Q. 73 - Calculus

Question

Describe the intervals on which each function f is continuous. At each point where f fails to be

continuous, use limits to determine the type of discontinuity

and any left- or right-continuity.

$f\left(x\right)=\left\{\begin{array}{l}{x}^2-3x-1, if x\ne -2\\ -3, if x=-2\end{array}\right.$

Step-by-Step Solution

Answer

The interval on which the function is continuous is -2

Step 1. Given information

The given function is $f\left(x\right)=\left\{\begin{array}{l}{x}^2-3x-1, if x\ne -2\\ -3, if x=-2\end{array}\right.$

Step 2. calculating left and right limits

$$\begin{gathered} \underset{x\ne -2}{\lim }f\left(x\right)=\underset{x\ne -2}{\lim }{x}^2-3x-1 \\ ={\left(-2\right)}^2-3\left(-2\right)-1 \\ =4+6-1 \\ =9 \end{gathered}$$

$$\begin{gathered} \underset{x=-2}{\lim }f\left(x\right)=\underset{x=-2}{\lim }{x}^2-3x-3 \\ ={\left(-2\right)}^2-3\left(-2\right)-3 \\ =4+6-3 \\ =7 \end{gathered}$$

since both left and right limits both exists but not equal

hence it is discontinuous at $-2$ 

Step 3. The solution

The interval of function at which it is discontinuous is $-2$