The minimum value of the twice differentiable function
$f (x) = \int_{0}^{x} e^{x - t} f^{'} (t) d t - (x^{2} - x + 1) e^{x}, x \in R,$ is:

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

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

$- 2 \sqrt{e}$

$- \sqrt{e}$

answer is A.

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Detailed Solution

$f (x) = e^{x} \cdot \int_{0}^{x} \frac{f^{'} (t)}{e^{t}} d t - (x^{2} - x + 1) e^{x} f^{'} (x) = e^{x} \cdot \int_{0}^{x} \frac{f^{'} (t)}{e^{t}} d t + e^{x} \cdot \frac{f^{'} (x)}{e^{x}} - (x^{2} - x + 1) e^{x} - (2 x - 1) e^{x} f^{'} (x) = e^{x} \cdot \int_{0}^{x} \frac{f^{'} (t)}{e^{t}} d t - (x^{2} - x + 1) e^{x} - [(2 x - 1) \cdot e^{x}] + f^{'} (x) f (x) = (2 x - 1) \cdot e^{x} f^{'} (x) = (2 x - 1) \cdot e^{x} + 2 e^{x} \Rightarrow f^{'} (x) = (2 x + 1) \cdot e^{x} F o r m a x i m a o r m i n i m a f^{'} (x) = 0 \Rightarrow x = - \frac{1}{2} a t x = - 1 / 2 t h e f u n c t i o n h a s m i n i m a \Rightarrow f (x) = e^{x} (2 x - 1) ∴ f (- \frac{1}{2}) = \frac{- 2}{\sqrt{e}}$