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Using mean value theorem we know that,

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Here, $f(x)=x(x+3)e^{-\frac{x}{2}}$

$\implies f'(x)=e^{-\frac{x}{2}}(2x+3)+(x^{2}+3x)e^{-\frac{x}{2}}*(-\frac{1}{2})$

$\implies f'(x)=e^{-\frac{x}{2}}*\frac{x+6-x^{2}}{2}$

$\implies f'(c)=e^{-\frac{c}{2}}*\frac{c+6-c^{2}}{2}$

Here, the range $[a,b]$ is given as $[-3,0].$

$f(a)=f(-3)=0$

$f(b)=f(0)=0$

So,

$f'(c)=\frac{f(b)-f(a)}{b-a}$

$\implies f'(c)=\frac{f(0)-f(-3)}{0-(-3)}$

$\implies e^{-\frac{c}{2}}*\frac{(c+6-c^{2})}{2}=0$

now, $e^{-\frac{x}{2}}\neq 0$

So, $\frac{c+6-c^{2}}{2}=0$

$\implies c^{2}-c-6=0$

$\implies c^{2}-(3-2)c-6=0$

$\implies (c-3)(c+2)=0$

$\implies c=3$ or $c=-2$

now value of $c$ has to be in the range $[-3,0]$ so $c=3$ is not possible..

So, the value of  $c$ is -2.

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