Given information
We are given with following Piecewise function
\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{x + 2}&{x < 0}\\{{e^x}}&{0 \le x \le 1}\\{2 - x}&{x > 1}\end{array}} \right.\]We have to find the numbers at which function is discontinuous.


Step 1:
The given piecewise functions are polynomial and exponential. Both are continuous function. So we just need to check the continuity at break points $x=0$ and $x=1$

Step 2: Continuity at x=0
The Left hand limit\[\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {x + 2} \right)\\ = \left( {0 + 2} \right)\\ = 2\end{array}\]The Right hand limit\[\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {{e^x}} \right)\\ = \left( {{e^0}} \right)\\ = 1\end{array}\]Since left hand limit is not equal to right hand limit. Function is discontinuous at x=0. Therefore, $x=0$

Step 3: Continuity at x=1
The Left hand limit\[\begin{array}{l}\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {{e^x}} \right)\\ = \left( {{e^1}} \right)\\ = e\end{array}\]The Right hand limit\[\begin{array}{l}\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {2 - x} \right)\\ = \left( {2 - 1} \right)\\ = 1\end{array}\]Since left hand limit is not equal to right hand limit. Function is discontinuous at x=0. Therefore, $x=1$The plot of the function is:

https://imgur.com/IAxLn9v



Discontinous points
$x=1$ and $x=0$