$f(x) = x(x + 3){e^{ - x/2}}$ in [-3,0]

We have, $f(x) = x(x + 3){e^{ - x/2}}$
(i) $f(x)$ is a continuous function. [since, it is a combination of polynomial functions $x(x + 3)$ and an exponential function ${e^{ - x/2}}$ which are continuous functions]

So, $f(x) = x(x + 3){e^{ - x/2}}$ is continuous in [-3,0]

.
(ii) therefore,${f^\prime }(x) = \left( {{x^2} + 3x} \right) \cdot \frac{d}{{dx}}{e^{ - x/2}} + {e^{ - x/2}} \cdot \frac{d}{{dx}}\left( {{x^2} + 3x} \right)$

$= \left( {{x^2} + 3x} \right) \cdot {e^{ - x/2}} \cdot \left( { - \frac{1}{2}} \right) + {e^{ - x/2}} \cdot (2x + 3)$
$= {e^{ - x/2}}\left[ {2x + 3 - \frac{1}{2} \cdot \left( {{x^2} + 3x} \right)} \right]$

$= {e^{ - x/2}}\left[ {\frac{{4x + 6 - {x^2} - 3x}}{2}} \right]$
$= {e^{ - x/2}} \cdot \frac{1}{2}\left[ { - {x^2} + x + 6} \right]$

$= \frac{{ - 1}}{2}{e^{ - x/2}}\left[ {{x^2} - x - 6} \right]$

$= \frac{{ - 1}}{2}{e^{ - x/2}}\left[ {{x^2} - 3x + 2x - 6} \right]$
$= \frac{{ - 1}}{2}{e^{ - x/2}}[(x + 2)(x - 3)]$ which exists in (-3,0)

Hence, $f(x)$ is differentiable in (-3,0) .

(iii) therefore,$f( - 3) = - 3( - 3 + 3){e^{ - 3/2}} = 0$
and $f(0) = 0(0 + 3){e^{ - 0/2}} = 0$

$\Rightarrow$ $f( - 3) = f(0)$

Since, conditions of Rolle's theorem are satisfied.

Hence, there exists a real number $c$ such that $f'(c) = 0$
$\Rightarrow$ $- \frac{1}{2}{e^{ - c/2}}(c + 2)(c - 3) = 0$
$\Rightarrow$ $c = - 2,3,$ where $- 2 \in ( - 3,0)$

Therefore, Rolle's theorem has been verified.