Answer
The function $f$ is discontinuous at $x=0$
The function $f$ is discontinuous at $x=1$
$f$ is continuous from the right at $x = 0$
$f$ is continuous from the left at $x = 1$
Work Step by Step
$f(0) = e^0 = 1$
$f(1) = e^1 = e$
We can check if the function is continuous at $x=0$:
$\lim\limits_{x \to 0^-}f(x) = (0)+2 = 2 \neq f(0)$
$\lim\limits_{x \to 0^+}f(x) = e^0 = 1 = f(0)$
$\lim\limits_{x \to 0^-}f(x) \neq \lim\limits_{x \to 0^+}f(x)$
$\lim\limits_{x \to 0}f(x)$ does not exist.
The function is discontinuous at $x=0$
However, since $\lim\limits_{x \to 0^+}f(x) = f(0)$, $f$ is continuous from the right at $x = 0$
We can check if the function is continuous at $x=1$:
$\lim\limits_{x \to 1^-}f(x) = e^1 = e = f(1)$
$\lim\limits_{x \to 1^+}f(x) = 2-(1) = 1 \neq f(1)$
$\lim\limits_{x \to 1^-}f(x) \neq \lim\limits_{x \to 1^+}f(x)$
$\lim\limits_{x \to 1}f(x)$ does not exist.
The function is discontinuous at $x=1$
However, since $\lim\limits_{x \to 1^-}f(x) = f(1)$, $f$ is continuous from the left at $x = 1$
