If f(x)=|x-3|,then f^{prime}(3) is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given,$f(x)=|x-3|$.We can redefine the function as:$f(x) = \begin{cases} 3-x, & x  3 \end{cases}$To check the differe

Vedclass · Accepted Answer

does not exist

Vedclass · Answer

(D) Given,$f(x)=|x-3|$.We can redefine the function as:$f(x) = \begin{cases} 3-x, & x  3 \end{cases}$To check the differentiability at $x=3$,we find the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$:$LHD = \lim_{h 	o 0^-} \frac{f(3+h)-f(3)}{h} = \lim_{h 	o 0^-} \frac{|3+h-3|-0}{h} = \lim_{h 	o 0^-} \frac{|h|}{h} = \lim_{h 	o 0^-} \frac{-h}{h} = -1$.$RHD = \lim_{h 	o 0^+} \frac{f(3+h)-f(3)}{h} = \lim_{h 	o 0^+} \frac{|3+h-3|-0}{h} = \lim_{h 	o 0^+} \frac{|h|}{h} = \lim_{h 	o 0^+} \frac{h}{h} = 1$.Since $LHD 
eq RHD$,the derivative $f^{\prime}(3)$ does not exist.

If $f(x)=|x-3|$,then $f^{\prime}(3)$ is

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) = $

If $f(x)=|x|+|\sin x|$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,then its left hand derivative at $x=0$ is

Which of the following is differentiable at $x=0$?

The function given by $y = ||x| - 1|$ is differentiable for all real numbers except the points

Which one of the following functions is continuous everywhere in its domain but has at least one point where it is not differentiable?

Vedclass Test Series

Exam Paper Generator

Online Exam Module