Question

The minimum value of the twice differentiable function $f(x)=\underset{0}{\overset{x}{\int }}{e}^{x-1}{f}^{\prime }(t)dt-\left(x^2-x+1\right)e^x,x\in R$, is :

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

Answer 1

Answer: (A)

$f(x)=e^x\cdot \underset{0}{\overset{x}{\int }}\frac{f^{\prime }(t)}{e^t}dt$
$f^{\prime }(x)=e^x\cdot \underset{0}{\overset{x}{\int }}\frac{f^{\prime }(t)}{e^t}dt+e^x\cdot \frac{f^{\prime }(x)}{e^x}$
$-\left[(2x-1)\cdot e^x+\left(x^2-x+1\right)\cdot e^x\right]$
$\underset{0}{\overset{x}{\int }}\frac{f^{\prime }(t)}{e^t}dt=x^2+x$
$\frac{f^{\prime }(x)}{e^x}=2x+1$
$f^{\prime }(x)=(2x+1)\cdot e^x$$<br/>f^{\prime }(x)=0\Rightarrow x=-\frac{1}{2}$
$f(x)=(2x+1)\cdot e^x-2e^x+C$

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