If f(x)=|x-2|, x in[0,4] then the Rolle's the | vedclass

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

Question

(A) For Rolle's Theorem to be applicable to a function $f(x)$ on $[a, b]$,three conditions must be satisfied: $1$. $f(x)$ is continuous on $[a, b]$. $

Vedclass · Accepted Answer

The function is not differentiable at every point in the $(0,4)$.

Vedclass · Answer

(A) For Rolle's Theorem to be applicable to a function $f(x)$ on $[a, b]$,three conditions must be satisfied: $1$. $f(x)$ is continuous on $[a, b]$. $2$. $f(x)$ is differentiable on $(a, b)$. $3$. $f(a) = f(b)$. Given $f(x) = |x-2|$ on $[0, 4]$: Calculate $f(0) = |0-2| = |-2| = 2$. Calculate $f(4) = |4-2| = |2| = 2$. Wait,$f(0) = 2$ and $f(4) = 2$,so $f(0) = f(4)$. However,check differentiability: $f(x) = |x-2|$ is not differentiable at $x = 2$,which lies in the interval $(0, 4)$. Since the function is not differentiable at $x = 2$,Rolle's Theorem cannot be applied.

If $f(x)=|x-2|, x \in[0,4]$ then the Rolle's theorem cannot be applied to the function because

Similar Questions

For all $x \in [0, 2024]$,assume that $f(x)$ is differentiable,$f(0) = -2$,and $f^{\prime}(x) \geq 5$. Then the least possible value of $f(2024)$ is:

If $f$ and $g$ are differentiable functions in $[0, 1]$ satisfying $f(0) = 2$,$g(1) = 2$,$g(0) = 0$,and $f(1) = 6$,then for some $c \in (0, 1)$:

Let $f: D \rightarrow R$ where $D=[0,1] \cup [2,4]$ be defined by $f(x)=\begin{cases} x, & \text{if } x \in [0,1] \\ 4-x, & \text{if } x \in [2,4] \end{cases}$. Then,

In the Mean Value Theorem,$f'(c) = \frac{f(b) - f(a)}{b - a}$. If $a = 0$,$b = \frac{1}{2}$,and $f(x) = x(x - 1)(x - 2)$,then the value of $c$ is:

The function $f(x) = x^3 - 6x^2 + ax + b$ satisfies the conditions of Rolle's theorem in $[1, 3]$. Then the values of $a$ and $b$ are respectively

Vedclass Test Series

Exam Paper Generator

Online Exam Module