Examine the differentiability of the function $f(x) = \begin{cases} 1 + x, & \text{if } x \leq 2 \\ 5 - x, & \text{if } x > 2 \end{cases} \text{ at } x = 2$.

Not differentiable at $x = 2$ because $\text{LHD} = 1$ and $\text{RHD} = -1$.

The correct answer is Option (2) → Not differentiable at $x = 2$ because $\text{LHD} = 1$ and $\text{RHD} = -1$. ##

We have, $f(x) = \begin{cases} 1 + x, & \text{if } x \leq 2 \\ 5 - x, & \text{if } x > 2 \end{cases} \text{ at } x = 2$

For differentiability at $x = 2$,

$Lf'(2) = \lim\limits_{x \to 2^-} \frac{f(x) - f(2)}{x - 2} = \lim\limits_{x \to 2^-} \frac{(1 + x) - (1 + 2)}{x - 2}$

Put $x = 2 - h$

$= \lim\limits_{h \to 0} \frac{(1 + 2 - h) - 3}{2 - h - 2} = \lim\limits_{h \to 0} \frac{-h}{-h} = 1$

$Rf'(2) = \lim\limits_{x \to 2^+} \frac{f(x) - f(2)}{x - 2} = \lim\limits_{x \to 2^+} \frac{(5 - x) - 3}{x - 2}$

Put $x = 2 + h$

$= \lim\limits_{h \to 0} \frac{5 - (2 + h) - 3}{2 + h - 2} = \lim\limits_{h \to 0} \frac{5 - 2 - h - 3}{h} = \lim\limits_{h \to 0} \frac{-h}{h}$

$= -1$

$∵Lf'(2) \neq Rf'(2)$

So, $f(x)$ is not differentiable at $x = 2$.