Question

f(x)= {1 + x if x ≤ 2 5 − x if x > 2 is not differentiable at x= 2.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:f(x) = \begin{cases} &\sf{x + 1 \: \: when \: x \leqslant 2} \\ &\sf{5 - x \: \: when \: x > 2} \end{cases}\end{gathered}\end{gathered}$

Now, to check differentiability at x = 2, we have to check Left Hand Derivative and Right Hand Derivative.

Consider LHD at x = 2

$\rm :\longmapsto\:LHD = \displaystyle\lim_{x \to 2^-}\dfrac{f(x) - f(2)}{x - 2}$

$\rm \: \: \: = \displaystyle\lim_{x \to 2^-}\dfrac{x + 1 - (2 + 1)}{x - 2}$

$\rm \: \: \: = \displaystyle\lim_{x \to 2^-}\dfrac{x + 1 - 2 - 1}{x - 2}$

$\rm \: \: \: = \displaystyle\lim_{x \to 2^-}\dfrac{x - 2}{x - 2}$

$\rm \: = \: \:1$

$\red{\boxed{\bf\implies \: \bf{ \: LHD = 1 \quad}}}$

Now, Consider RHD at x = 2

$\rm :\longmapsto\: RHD= \displaystyle\lim_{x \to 2^ + }\dfrac{f(x) - f(2)}{x - 2}$

$\rm \: \: = \displaystyle\lim_{x \to 2^ + }\dfrac{5 - x- (2 + 1)}{x - 2}$

$\rm \: \: = \displaystyle\lim_{x \to 2^ + }\dfrac{5 - x- 2 - 1}{x - 2}$

$\rm \: \: = \displaystyle\lim_{x \to 2^ + }\dfrac{5 - x- 3}{x - 2}$

$\rm \: \: = \displaystyle\lim_{x \to 2^ + }\dfrac{2 - x}{x - 2}$

$\rm \: \: = \displaystyle\lim_{x \to 2^ + }\dfrac{ - (x - 2)}{x - 2}$

$\rm \: = \: \: - 1$

$\red{\boxed{\bf\implies \: \bf{ \: RHD = - \: 1 \quad}}}$

So, we concluded that

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf{ \: LHD \: \ne \: RHD}}}$

So, f(x) is not differentiable.

Additional Information :-

1. Every differentiable function is always continuous but every continuous function may or may not be differentiable.

For example, f(x) = | x - 1 | is continuous at x = 1 but fails to differentiable at x = 1.