Let$f(x) = \left\{ \begin{array}{r} 1,\ \ \ \ \ \ \ \forall\ \ \ \ \ \ \ \ \ x

Question

Let$f(x) = \left\{ \begin{array}{r} 1,\ \ \ \ \ \ \ \forall\ \ \ \ \ \ \ \ \ x < 0 \ 1 + \sin{x,\ \ \forall\ \ \ \ \ 0 \leq x \leq \pi/2} \end{array} ight.\$, then what is the value of $f^{'}(x)$ at $x = 0$?

Dr. Abubacker · Accepted Answer

(d)
LHD$= \lim_{h ightarrow 0}\frac{f(0 - h) - f(0)}{- h}$
$= \lim_{h ightarrow 0}\frac{1 - 1}{- h} = 0$
RHD$= \lim_{h ightarrow 0}\frac{f(0 + h) - f(0)}{h}$
$= \lim_{h ightarrow 0}\frac{1 + \sin{(0 + h) - 1}}{h} = \lim_{h ightarrow 0}\frac{\sin h}{h} = 1$
$\Rightarrow$ LHD$
eq$RHD

Q. Let$f(x) = \left\{ \begin{array}{r} 1,\ \ \ \ \ \ \ \forall\ \ \ \ \ \ \ \ \ x < 0 \\ 1 + \sin{x,\ \ \forall\ \ \ \ \ 0 \leq x \leq \pi/2} \end{array} \right.\$, then what is the value of $f^{'}(x)$ at $x = 0$?

Explanation

Reviewed by Dr. Abubacker

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