The minimum value of the twice differentiable function $$f(x) = \int_0^x e^{x-t} f'(t) dt - (x^2 - x + 1)e^x$$, $$x \in \mathbb{R}$$, is

We are given $$f(x) = \int_0^x e^{x-t} f'(t)\, dt - (x^2 - x + 1)e^x$$.

Since $$\int_0^x e^{x-t} f'(t)\, dt = e^x \int_0^x e^{-t} f'(t)\, dt$$, it follows that $$f(x) = e^x \int_0^x e^{-t} f'(t)\, dt - (x^2 - x + 1)e^x$$.

Differentiating both sides gives $$f'(x) = e^x \int_0^x e^{-t} f'(t)\, dt + e^x \cdot e^{-x} f'(x) - (2x - 1)e^x - (x^2 - x + 1)e^x$$, which simplifies to $$f'(x) = e^x \int_0^x e^{-t} f'(t)\, dt + f'(x) - (x^2 + x)e^x$$. This implies $$e^x \int_0^x e^{-t} f'(t)\, dt = (x^2 + x)e^x$$, so $$\int_0^x e^{-t} f'(t)\, dt = x^2 + x$$.

Differentiating the relation $$\int_0^x e^{-t} f'(t)\, dt = x^2 + x$$ with respect to x yields $$e^{-x} f'(x) = 2x + 1$$, hence $$f'(x) = (2x + 1)e^x$$.

Substituting this expression into the original formula leads to $$f(x) = e^x(x^2 + x) - (x^2 - x + 1)e^x = (2x - 1)e^x$$.

To find the minimum, we compute $$f'(x) = 2e^x + (2x - 1)e^x = (2x + 1)e^x$$ and set it to zero, which gives $$x = -\frac{1}{2}$$ since $$e^x \neq 0$$. Moreover, $$f''(x) = (2x + 3)e^x$$ and at $$x = -\frac{1}{2}$$ we have $$f''\bigl(-\frac{1}{2}\bigr) = 2e^{-1/2} > 0$$, confirming a minimum. Evaluating at this point yields $$f\left(-\frac{1}{2}\right) = (2 \cdot (-\frac{1}{2}) - 1)e^{-1/2} = (-2)e^{-1/2} = -\frac{2}{\sqrt{e}}$$.

The answer is Option A.