## Calculus: Early Transcendentals 8th Edition

The function is discontinuous at $x=4$ $f$ is continuous from the left at $x = 4$
$f(1) = 2^1 = 2$ $f(4) = 3-(4) = -1$ 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 $x = 4$