Q. 43 - 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.

$f\left(x\right) = \left\{\begin{array}{c}x+1, if x<1\\ 3x-1, if1⩽x<2 \\ x+2 , if x⩾2\end{array}\right. .$

Step-by-Step Solution

Answer

$$\begin{gathered} The function is discontinuous at x=2 only and this is jump discontinuity \\ and the function is right continuous at x=2. \end{gathered}$$

Step 1. Given Information.

Given the function:

$f\left(x\right) = \left\{\begin{array}{c}x+1, if x<1\\ 3x-1, if1⩽x<2 \\ x+2 , if x⩾2\end{array}\right. and its break points which are x=1,2.$

Step 2. Finding the limits at the break points.

$$\begin{gathered} At x=1, \\ LHL = \underset{x\to 1^{-}}{\lim }f\left(x\right) = \underset{x\to 1^{-}}{\lim } x+1 = 1+1 = 2. \\ RHL = \underset{x\to 1^{+}}{\lim }f\left(x\right) = \underset{x\to 1^{+}}{\lim } 3x-1 = 3-1 = 2. \\ f\left(1\right) = 3\left(1\right)-1 = 3-1 = 2. \\ Since, \\ LHL = RHL = f\left(1\right). \\ This means it is continuous at x=1. \end{gathered}$$

$$\begin{gathered} Now at x=2, \\ LHL = \underset{x\to 2^{-}}{\lim } f\left(x\right) = \underset{x\to 2^{-}}{\lim } 3x-1 = 6-1 = 5. \\ RHL = \underset{x\to 2^{+}}{\lim } f\left(x\right) = \underset{x\to 2^{+}}{\lim } x+2 = 2+2 = 4. \\ f\left(2\right) = x+2 = 2+2 = 4. \\ So, \\ RHL = f\left(2\right) \ne LHL. \end{gathered}$$

Step 3. Finding the type of discontinuity.

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

Now we know $RHL \ne LHL and RHL = f\left(2\right)$ this means both left and righthand limit exists but they are not equal so this is jump discontinuity.

And also, RHL = f(2) this means this function is right continuous.