Q. 42 - Calculus

Question

In Exercises 39–44, use Theorem 1.16 and left and right limits to determine whether each function f is continuous at its break point(s). For each discontinuity of f, describe the type of discontinuity and any one-sided discontinuity.

$f\left(x\right) = \left\{\begin{array}{c}\sqrt{-x} , if x<0\\ 2 , if x=0\\ \sqrt{x} , if x>0\end{array}\right. .$

Step-by-Step Solution

Answer

$$\begin{gathered} The given function is not continuous at x=0, it has point discontinuity \\ and it is neither left nor right continuous. \end{gathered}$$

Step 1. Given Information.

Given the function: 

$f\left(x\right) = \left\{\begin{array}{c}\sqrt{-x} , if x<0\\ 2 , if x=0\\ \sqrt{x} , if x>0\end{array}\right. and it has it\text{'}s break point at x=0.$

Step 2. Finding the limits at the break point.

$$\begin{gathered} At x = 0, \\ LHL = \underset{x\to 0^{-}}{\lim } f\left(x\right) = \underset{x\to 0^{-}}{\lim } \sqrt{-x} = \sqrt{-0} = 0. \\ RHL = \underset{x\to 0^{+}}{\lim } f\left(x\right) = \underset{x\to 0^{+}}{\lim } \sqrt{x} = \sqrt{0} = 0. \\ f\left(0\right) = 2. \\ Now, \\ LHL = RHL \ne f\left(0\right). \end{gathered}$$

Step 3. Finding the type of discontinuity.

Since the function is discontinuous at $x=0$ from Step 2.

Now we know LHL = RHL, none is equal to f(0) and both exist then this type of discontinuity is point discontinuity.

And also, $LHL \ne f\left(0\right) and also RHL \ne f\left(0\right)$So, it is neither left continuous nor right continuous.