Q. 41 - 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.

$f\left(x\right) = \left\{\begin{array}{c}{x}^2 , if x<2\\ 4 , if x=2\\ 2x+1 , if x>2\end{array} .\right.$

Step-by-Step Solution

Answer

$$\begin{gathered} The given function is not continuous at x=2, \\ it has jump discontiuity and it is left continuous. \end{gathered}$$

Step 1. Given Information.

Given the function:

$f\left(x\right) = \left\{\begin{array}{c}{x}^2 , if x<2\\ 4 , if x=2\\ 2x+1 , if x>2\end{array} and it has it\text{'}s break point at x=2.\right.$

Step 2. Finding the limits at the break point.

$$\begin{gathered} At x=2, \\ LHL = \underset{x\to 2^{-}}{\lim } f\left(x\right) = \underset{x\to 2^{-}}{\lim } x^2 = 2^2 = 4. \\ RHL = \underset{x\to 2^{+}}{\lim } f\left(x\right) = \underset{x\to 2^{+}}{\lim } 2x+1 = 2\left(2\right)+1 = 5. \\ f\left(2\right) = 4. \\ So, LHL = f\left(2\right) \ne RHL. \end{gathered}$$

Step 3. Finding the type of discontinuity.

Since the function is discontinuous only at $x=2,$ 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, LHL = f(2) this means this function is left continuous.