Question

Describe the intervals on which each function f is continuous.

$f\left(x\right)=\left\{\begin{array}{l}{x}^2+1ifx\le 0\\ 1-xifx>0\end{array}\right.$

Short answer

Discontinuous at $x=1$

Step-by-step solution

Step 1. Given information is

$f\left(x\right)=\left\{\begin{array}{l}{x}^2+1ifx\le 0\\ 1-xifx>0\end{array}\right.$

Step 2. Calculating left and right limits, we get

$$\begin{gathered} \underset{x\to 1^{-}}{\lim }g\left(x\right)=\underset{x\to 1^{-}}{\lim }(x^2+1)=1+1=2 \\ \underset{x\to 1^{+}}{\lim }g\left(x\right)=\underset{x\to 1^{+}}{\lim }(1-x)=1-1=0 \end{gathered}$$

Step 3. Conclusion, 

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

hence it is discontinuous at$x=1$