The value of c for which Rolle's theorem hold | vedclass

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

Question

(C) Given the function $f(x) = x^3 - 3x^2 + 2x$ on the interval $[0, 2]$.Since $f(x)$ is a polynomial,it is continuous on $[0, 2]$ and differentiable

Vedclass · Accepted Answer

$1 \pm \frac{1}{\sqrt{3}}$

Vedclass · Answer

(C) Given the function $f(x) = x^3 - 3x^2 + 2x$ on the interval $[0, 2]$.Since $f(x)$ is a polynomial,it is continuous on $[0, 2]$ and differentiable on $(0, 2)$.Also,$f(0) = 0^3 - 3(0)^2 + 2(0) = 0$ and $f(2) = 2^3 - 3(2)^2 + 2(2) = 8 - 12 + 4 = 0$.Since $f(0) = f(2)$,Rolle's theorem states there exists at least one $c \in (0, 2)$ such that $f'(c) = 0$.$f'(x) = 3x^2 - 6x + 2$.Setting $f'(c) = 0$ gives $3c^2 - 6c + 2 = 0$.Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$c = \frac{6 \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)} = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6} = \frac{6 \pm 2\sqrt{3}}{6} = 1 \pm \frac{1}{\sqrt{3}}$.Both values $1 + \frac{1}{\sqrt{3}}$ and $1 - \frac{1}{\sqrt{3}}$ lie in the interval $(0, 2)$.

The value of $c$ for which Rolle's theorem holds for the function $f(x)=x^3-3x^2+2x$ in the interval $[0,2]$ is:

Similar Questions

If the function $f(x)=x^3+b x^2+c x-6$ satisfies all the conditions of Rolle's theorem in $[1,3]$ and $f^{\prime}\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}\right)=0$,then $b c=$

Which of the following functions satisfies the conditions of Rolle's theorem on the given interval?

Let $f(x)$ be a differentiable function in $[0, 2]$,$f(0) = 0$ and $f'(x) \le \frac{1}{2}$ for all $x \in [0, 2]$. Then:

If for $f(x) = 2x - x^2$,Lagrange's Mean Value Theorem satisfies in $[0, 1]$,then the value of $c \in [0, 1]$ is

Let $f$ be a function that is derivable on the interval $[0, 1]$. Then, which of the following statements is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module