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

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(D) To find $f'(2)$,we must check the left-hand derivative $(Lf'(2))$ and the right-hand derivative $(Rf'(2))$ at $x = 2$.$1$. Left-hand derivative $(

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Does not exist

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

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

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