$f(x) = {x^3} - 2{x^2} - x + 3{\mathop{\rm in}\nolimits} [0,1]$

We have, $f(x) = {x^3} - 2{x^2} - x + 3$ in [0,1]

(i) Since, $f(x)$ is a polynomial function.

Hence, $f(x)$ is continuous in [0,1] .

(ii) ${f^\prime }(x) = 3{x^2} - 4x - 1,$ which exists in (0,1)
Hence, $f(x)$ is differentiable in (0,1) .

Since, conditions of mean value theorem are satisfied.

Therefore, by mean value theorem $\exists c \in (0,1)$, such that
${f^\prime }(c) = \frac{{f(1) - f(0)}}{{1 - 0}}$

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

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

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

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

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

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

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

Hence, the mean value theorem has been verified.