Rolle's theorem is not applicable to the func | vedclass

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

Question

(B) The function is defined as $f(x) = \begin{cases} -x, & 	ext{if } -1 \le x < 0 \ x, & 	ext{if } 0 \le x \le 1 \end{cases}$.First,we check the co

Vedclass · Accepted Answer

$f$ is not differentiable on $(-1, 1)$

Vedclass · Answer

(B) The function is defined as $f(x) = \begin{cases} -x, & 	ext{if } -1 \le x First,we check the conditions for Rolle's theorem:$1$. $f(x)$ is continuous on $[-1, 1]$.$2$. $f(-1) = |-1| = 1$ and $f(1) = |1| = 1$,so $f(-1) = f(1)$.$3$. We check for differentiability at $x = 0$:Right-hand derivative: $Rf'(0) = \lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{|h|}{h} = \lim_{h 	o 0^+} \frac{h}{h} = 1$.Left-hand derivative: $Lf'(0) = \lim_{h 	o 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^-} \frac{|h|}{h} = \lim_{h 	o 0^-} \frac{-h}{h} = -1$.Since $Rf'(0) 
eq Lf'(0)$,the function is not differentiable at $x = 0$,which lies in the interval $(-1, 1)$.Therefore,Rolle's theorem is not applicable because the function is not differentiable on $(-1, 1)$.

Rolle's theorem is not applicable to the function $f(x) = |x|$ defined on $[-1, 1]$ because

Similar Questions

The value of $c$ for Lagrange's Mean Value Theorem for $f(x) = \sqrt{25-x^2}$ on the interval $[1, 5]$ is

If $a + b + c = 0$,then how many roots does the equation $3ax^2 + 2bx + c = 0$ have in the interval $(0, 1)$?

Let $f$ be any continuous function on $[0,2]$ and twice differentiable on $(0,2)$. If $f(0)=0, f(1)=1$ and $f(2)=2$,then

If $f(x) = \sin^2 x + x \sin 2x \log x$,then $f(x) = 0$ has

Let $f(x)$ be a differentiable function in $[2,7]$. If $f(2)=3$ and $f^{\prime}(x) \leq 5$ for all $x$ in $(2,7)$, then the maximum possible value of $f(x)$ at $x=7$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module