Q. 39 - 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.

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

Step-by-Step Solution

Answer

$$\begin{gathered} The given function is continuous at all points in its domain \\ and it is both left and right continuous at x=3. \end{gathered}$$

Step 1. Given Information.

Given the function:

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

Step 2. Finding the limits at the break point.

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

Step 3. Finding the type of discontinuity.

Since the function is not discontinuous anywhere so no question arises for the type of discontinuity and as it is continuous all over in its domain therefore, it is both left continuous and right continuous at x=3.