If f(x) = begin{cases} x + 2, & -1

Question

(D) Given $f(x) = \begin{cases} x + 2, & -1  3 \end{cases}$ and $f(3) = 5$. To check differentiability at $x = 3$

Vedclass · Accepted Answer

Does not exist

Vedclass · Answer

(D) Given $f(x) = \begin{cases} x + 2, & -1  3 \end{cases}$ and $f(3) = 5$. To check differentiability at $x = 3$,we calculate the Left Hand Derivative $(LHD)$ and Right Hand Derivative $(RHD)$. $LHD = \lim_{x 	o 3^-} \frac{f(x) - f(3)}{x - 3} = \lim_{h 	o 0} \frac{f(3 - h) - f(3)}{-h}$ $= \lim_{h 	o 0} \frac{(3 - h + 2) - 5}{-h} = \lim_{h 	o 0} \frac{5 - h - 5}{-h} = \lim_{h 	o 0} \frac{-h}{-h} = 1$. $RHD = \lim_{x 	o 3^+} \frac{f(x) - f(3)}{x - 3} = \lim_{h 	o 0} \frac{f(3 + h) - f(3)}{h}$ $= \lim_{h 	o 0} \frac{(8 - (3 + h)) - 5}{h} = \lim_{h 	o 0} \frac{8 - 3 - h - 5}{h} = \lim_{h 	o 0} \frac{-h}{h} = -1$. Since $LHD 
eq RHD$,the derivative $f'(3)$ does not exist.

If $f(x) = \begin{cases} x + 2, & -1 < x < 3 \\ 5, & x = 3 \\ 8 - x, & x > 3 \end{cases}$,then at $x = 3$,$f'(x) = $

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