# Chapter 2 - Differentiation - 2.1 Exercises - Page 105: 89

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.

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