Chapter 2.3, Problem 22E

(a)

To determine

To find: The points of discontinuity of $f\left(x\right)=\{\begin{array}{l}1-x^2,x\ne -1\\ 2,x=-1\end{array}$ .

Answer to Problem 22E

The point of discontinuity of $f\left(x\right)=\{\begin{array}{l}1-x^2,x\ne -1\\ 2,x=-1\end{array}$ is $x=-1$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=\{\begin{array}{l}1-x^2,x\ne -1\\ 2,x=-1\end{array}$ .

Calculation:

Check the points of discontinuity of the function $f\left(x\right)=\{\begin{array}{l}1-x^2,x\ne -1\\ 2,x=-1\end{array}$ with the help of graph.

To graph a function $f\left(x\right)=\{\begin{array}{l}1-x^2,x\ne -1\\ 2,x=-1\end{array}$ , follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press $\boxed{\text{Y}=}$ key and enter right hand side of the equation $f\left(x\right)=1-x^2$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{1}\boxed{-}\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{2}\boxed{(-)}\boxed{1}\boxed{)}$ . Now press the $\boxed{\text{ENTER}}$ button and enter right hand side of the equation $f\left(x\right)=2$ after the symbol ${\text{Y}}_2$ . Enter the keystrokes $\boxed{(}\boxed{2}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{1}\boxed{(-)}\boxed{1}\boxed{)}$ .

The display will show the equation,

  $\begin{array}{l}{\text{Y}}_{\text{1}}=\left(1-\text{X}^2\right)/\left(\text{X}\ne -1\right)\\ {\text{Y}}_2=\left(2\right)/\left(\text{X}=-1\right)\end{array}$

Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of given function in standard window as shown in Figure (1).

  

Figure (1)

As observed from the graph, the function $f\left(x\right)$ is continuous for all the points $x<-1$ and also for all the points $x>-1$ .

Now check the continuity for $x=-1$ .

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

As observed from 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 $0$ .

As the left hand limit of $f\left(x\right)$ is equal to the right hand limit at $x=-1$ . So, the value of $\underset{x\to -1}{\lim }f\left(x\right)$ is equal to $0$ .

As the value of $f\left(x\right)$ at the point $x=-1$ is equal to $2$ . So, the value of $f\left(-1\right)=2$ .

As the value of $\underset{x\to -1}{\lim }f\left(x\right)$ is not equal to the value of $f\left(-1\right)$ . So, the function is discontinuous at $x=-1$ .

Therefore, the point of discontinuity of $f\left(x\right)=\{\begin{array}{l}1-x^2,x\ne -1\\ 2,x=-1\end{array}$ is $x=-1$ .

(b)

To determine

To check: Whether the discontinuity of $f\left(x\right)=\{\begin{array}{l}1-x^2,x\ne -1\\ 2,x=-1\end{array}$ is removable or not, explain.

Answer to Problem 22E

The discontinuity of $f\left(x\right)=\{\begin{array}{l}1-x^2,x\ne -1\\ 2,x=-1\end{array}$ at $x=-1$ is removable as the limit value of $f\left(x\right)$ at $x=-1$ and $f\left(-1\right)$ are not equal.

Explanation of Solution

Given information:

The function is $f\left(x\right)=\{\begin{array}{l}1-x^2,x\ne -1\\ 2,x=-1\end{array}$ .

Calculation:

As calculated in part(a), the value of $\underset{x\to -1}{\lim }f\left(x\right)$ is equal to $0$ and the value of $f\left(-1\right)$ is equal to $2$ .

So, the function $f\left(x\right)$ can be made continuous be setting the value of $f\left(-1\right)$ .

Therefore, the discontinuity of $f\left(x\right)=\{\begin{array}{l}1-x^2,x\ne -1\\ 2,x=-1\end{array}$ at $x=-1$ is removable.