Show that the function $f$ defined as follows,

Solution:

Given:

$f(x)= \begin{cases}3 x-2, & 02\end{cases}$

First, we will show that $f(x)$ is continuos at $x=2$.

We have,

$(\mathrm{LHL}$ at $x=2)$

$=\lim _{x \rightarrow 2^{-}} f(x)$

$=\lim _{h \rightarrow 0} f(2-h)$

$=\lim _{h \rightarrow 0} 2(2-h)^{2}-(2-h)$

$=\lim _{h \rightarrow 0}\left(8+2 h^{2}-8 h-2+h\right)$

$=6$

(RHL at x = 2) 

$=\lim _{x \rightarrow 2^{+}} f(x)$

$=\lim _{h \rightarrow 0} f(2+h)$

$=\lim _{h \rightarrow 0} 5(2+h)-4$

$=\lim _{h \rightarrow 0}(10+5 h-4)$

$=6$

and $f(2)=2 \times 4-2=6$

Thus, $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=f(2)$.

Hence the function is continuous at x=2.

Now, we will check whether the given function is differentiable at x = 2.

We have,

(LHD at x = 2)

$\lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}$

$=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}$

$=\lim _{h \rightarrow 0} \frac{2 h^{2}-7 h+6-6}{-h}$

$=\lim _{h \rightarrow 0}-2 h+7$

$=7$

(RHD at x = 2)

$\lim _{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2}$

$=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$

$=\lim _{h \rightarrow 0} \frac{10+5 h-4-6}{h}$

$=5$

Thus, LHD at x=2 ≠ RHD at x = 2.

Hence, function is not differentiable at x = 2.