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 :

<p>$f(x) = \int\limits_0^x {{e^{x - t}}f'(t)dt - ({x^2} - x + 1){e^x}}$</p> <p>$f(x) = {e^x}\int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1){e^x}}$</p> <p>${e^{ - x}}f(x) = \int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1)}$</p> <p>Differentiate on both side</p> <p>${e^{ - x}}f'(x) + ( - f(x){e^{ - x}}) = {e^{ - x}}f'(x) - 2x + 1$</p> <p>$f(x) = {e^x}(2x - 1)$</p> <p>$f'(x) = {e^x}(2) + {e^x}(2x - 1)$</p> <p>$= {e^x}(2x + 1)$</p> <p>$x = - {1 \over 2}$</p> <p>$f''(x) = {e^x}(2) + (2x + 1){e^x}$</p> <p>$= {e^x}(2x + 3)$</p> <p>For $x = - {1 \over 2}\,\,f''(x) > 0$</p> <p>$\Rightarrow$ Maxima</p> <p>$\therefore$ Max. $= {e^{ - {1 \over 2}}}( - 1 - 1)$</p> <p>$\therefore$ $- {2 \over {\sqrt e }}$</p>