$f(x)=x^3-5x^2-3x \qquad[1,3]$
Step 1:
$f(x)$ is a polynomial function .
It is continuous in $[1,3]$ and differentiable in $(1,3) $
The conditions for Lagranges Theorem are satisfied.
Step 2:
To find $c \in (1,3)$
Such that $f'(c)=\large\frac{f(3)-f(1)}{3-1}$
$f'{x}=3x^2-10 x -3$
$f'(c)=3c^2-10c-3$
$\large\frac{f(3)-f{(1)}}{3-1}=\large\frac{27-45-9-(1-5-3)}{2}=\frac{-27+7}{2}$
$3c^2-10 c-3=-10$
$3c^2-10 c+7=0 \qquad c=\large\frac{10 \pm \sqrt {100-84}}{6}=\frac{10 \pm 4}{6}=\frac{14}{6},$$1$
$\large\frac{14}{6} =\frac{7}{3} \in $$(1,3)$ is the required 'c'
Lagrange's Theorem is verified