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

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

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

Let $f(x) = x |\sin x|$,$x \in R$. Then,

If $f(x) = \begin{cases} e^x + a & \text{for } x < 0 \\ x - 3 & \text{for } x \geqslant 0 \end{cases}$ is differentiable at $x = 0$,then $a$ equals:

The set of points where $f(x) = \frac{4x}{5 + 6|x|}$ is differentiable is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module