Q. 40 - 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.

$f\left(x\right) = \left\{\begin{array}{l}(x-3) , if x<0\\ -(x-3) , if x⩾0\end{array} .\right.$

Step-by-Step Solution

Answer

$$\begin{gathered} The given function is not continuous at x=0, \\ it has jump discontinuity and it is right continuous. \end{gathered}$$

Step 1. Given Information.

Given the function;

$f\left(x\right) = \left\{\begin{array}{l}(x-3) , if x<0\\ -(x-3) , if x⩾0\end{array} and it has it\text{'}s break point at x=0.\right.$

Step 2. Finding the limits at the break point.

$$\begin{gathered} At x=0, \\ LHL = \underset{x\to 0^{-}}{\lim } f\left(x\right) = \underset{x\to 0^{-}}{\lim } (x-3) = 0-3 =-3. \\ RHL = \underset{x\to 0^{+}}{\lim } f\left(x\right) = \underset{x\to 0^{+ }}{\lim } -(x-3) = -(0-3) = 3. \\ f\left(0\right) = -(x-3) = -(0-3) = 3. \\ So, \\ RHL = f\left(0\right) \ne LHL. \end{gathered}$$

Step 3. Finding the type of discontinuity.

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

Now we know $LHL\ne RHL$ this means both left and righthand limit exists but they are not equal so this is jump discontinuity.

And also, RHL = f(0) this means this function is right continuous.