The minimum value of the twice differentiable function f x = ∫ 0 x e x - t f ' t d t - x 2 - x + 1…

Given,

$f\left(x\right)={\int }_0^x{e}^{x-t}{f}^{\text{'}}\left(t\right)dt-\left(x^2-x+1\right)e^x$

$\Rightarrow f\left(x\right)=e^x{\int }_0^x{e}^{-t}{f}^{\text{'}}\left(t\right)dt-\left(x^2-x+1\right)e^x$

$\Rightarrow e^{-x}f\left(x\right)={\int }_0^x{e}^{-t}{f}^{\text{'}}\left(t\right)dt-\left(x^2-x+1\right)$

Differentiate on both side w.r.t $x$ we get,

$e^{-x}{f}^{\text{'}}\left(x\right)+\left(-f\left(x\right)e^{-x}\right)=e^{-x}{f}^{\text{'}}\left(x\right)-2x+1$

$\Rightarrow f\left(x\right)=e^x\left(2x-1\right)$

$\Rightarrow f^{\text{'}}\left(x\right)=e^x\left(2\right)+e^x\left(2x-1\right)$

$\Rightarrow f^{\text{'}}\left(x\right)=e^x\left(2x+1\right)......\left(1\right)$

Now finding critical point we get,

$2x+1=0\Rightarrow x=-\frac{1}{2}$

Now differentiating equation $\left(1\right)$ to check maxima and minima we get,

$f^{"}\left(x\right)=e^x\left(2\right)+\left(2x+1\right)e^x$

$\Rightarrow f^{"}\left(x\right)=e^x\left(2x+3\right)$

For $x=-\frac{1}{2}, f^{"}\left(\frac{-1}{2}\right)>0$, so it will give point of  minima,

Now minimum value will be given by, $f\left(\frac{-1}{2}\right)=e^{-\frac{1}{2}}\left(-1-1\right)=-\frac{2}{\sqrt{e}}$