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Answer: f(x) is differentiable at x = 3 and f ` (3) = 12.

Hint: For differentiability, LHD = RHD where LHD is left hand derivative and RHD is right hand derivative. Also, LHD and RHD should exist in given limit.

Given: $f\left ( x \right )=$$\left\{\begin{array}{c} 12 x-13, \text { if } x \leq 3 \\ 2 x^{2}+5, \text { if } x>3 \end{array}\right.$

Solution:

As we know

$f'\left ( x \right )=\lim_{h\rightarrow 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}$

Now, we have to check differentiability of given function at x = 3:

$LHD at x=3:\lim_{x\rightarrow 3^{-}}\frac{f\left ( x \right )-f\left ( 3 \right )}{x-3}$

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{f(3-h)-f(3)}{3-h-3} \\ &=\lim _{h \rightarrow 0} \frac{[12(3-h)-13]-[12(3)-13]}{-h} \\ &=\lim _{h \rightarrow 0} \frac{36-12 h-13-36+13}{-h} \\ &=\lim _{h \rightarrow 0} \frac{-12 h}{-h} \\ &=12 \end{aligned}

\begin{aligned} &\text { RHD at } \mathrm{x}=0: \lim _{x \rightarrow 3^{+}} \frac{f(x)-f(3)}{x-3} \\ &=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{3+h-3} \\ &=\lim _{h \rightarrow 0} \frac{\left[2(3+h)^{2}+5\right]-[12(3)-13]}{h} \\ &=\lim _{h \rightarrow 0} \frac{18+12 h+2 h^{2}+5-36+13}{h} \\ &=\lim _{h \rightarrow 0} \frac{2 h^{2}+12 h}{h} \\ &=\lim _{h \rightarrow 0} \frac{h(2 h+12)}{h} \\ &=\lim _{h \rightarrow 0} 2 h+12 \\ &=12 \end{aligned}

(LHD at x = 3) = (RHD at x = 3)

Hence, f(x) is differentiable at x = 3.

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