The set of points where the function f(x)=|x- | vedclass

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

Question

(B) The function $f(x) = |x-1| e^{x}$ is a product of two functions: $g(x) = |x-1|$ and $h(x) = e^{x}$.We know that $e^{x}$ is differentiable for all

Vedclass · Accepted Answer

$R-\{1\}$

Vedclass · Answer

(B) The function $f(x) = |x-1| e^{x}$ is a product of two functions: $g(x) = |x-1|$ and $h(x) = e^{x}$.We know that $e^{x}$ is differentiable for all $x \in R$.The function $g(x) = |x-1|$ is continuous everywhere but is not differentiable at $x = 1$ because the left-hand derivative and right-hand derivative at $x = 1$ are not equal.Specifically,at $x = 1$,the left-hand derivative is $-e^{1} = -e$ and the right-hand derivative is $e^{1} = e$.Since the product of a non-differentiable function and a non-zero differentiable function at a point is non-differentiable at that point,$f(x)$ is not differentiable at $x = 1$.Therefore,the set of points where $f(x)$ is differentiable is $R - \{1\}$.

The set of points where the function $f(x)=|x-1| e^{x}$ is differentiable,is

Similar Questions

If $f: R \rightarrow R$ is defined as $f(x)=|x+1|+|x-1|$, then $f(x)$ is

If $f(x) = a|\sin x| + be^{|x|} + c|x|^3$,where $a, b, c \in \mathbb{R}$,is differentiable at $x = 0$,then:

$f(x) = |\log_e |x||$ is differentiable at

The number of points,where the function $f: R \rightarrow R, f(x) = |x-1| \cos |x-2| \sin |x-1| + (x-3)|x^2-5x+4|$ is $NOT$ differentiable,is:

Let $S = \{t \in R : f(x) = |x-\pi|(e^{|x|}-1)\sin|x| \text{ is not differentiable at } t\}$. Then the set $S$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module