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  4. The minimum value of the twice differentiable function f(x)=∫ limits0x ex-1 f prime(t) d t-(x2-x+1) ex, x ∈ R, is :

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Q. The minimum value of the twice differentiable function $f(x)=\int\limits_0^x e^{x-1} f^{\prime}(t) d t-\left(x^2-x+1\right) e^x, x \in R$, is :

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Solution:

$ f(x)=e^x \cdot \int\limits_0^x \frac{f^{\prime}(t)}{e^t} d t $
$ f^{\prime}(x)=e^x \cdot \int\limits_0^x \frac{f^{\prime}(t)}{e^t} d t+e^x \cdot \frac{f^{\prime}(x)}{e^x}$
$-\left[(2 x-1) \cdot e^x+\left(x^2-x+1\right) \cdot e^x\right]$
$\int\limits_0^x \frac{f^{\prime}(t)}{e^t} d t=x^2+x $
$ \frac{f^{\prime}(x)}{e^x}=2 x+1 $
$ f^{\prime}(x)=(2 x+1) \cdot e^x $$
f^{\prime}(x)=0 \Rightarrow x=-\frac{1}{2}$
$ f(x)=(2 x+1) \cdot e^x-2 e^x+C$
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$-1=1-2+C $
$ C=0 $
$ f(x)=e^x(2 x-1) $
$ f\left(-\frac{1}{2}\right)=\frac{-2}{\sqrt{e}}$