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

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(C) For Rolle's theorem to be satisfied,we must have $f(-3) = f(0)$.$f(-3) = (-3)(-3+3) e^{3/2} = 0$.$f(0) = (0)(0+3) e^{0} = 0$.Since $f(-3) = f(0) =

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$-2$

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(C) For Rolle's theorem to be satisfied,we must have $f(-3) = f(0)$.$f(-3) = (-3)(-3+3) e^{3/2} = 0$.$f(0) = (0)(0+3) e^{0} = 0$.Since $f(-3) = f(0) = 0$,Rolle's theorem applies.We need to find $c \in (-3, 0)$ such that $f'(c) = 0$.Given $f(x) = (x^2 + 3x) e^{-\frac{x}{2}}$.Using the product rule: $f'(x) = (2x + 3) e^{-\frac{x}{2}} + (x^2 + 3x) e^{-\frac{x}{2}} \left(-\frac{1}{2}\right)$.$f'(x) = e^{-\frac{x}{2}} \left( 2x + 3 - \frac{x^2}{2} - \frac{3x}{2} \right)$.$f'(x) = e^{-\frac{x}{2}} \left( -\frac{1}{2}x^2 + \frac{1}{2}x + 3 \right)$.Setting $f'(c) = 0$ implies $-\frac{1}{2}c^2 + \frac{1}{2}c + 3 = 0$.Multiplying by $-2$,we get $c^2 - c - 6 = 0$.Factoring the quadratic: $(c - 3)(c + 2) = 0$.Thus,$c = 3$ or $c = -2$.Since $c$ must be in the interval $(-3, 0)$,we choose $c = -2$.

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

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