Chapter 2.3, Problem 20E

(a)

To determine

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

Answer to Problem 20E

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

Explanation of Solution

Given information:The function is $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ 2,x=2\\ \frac{x}{2},x>2\end{array}$ .

Calculation:

The function corresponding to all the values of $x<2$ is $f\left(x\right)=3-x$ .

As the function $f\left(x\right)=3-x$ is a polynomial function. So, the function $f\left(x\right)$ is continuous for all the values of $x<2$ .

The function corresponding to all the values of $x>2$ is $f\left(x\right)=\frac{x}{2}$ .

As the function $f\left(x\right)=\frac{x}{2}$ is a polynomial function. So, the function $f\left(x\right)$ is continuous for all the values of $x>2$ .

Now for the continuity of $f\left(x\right)$ at $x=2$ the condition is $\underset{x\to 2}{\lim }f\left(x\right)=f\left(2\right)$ .As the function corresponding to all the values of $x<2$ is $f\left(x\right)=3-x$ .

Find the left hand limit of $f\left(x\right)$ at $x=2$ .

  $\begin{array}{c}\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }\left(3-x\right)\\ =3-2\\ =1\end{array}$

So, the left hand limit of $f\left(x\right)$ at $x=2$ is equal to $1$ .

As the function corresponding to all the values of $x>2$ is $f\left(x\right)=\frac{x}{2}$ .

Find the right hand limit of $f\left(x\right)$ at $x=2$ .

  $\begin{array}{c}\underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }\left(\frac{x}{2}\right)\\ =\frac{2}{2}\\ =1\end{array}$

So, the right hand limit of $f\left(x\right)$ at $x=2$ is equal to $1$ .

As the left hand limit and right hand limit of $f\left(x\right)$ are equal. So, the value of $\underset{x\to 2}{\lim }f\left(x\right)$ is equal to $1$ .

The value of function $f\left(x\right)$ is equal to $2$ for $x=2$ . So, the value of $\underset{x\to 2}{\lim }f\left(x\right)$ is not equal to $f\left(2\right)$ .

So, the function is discontinuous at $x=2$ .

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

(b)

To determine

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

Answer to Problem 20E

The discontinuity of $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ 2,x=2\\ \frac{x}{2},x>2\end{array}$ at $x=2$ is removable as the limit value at $x=2$ is not equal to $f\left(2\right)$ .

Explanation of Solution

Given information:The function is $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ 2,x=2\\ \frac{x}{2},x>2\end{array}$ .

Calculation:

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

So, the value of $\underset{x\to 2}{\lim }f\left(x\right)$ is not equal to $f\left(2\right)$ .

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