The function f(x) = x(x + 3)e^{-frac{1}{2}x} | vedclass

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(B) Given $f(x) = (x^2 + 3x)e^{-\frac{1}{2}x}$.Find the derivative $f'(x)$ using the product rule:$f'(x) = (2x + 3)e^{-\frac{1}{2}x} + (x^2 + 3x)e^{-\

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-$2$

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(B) Given $f(x) = (x^2 + 3x)e^{-\frac{1}{2}x}$.Find the derivative $f'(x)$ using the product rule:$f'(x) = (2x + 3)e^{-\frac{1}{2}x} + (x^2 + 3x)e^{-\frac{1}{2}x} \cdot (-\frac{1}{2})$$f'(x) = e^{-\frac{1}{2}x} [2x + 3 - \frac{1}{2}x^2 - \frac{3}{2}x]$$f'(x) = e^{-\frac{1}{2}x} [-\frac{1}{2}x^2 + \frac{1}{2}x + 3]$Set $f'(c) = 0$:$-\frac{1}{2}c^2 + \frac{1}{2}c + 3 = 0$Multiply by $-2$:$c^2 - c - 6 = 0$$(c - 3)(c + 2) = 0$$c = 3$ or $c = -2$.Since $c$ must lie in the open interval $(-3, 0)$, we choose $c = -2$.

The function $f(x) = x(x + 3)e^{-\frac{1}{2}x}$ satisfies all the conditions of Rolle's theorem in $[-3, 0]$. Find the value of $c$ such that $f'(c) = 0$.

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