Question

If $f\left(x\right)=x^x,x>0$, then

• A. $f\left(x\right)$ is increasing on $(1/e,\infty )$
• B. $f\left(x\right)$ is decreasing on $(0,1/e)$
• C. $x=1/e^2$ is point of local minima
• D. local minimum value of $f\left(x\right)=e^{-1/e}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $f\left(x\right)$ is increasing on $(1/e,\infty )$
B $f\left(x\right)$ is decreasing on $(0,1/e)$
D local minimum value of $f\left(x\right)=e^{-1/e}$

$f\left(x\right)=x^x,x>0$
$f^{\prime }\left(x\right)=x^x(1+\log x)$
For $f^{\prime }\left(x\right)>0,x^x(1+\log x)>0$
$1+\log x>0\Rightarrow x>e^{-1}$
$\Rightarrow x\in (1/e,\infty )$

For $f^{\prime }\left(x\right)<0,x^x(1+\log x)<0$
$1+\log x<0\Rightarrow x<e^{-1}$
$\Rightarrow x\in (0,1/e)$

For local minimum, $f^{\prime }\left(x\right)=0$
$\Rightarrow x=\frac{1}{e}$
$f\left(1/e\right)=e^{-\frac{1}{e}}$