Question

If f (x) = |x − 2| write whether f' (2) exists or not.

Answer 1

Given:  

`f(x) = |x -2| = {(x-2,x>2,),(-x+2 , x le 2,):}`

Now,
(LHD at x = 2)

\[\lim_{x \to 2^-} \frac{f(x) - f(2)}{x - 2} \]
\[ = \lim_{h \to 0} \frac{f(2 - h) - f(2)}{2 - h - 2} \]
\[ = \lim_{h \to 0} \frac{( - 2 + h + 2) - 0}{- h} \]
\[ = - 1\]

(RHD at x = 2)

\[\lim_{x \to 2^+} \frac{f(x) - f(2)}{x - 2} \]
\[ = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{2 + h - 2} \]
\[ = \lim_{h \to 0} \frac{2 + h - 2 - 0}{h}\]
\[ = 1\]

Thus, (LHD at x = 2) ≠ (RHD at x = 2)

Hence, \[\lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} = f'(2)\] does not exist.