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Solution:

Given: $f(x)= \begin{cases}12 x-13, & x \leq 3 \\ 2 x^{2}+5, & x>3\end{cases}$

We have to show that the given function is differentiable at x = 3.

$(\mathrm{LHD}$ at $x=3)=\lim _{x \rightarrow 3^{-}} \frac{f(x)-f(3)}{x-3}$'

$=\lim _{x \rightarrow 3} \frac{12 x-13-23}{x-3}$

$=\lim _{x \rightarrow 3} \frac{12 x-36}{x-3}$

$=\lim _{x \rightarrow 3} \frac{12(x-3)}{x-3}$

$=\lim _{x \rightarrow 3} 12$

 

$=12$

(RHD at $x=3$ ) $=\lim _{x \rightarrow 3^{+}} \frac{f(x)-f(3)}{x-3}$

$=\lim _{x \rightarrow 3} \frac{2 x^{2}+5-23}{x-3}$

$=\lim _{x \rightarrow 3} \frac{2 x^{2}-18}{x-3}$

$=\lim _{x \rightarrow 3} \frac{2\left(x^{2}-9\right)}{x-3}$

$=\lim _{x \rightarrow 3} 2(x+3)$

$=2 \times 6$

 

$=12$

Thus, $(\mathrm{LHD}$ at $x=3)=(\mathrm{RHD}$ at $x=3)=12$.

So, $f(x)$ is differentiable at $x=3$ and $f^{\prime}(3)=12$.