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

Question

(D) To find $f'(0)$,we must check the left-hand derivative $(Lf'(0))$ and the right-hand derivative $(Rf'(0))$.$Rf'(0) = \lim_{h 	o 0^+} \frac{f(0+h)

Vedclass · Accepted Answer

Does not exist

Vedclass · Answer

(D) To find $f'(0)$,we must check the left-hand derivative $(Lf'(0))$ and the right-hand derivative $(Rf'(0))$.$Rf'(0) = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{(1 + \sin h) - 1}{h} = \lim_{h 	o 0^+} \frac{\sin h}{h} = 1$.$Lf'(0) = \lim_{h 	o 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h 	o 0^+} \frac{1 - 1}{-h} = 0$.Since $Lf'(0) 
eq Rf'(0)$,the derivative $f'(0)$ does not exist.

If $f(x) = \begin{cases} 1, & x < 0 \\ 1 + \sin x, & 0 \le x < \frac{\pi}{2} \end{cases}$,then $f'(0) = $

Similar Questions

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

In the interval $[0, 3]$,the function $f(x) = |x - 1| + |x - 2|$ is

The function $f(x) = (x - a)^2 \cos \frac{1}{(x-a)}$ for $x \neq a$ and $f(a) = 0$,is

If $f(x) = \begin{cases} \frac{x^2 \ln \cos x}{\ln(1 + x^2)}, & x \neq 0 \\ 0, & x = 0 \end{cases}$,then $f(x)$ is

The number of points at which the function,$f(x) = |x - 0.5| + |x - 1| + \tan x$ does not have a derivative in the interval $(0, 2)$ is :

Vedclass Test Series

Exam Paper Generator

Online Exam Module