Verify Mean Value Theorem, if $f(x) = {x^3} - 5{x^2} - 3x$ in the interval [a, b], where a $=$ 1 and b $=$ 3.

We have, $f(x) = {x^3} - 5{x^2} - 3x$, $x \in (1,3)$ , which is a polynomial function, so (i) It is continuous in [1, 3].

(ii) It is derivable in (1, 3). So, all conditions of mean value theorem are verified.

Hence thereexists, at least onec, such that
$f'(c) = \cfrac{{f(3) - f(1)}}{{3 - 1}} \Rightarrow 3{c^2} - 10c - 3 = \cfrac{{ - 27 - ( - 7)}}{2}$

$\Rightarrow$ $3{c^2} - 10c - 3 = - 10$ $\Rightarrow$ $3{c^2} - 10c + 7 = 0$

$\Rightarrow$ $c = \cfrac{{10 \pm \sqrt {100 - 84} }}{6} = \cfrac{{10 \pm 4}}{6}$

$\Rightarrow$ $c = \cfrac{7}{3},1$
But, $c = \cfrac{7}{3} \in (1,3)$
So, mean value theorem is verified.