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Function $f (x) = x (x + 3) e^{- x / 2}$ satisfy all condition of Rolle’s theorem in the interval $[- 3, 0]$ , then the value of $c =$

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–1

–3/2

–2

answer is D.

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Detailed Solution

$f' (c) = 0$

$\Rightarrow \frac{1}{2} (- c^{2} + c + 6) e^{- c / 2} = 0$

$\Rightarrow (c + 2) (c - 3) = 0$

$\Rightarrow c = - 2, 3$

But $c = 3 \in] - 3, 0 [$

$∴ c = - 2$ .

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