The minimum value of the twice differentiable function f(x) = ∫ _ 0...

Given,

$f\left(x\right) = {∫}_{0}^{x}{e}^{x - t}{f}^{'}\left(t\right)dt - \left({x}^{2} - x + 1\right){e}^{x}$

$⇒ f\left(x\right) = {e}^{x}{∫}_{0}^{x}{e}^{- t}{f}^{'}\left(t\right)dt - \left({x}^{2} - x + 1\right){e}^{x}$

$⇒ {e}^{- x}f\left(x\right) = {∫}_{0}^{x}{e}^{- t}{f}^{'}\left(t\right)dt - \left({x}^{2} - x + 1\right)$

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

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

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

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

$⇒ {f}^{'}\left(x\right) = {e}^{x}$\left(2x + 1\right) . . . . . . \left(1\right)$</p><p>Now finding critical point we get,</p><p>$2x + 1 = 0 ⇒ x = - \dfrac{1}{2}$</p><p>Now differentiating equation \left(1\right)$ to check maxima and minima we get,

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

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

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

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