Q.The function $f(x) = x(x+3)e^{-x/2}$ satisfies all the conditions of Rolle's theorem on $[-3, 0]$. Find the value of $c$ such that $f'(c) = 0$.

Rolle's Theorem states that if a real-valued function $f$ is continuous on the closed interval $[a,b]$, differentiable on the open interval $(a,b)$, and $f(a)=f(b)$, then there exists at least one point $c$ in $(a,b)$ where $f'(c)=0$. Geometrically, if a smooth curve starts and ends at the same height, somewhere in between it must have a point where the tangent is horizontal (parallel to the X-axis) — the curve must turn around. To apply the theorem to a given function on a given interval: first check the three hypotheses (the function must be continuous on the closed interval and differentiable on the open interval — true automatically for polynomials, and for combinations of standard continuous/differentiable functions such as $e^x$, $\sin x$, $\cos x$, $\log x$ wherever they are individually defined and smooth; and $f(a)$ must equal $f(b)$, checked by direct substitution). If any hypothesis fails (as with a function that is not differentiable at an interior point, or one where the two endpoint values differ), Rolle's Theorem does not apply. If all hypotheses hold, differentiate $f$, set $f'(x)=0$, solve for $x$, and confirm the solution(s) $c$ actually lie strictly …