Explanation

(c)
LHL$= \lim_{x \rightarrow 2^{-}}{f(x) = \lim_{h \rightarrow 0}{1 + (2 - h) = 3}}$
RHL$= \lim_{x \rightarrow 2^{+}}{f(x) = \lim_{h \rightarrow 0}{5 - (2 + h) = 3,\ \ \ \ f(2) = 3}}$
Hence, $f$ is continuous at $x = 2$
Now, $Rf^{''}(2) = \lim_{h \rightarrow 0}\ \frac{f(2 + h) - f(2)}{h}$
$= \lim_{h \rightarrow 0}\frac{5 - (2 + h) - 3}{h} = - 1$
$Lf^{''}(2) = \lim_{h \rightarrow 0}\frac{f(2 - h) - f(2)}{- h}$
$= \lim_{h \rightarrow 0}\ \frac{1 + (2 - h) - 3}{- h} = 1$
$\therefore\ \ Rf^{''}(2) \neq Lf^{''}(2)$
$\therefore\ \ f$ is not differentiable at $x = 2$
Alternate
[image:5_image1.emf]
It is clear from the graph that $f(x)$ is continuous everywhere also it is differentiable everywhere except at $x = 2$