .

So, the point of discontinuity for $f\left(x\right)$ at $x=1$ is non removable.

Therefore, the discontinuity of $y=f\left(x\right)$ at $x=0$ is removable and the discontinuity $x=1$ is non removable.