Question

The function $f\left(x\right)=x{e}^{(x-x^2)}$ attains the maximum value at $x=?$

• A. -1
• B. 1/2
• C. -1/2
• D. 1

Answer 1

Answer: (D)

1

$f\left(x\right)=x{e}^{x-x^2}$
$f^{\prime }\left(x\right)=e^{x-x^2}+x{e}^{x-x^2}(1-2x)=e^{x-x^2}(1+x-2x^2)$
$f^{\prime \prime }\left(x\right)=\left(\right(1+x-2x^2\left)\right(1-2x)+(1-4x\left)\right)e^{x-x^2}$
For minima or maxima of $f\left(x\right)$
$f^{\prime }\left(x\right)=0=e^{x-x^2}(1+x-2x^2)\Rightarrow x=-1/2,1$
clearly $f^{\prime \prime }\left(1\right)<0$ and $f^{\prime \prime }\left(1\right)>0$ hence at $x=1$, $f$ will achieve its maximum value
which is $1$