$f(x) = x{(x - 1)^2}$ in [0,1]

We have, $f(x) = x{(x - 1)^2}$ in [0,1]

As we know that, Rolle's theorem states that, if $f$ be any real valued function, defined in the closed interval [a, b], such that

(i) $f$ is continuous on [a, b].

(ii) $f$ is differentiable (a,b)

.(iii) $f(a) = f(b)$.

Then, there exists a real number ${\rm{c}}$ in the open interval $(a,b)$, such that ${f^\prime }(c) = {0^\prime }$. Here, we shall verify the Rolle's theorem for the given function.

(i) Since, $f(x) = x{(x - 1)^2}$ is a polynomial function.
So, it is continuous in [0,1] .

(ii) Now, ${f^\prime }(x) = x \cdot \frac{d}{{dx}}{(x - 1)^2} + {(x - 1)^2}\frac{d}{{dx}}x$

$= x \cdot 2(x - 1) \cdot 1 + {(x - 1)^2}$

$= 2{x^2} - 2x + {x^2} + 1 - 2x$

$= 3{x^2} - 4x + 1$ which exists in (0,1) .

So, $f(x)$ is differentiable in (0,1) .

(iii) Now, $f(0) = 0$ and $f(1) = 0 \Rightarrow f(0) = f(1)$

$f$ satisfies the above conditions of Rolle's theorem.

Hence, by Rolle's theorem there exists c $\in (0,1)$ such that
${f^\prime }(c) = 0$

$\Rightarrow$ $3{c^2} - 4c + 1 = 0$

$\Rightarrow$ $3{c^2} - 3c - c + 1 = 0$

$\Rightarrow$ $3c(c - 1) - 1(c - 1) = 0$

$\Rightarrow$ $(3c - 1)(c - 1) = 0$

$\Rightarrow$ $c = \frac{1}{3},1 \Rightarrow \frac{1}{3} \in (0,1)$

Thus, we see that there exists a real number $c$ in the open interval (0,1) .

Hence, Rolle's theorem has been verified.