Question

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

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

Answer 1

Answer: (B)

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

Here, we observe that
(a) $f\left(x\right)$ is polynomial, so it is continuous in the interval $[0,2]$.
(b) $f^{\prime }\left(x\right)=3x^2-6x+2$ exists for all $xϵ(0,2)$.
So, $f\left(x\right)$ is differentiable for all $xϵ(0,2)$ and
(c) $f\left(0\right)=0,f\left(2\right)=2^3-3(2{)}^2+2(2)=0$
$\Rightarrow f\left(0\right)=f\left(2\right)$
Thus, all the three conditions of Rolle's theorem are satisfied.
So, there must exisst $cϵ[0,2]$ such that $f^{\prime }\left(c\right)=0$
$\Rightarrow f^{\prime }\left(c\right)=3c^2-6c+2=0$
$\Rightarrow c=1\pm \frac{1}{\sqrt{3}}ϵ[0,2]$.