Chapter 2.3, Problem 21E

(a)

To determine

To find:The points of discontinuity of $f\left(x\right)=\{\begin{array}{l}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}$ .

Answer to Problem 21E

The point of discontinuity of $f\left(x\right)=\{\begin{array}{l}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}$ is $x=1$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=\{\begin{array}{l}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}$ .

Calculation:

Check the points of discontinuity of the function $f\left(x\right)=\{\begin{array}{l}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}$ with the help of graph.

To graph a function $f\left(x\right)=\{\begin{array}{l}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}$ , 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 $f\left(x\right)=\frac{1}{x-1}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{1}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{-}\boxed{1}\boxed{)}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{5}\boxed{1}\boxed{)}$ . Now press the $\boxed{\text{ENTER}}$ buttonand enter right hand side of the equation $f\left(x\right)=x^3-2x+5$ after the symbol ${\text{Y}}_2$ . Enter the keystrokes $\boxed{(}\boxed{\text{X}}\boxed{^}\boxed{3}\boxed{-}\boxed{2}\boxed{\text{X}}\boxed{+}\boxed{5}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{4}\boxed{1}\boxed{)}$ .

The display will show the equation,

  $\begin{array}{l}{\text{Y}}_{\text{1}}=\left(1/\left(\text{X}-1\right)\right)/\left(\text{X}<1\right)\\ {\text{Y}}_2=\left(\text{X}^3-2\text{X}+5\right)/\left(\text{X}\ge 1\right)\end{array}$

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

As observed from graph, the value of $f\left(x\right)$ approaches to $4$ as $x$ approaches to $1$ from the right side. So, the right hand limit of $f\left(x\right)$ at $x=1$ is $4$ .

As 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 $f\left(x\right)$ is discontinuous at the point $x=1$ .

Therefore, the point of discontinuity of $f\left(x\right)=\{\begin{array}{l}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}$ is $x=1$ .

(b)

To determine

To check:Whether the discontinuity of $f\left(x\right)=\{\begin{array}{l}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}$ is removable or not, explain.

Answer to Problem 21E

The discontinuity of $f\left(x\right)=\{\begin{array}{l}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}$ at $x=1$ isnon removable as the left hand limit and right hand limit at $x=1$ are not equal.

Explanation of Solution

Given information:

The function is $f\left(x\right)=\{\begin{array}{l}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}$ .

Calculation:

As calculated in part(a), the value of $\underset{x\to 1^{-}}{\lim }f\left(x\right)$ is equal to $-\infty$ and the value of $\underset{x\to 1^{+}}{\lim }f\left(x\right)$ is equal to $4$ .

So, the value of $\underset{x\to 1}{\lim }f\left(x\right)$ doesn’t exist.

Therefore, the discontinuity of $f\left(x\right)=\{\begin{array}{l}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}$ at $x=1$ is non removable.