Verify Rolle's theorem for the function $f\left(x\right)=x^3-3x^2+2x$ in the interval $[0,2]$.

The given function is $f\left(x\right) = x^3-3x^2+2x$ and we are supposed to verify Rolle's theorem for it in the interval $[0,2]$.

For Rolle's theorem to be applicable, a function must be continuous and differentiable in a given interval and have equal values at the extremities of the interval.

Here, for the given function, we can say,

(a) $f\left(x\right)$ is a polynomial and since polynomials are always continuous over the set of real numbers, hence it is continuous in $[0,2]$.

(b) $f^{\text{'}}\left(x\right)=3x^2-6x + \hspace{0.17em}2$, which exists for all $x\in (0,2)$. 

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

(c)Further, 

 $$\begin{gathered} f\left(0\right)=0 - 0 + 0 = 0 \\ f\left(2\right) = 8 - 12 + 4 = 0 \\ \Rightarrow f\left(2\right) = f\left(0\right) \end{gathered}$$

Thus, all the necessary conditions of Rolle's theorem are satisfied by $f\left(x\right)$ in the given interval.

Hence, there must exists some $c\in (0,2)$ such that $f^{\text{'}}\left(c\right)=0$.

$\Rightarrow f^{\text{'}}\left(c\right)=3c^2-6c + 2=0$.

Using quadratic formula to solve for $c$, we get,

$c= \frac{6 \pm \sqrt{36-24}}{6} = 1\pm \frac{1}{\sqrt{3}}$

where, $c=1\pm \frac{1}{\sqrt{3}}\in (0,2)$,

Thus, Rolle's theorem is verified for the given function in the given interval.