Let f(x) = begin{cases} x + 1, & text{when } | vedclass

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(D) To check if $f'(2)$ exists,we calculate the right-hand derivative $(Rf'(2))$ and the left-hand derivative $(Lf'(2))$ at $x = 2$.First,find $f(2)$

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Does not exist

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(D) To check if $f'(2)$ exists,we calculate the right-hand derivative $(Rf'(2))$ and the left-hand derivative $(Lf'(2))$ at $x = 2$.First,find $f(2)$ using the definition $f(x) = 2x - 1$ for $x \ge 2$:$f(2) = 2(2) - 1 = 3$.Right-hand derivative $(Rf'(2))$:$Rf'(2) = \lim_{h 	o 0^+} \frac{f(2 + h) - f(2)}{h} = \lim_{h 	o 0^+} \frac{2(2 + h) - 1 - 3}{h} = \lim_{h 	o 0^+} \frac{4 + 2h - 1 - 3}{h} = \lim_{h 	o 0^+} \frac{2h}{h} = 2$.Left-hand derivative $(Lf'(2))$:$Lf'(2) = \lim_{h 	o 0^+} \frac{f(2 - h) - f(2)}{-h} = \lim_{h 	o 0^+} \frac{(2 - h) + 1 - 3}{-h} = \lim_{h 	o 0^+} \frac{3 - h - 3}{-h} = \lim_{h 	o 0^+} \frac{-h}{-h} = 1$.Since $Rf'(2) 
eq Lf'(2)$,the derivative $f'(2)$ does not exist.

Let $f(x) = \begin{cases} x + 1, & \text{when } x < 2 \\ 2x - 1, & \text{when } x \ge 2 \end{cases}$,then $f'(2) = $

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