If the function f(x) = x(x + 3) e^{-x/2} sati | vedclass

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(C) Given function is $f(x) = (x^2 + 3x) e^{-x/2}$.According to Rolle's theorem,there exists at least one $c \in (-3, 0)$ such that $f'(c) = 0$.First,

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(C) Given function is $f(x) = (x^2 + 3x) e^{-x/2}$.According to Rolle's theorem,there exists at least one $c \in (-3, 0)$ such that $f'(c) = 0$.First,find the derivative $f'(x)$ using the product rule:$f'(x) = (2x + 3) e^{-x/2} + (x^2 + 3x) \cdot (-\frac{1}{2}) e^{-x/2}$$f'(x) = e^{-x/2} [2x + 3 - \frac{1}{2}(x^2 + 3x)]$Set $f'(c) = 0$:$2c + 3 - \frac{c^2 + 3c}{2} = 0$Multiply by $2$:$4c + 6 - c^2 - 3c = 0$$-c^2 + c + 6 = 0$$c^2 - c - 6 = 0$$(c - 3)(c + 2) = 0$So,$c = 3$ or $c = -2$.Since the interval is $[-3, 0]$,we must have $c \in (-3, 0)$.Therefore,$c = -2$ is the correct value.

If the function $f(x) = x(x + 3) e^{-x/2}$ satisfies Rolle's theorem in the interval $[-3, 0]$,then find the value of $c$.

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