Answer
Yes, the function is differentiable at $x = 2$.
Work Step by Step
Derivative from the left:
$\lim\limits_{x \to 2^{-}} f(x)$
$= \lim\limits_{x \to 2^{-}} \frac{f(x) - f(c)}{x - c}$
$= \lim\limits_{x \to 2^{-}} \frac{(\frac{1}{2}x + 1) - (\frac{1}{2}(2) + 1)}{x - 2}$
$= \lim\limits_{x \to 2^{-}} \frac{(\frac{1}{2}x + 1) - 2}{x - 2}$
$= \lim\limits_{x \to 2^{-}} \frac{\frac{1}{2}x - 1}{x - 2}$
$= \lim\limits_{x \to 2^{-}} \frac{\frac{1}{2}(x - 2)}{x - 2}$
$= \lim\limits_{x \to 2^{-}} \frac{1}{2}$
$= \frac{1}{2}$
Derivative from the right:
$\lim\limits_{x \to 2^{+}} f(x)$
$= \lim\limits_{x \to 2^{+}} \frac{f(x) - f(c)}{x - c}$
$= \lim\limits_{x \to 2^{+}} \frac{\sqrt {2x} - \sqrt {2(2)}}{x - 2}$
$= \lim\limits_{x \to 2^{+}} \frac{\sqrt {2x} - 2}{x - 2}$
$= \lim\limits_{x \to 2^{+}} (\frac{\sqrt {2x} - 2}{x - 2})(\frac{\sqrt {2x} + 2}{\sqrt {2x} + 2})$
$= \lim\limits_{x \to 2^{+}} \frac{2x - 4}{(x - 2)(\sqrt {2x} + 2)}$
$= \lim\limits_{x \to 2^{+}} \frac{2(x - 2)}{(x - 2)(\sqrt {2x} + 2)}$
$= \lim\limits_{x \to 2^{+}} \frac{2}{\sqrt {2x} + 2}$
$= \frac{2}{\sqrt {2(2)} + 2}$
$= \frac{2}{2 + 2}$
$= \frac{1}{2}$
Since $\lim\limits_{x \to 2^{-}} f(x) = \lim\limits_{x \to 2^{+}} f(x)$, the function $f(x)$ is differentiable at $x = 2$.
