Answer
Yes, the function is differentiable at $x = 2$.
Work Step by Step
Derivative from the left:
$\lim\limits_{x \to 2^{-}} f(x)$
$= \lim\limits_{x \to 2^{-}} \frac{f(x) - f(c)}{x - c}$
$= \lim\limits_{x \to 2^{-}} \frac{(x^{2} + 1) - (2^{2} + 1)}{x - 2}$
$= \lim\limits_{x \to 2^{-}} \frac{(x^{2} + 1) - 5}{x - 2}$
$= \lim\limits_{x \to 2^{-}} \frac{(x^{2} - 4)}{x - 2}$
$= \lim\limits_{x \to 2^{-}} \frac{(x + 2)(x - 2)}{x - 2}$
$= \lim\limits_{x \to 2^{-}} x + 2$
$= 4$
Derivative from the right:
$\lim\limits_{x \to 2^{+}} f(x)$
$=\lim\limits_{x \to 2^{+}} \frac{f(x) - f(c)}{x - c}$
$= \lim\limits_{x \to 2^{+}} \frac{(4x - 3) - (4(2) - 3)}{x - 2}$
$= \lim\limits_{x \to 2^{+}} \frac{(4x - 3) - 5}{x - 2}$
$= \lim\limits_{x \to 2^{+}} \frac{4x - 8}{x - 2}$
$= \lim\limits_{x \to 2^{+}} \frac{4(x - 2)}{x - 2}$
$= \lim\limits_{x \to 2^{+}} 4$
$= 4$
Since $\lim\limits_{x \to 2^{-}} f(x) = \lim\limits_{x \to 2^{+}} f(x)$, the function $f(x)$ is differentiable at x = 2.
