Answer
The function is discontinuous at $x=4$
$f$ is continuous from the left at $x = 4$
Work Step by Step
$f(1) = 2^1 = 2$
$f(4) = 3-(4) = -1$
We can check if the function is continuous at $x=1$:
$\lim\limits_{x \to 1^-}f(x) = 2^1 = 2 = f(1)$
$\lim\limits_{x \to 1^+}f(x) = 3-(1) = 2 = f(1)$
$\lim\limits_{x \to 1}f(x) = 2 = f(1)$
The function is continuous at $x = 1$
We can check if the function is continuous at $x=4$:
$\lim\limits_{x \to 4^-}f(x) = 3-(4) = -1 = f(4)$
$\lim\limits_{x \to 4^+}f(x) = \sqrt{4} = 2 \neq f(4)$
$\lim\limits_{x \to 4^-}f(x) \neq \lim\limits_{x \to 4^+}f(x)$
$\lim\limits_{x \to 4}f(x)$ does not exist.
Therefore, the function is discontinuous at $x=4$
Since $\lim\limits_{x \to 4^-}f(x) = f(4)$, $f$ is continuous from the left at $x = 4$
