Chapter 2.3, Problem 19E

(a)

To determine

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

Answer to Problem 19E

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

Explanation of Solution

Given information:The function is $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ \frac{x}{2}+1,x>2\end{array}$ .

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 $f\left(x\right)=\frac{x}{2}+1$ .

As the function $f\left(x\right)=\frac{x}{2}+1$ 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 $f\left(x\right)=\frac{x}{2}+1$ .

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

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

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

As the left hand limit and right hand limit of $f\left(x\right)$ are not equal. So, the value of $\underset{x\to 2}{\lim }f\left(x\right)$ doesn’t exist.

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

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

(b)

To determine

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

Answer to Problem 19E

The discontinuity of $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ \frac{x}{2}+1,x>2\end{array}$ at $x=2$ is non-removable as the left hand limit is not equal to the right hand limit.

Explanation of Solution

Given information:The function is $f\left(x\right)=\{\begin{array}{l}3-x,x<2\\ \frac{x}{2}+1,x>2\end{array}$ .

Calculation:

As calculated in part(a), the left hand limit of $f\left(x\right)$ at $x=2$ is equal to $1$ and the right hand limit of $f\left(x\right)$ at $x=2$ is equal to $2$ .

So, the left hand limit is not equal to the right hand limit of $f\left(x\right)$ at $x=2$ .

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