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

Let $f$ be defined on $D = R - \{-1, 1\}$ by $f(x) = \frac{|x|}{1 - |x|}$,then

If $f(x) = \max(|2-x|, 2-x^3)$ for $x \in R$,then which of the following is true?

Define a function $f: R \rightarrow R$ by $f(x) = \begin{cases} \frac{\sin x^2}{x}, & \text{for } x < 0 \\ x^2 + ax + b, & \text{for } x \geq 0 \end{cases}$. Suppose $f(x)$ is differentiable on $R$. Then,

$f(x) = \cos^{-1}(2x^2 - 1)$ is not differentiable at $x = a$,then $a$ equals:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module