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Question

The minimum value of the twice differentiable function $f(x)=\int\limits_{0}^{x} \mathrm{e}^{x-\mathrm{t}} f^{\prime}(\mathrm{t}) \mathrm{dt}-\left(x^{2}-x+1\right) \mathrm{e}^{x}$, $x \in \mathbf{R}$, is :

$-\frac{2}{\sqrt{\mathrm{e}}}$
$-2 \sqrt{\mathrm{e}}$
$-\sqrt{\mathrm{e}}$
$\frac{2}{\sqrt{\mathrm{e}}}$

Solution

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