Question

Describe the intervals on which each function f is continuous.

$f\left(x\right)=\left\{\begin{array}{l}3x+2ifx<-1\\ 5+4x^{3}ifx\ge -1\end{array}\right.$

Short answer

Discontinuous at$x=-1$

Step-by-step solution

Step 1. Given information is

$f\left(x\right)=\left\{\begin{array}{l}3x+2ifx<-1\\ 5+4x^{3}ifx\ge -1\end{array}\right.$

Step 2. Calculating left and right limits, we get

$$\begin{gathered} \underset{x\to 1^{-}}{\lim }g\left(x\right)=\underset{x\to 1^{-}}{\lim }(3x+2)=3(-1)+2=-3+2=-1 \\ \underset{x\to 1^{+}}{\lim }g\left(x\right)=\underset{x\to 1^{+}}{\lim }(5+4x^3)=5+4(-1)=1 \end{gathered}$$

Step 3. Conclusion 

Since the left and right limits, both exist but are not equal as $x\to 1$

hence it is discontinuous at$x=-1$