Question

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 :

• A. $-\frac{2}{\sqrt{ e }}$
• B. $-2 \sqrt{e}$
• C. $-\sqrt{ e }$
• D. $\frac{2}{\sqrt{e}}$

Answer 1

Answer: (A)

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

$-1=1-2+C$
$C=0$
$f(x)=e^x(2 x-1)$
$f\left(-\frac{1}{2}\right)=\frac{-2}{\sqrt{e}}$