Question

What are the values of $c$ for while Rolle's theorem for the functions$f\left(x\right)=x^3-3x^2+2x$ in the interval $\left[0,2\right]$ is verified?

• A. $c=\pm 1$
• B. $c=1\pm \frac{1}{\sqrt{3}}$
• C. $c=\pm 2$
• D. None of these

Answer 1

The correct option is B

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

Step 1: Analysing the given data:

Given function is $f\left(x\right)=x^3-3x^2+2x$.

Rolle's theorem states that if a function $f$ is continuous on the closed interval $\left[a,b\right]$ and differentiable on the open interval $\left(a,b\right)$ such that $f\left(a\right)=f\left(b\right)$, then $f\text{'}\left(x\right)=0$ for some $x$ with $a\le x\le b$

$f\left(x\right)$is a polynomial. So it is continuous in the interval $\left[0,2\right]$

$f\text{'}\left(x\right)=3x^2-6x+2$ exists for all $x\in \left[0,2\right]$

So, $f\left(x\right)$ is differentiable for all $x\in \left(0,2\right)$

$f\left(0\right)={\left(0\right)}^3-3{\left(0\right)}^2+2\left(0\right)$

$\Rightarrow f\left(0\right)=0$

$f\left(2\right)={\left(2\right)}^3-3{\left(2\right)}^2+2\left(2\right)$

$\Rightarrow f\left(2\right)=8-12+4$

$\Rightarrow f\left(2\right)=0$

$\therefore f\left(0\right)=f\left(2\right)$. All the conditions of Rolle's theorem are satisfied.

Step 2: Find the value of $c$

Since Rolle's theorem is satisfied, there must exist a $c\in \left[0,2\right]$ such that $f\text{'}\left(c\right)=0$

$f\text{'}\left(c\right)=3c^2-6c+2$

$\Rightarrow 3c^2-6c+2=0$

$\Rightarrow c=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4·3·2}}{2·3}$ $\left[\because Applyingformulaforquadraticequations\right]$

$\Rightarrow c=\frac{6\pm \sqrt{36-24}}{6}$

$\Rightarrow c=\frac{6\pm \sqrt{12}}{6}$

$\Rightarrow c=\frac{6\pm 2\sqrt{3}}{6}$

$\Rightarrow c=1\pm \frac{1}{\sqrt{3}}$

Hence, the correct option is B