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(C) Given the function $f(x) = x(x+3)e^{-x/2}$.Since $f(x)$ satisfies Rolle's theorem in $[-3, 0]$, there exists at least one $c \in (-3, 0)$ such tha

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(C) Given the function $f(x) = x(x+3)e^{-x/2}$.Since $f(x)$ satisfies Rolle's theorem in $[-3, 0]$, there exists at least one $c \in (-3, 0)$ such that $f'(c) = 0$.First, we find the derivative $f'(x)$ using the product rule:$f'(x) = \frac{d}{dx}[x(x+3)] \cdot e^{-x/2} + x(x+3) \cdot \frac{d}{dx}[e^{-x/2}]$$f'(x) = (2x+3)e^{-x/2} + (x^2+3x) \cdot \left(-\frac{1}{2}ight)e^{-x/2}$$f'(x) = e^{-x/2} \left[ 2x+3 - \frac{x^2+3x}{2} ight]$$f'(x) = e^{-x/2} \left[ \frac{4x+6-x^2-3x}{2} ight] = \frac{-x^2+x+6}{2} e^{-x/2}$Setting $f'(x) = 0$:$\frac{-(x^2-x-6)}{2} e^{-x/2} = 0$Since $e^{-x/2} 
eq 0$, we have $x^2-x-6 = 0$.$(x-3)(x+2) = 0$Thus, $x = 3$ or $x = -2$.Since the root must lie in the interval $(-3, 0)$, the valid root is $x = -2$.

If the function $f(x) = x(x+3)e^{-x/2}$ satisfies all the conditions of Rolle's theorem in $[-3, 0]$, then a root of $f'(x) = 0$ is

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