Question

Find the value of $x$ at which $f\left(x\right)=x^x$ attains the minimum value.

• A. $0$
• B. $e$
• C. $1/e$
• D. $\frac{1}{1+e}$

Answer 1

Answer: (C)

$1/e$

$y=x^x$
Or
$lny=xlnx$
Now differentiating with respect to x, we get
$y^{\prime }\frac{1}{y}=ln\left(x\right)+1$
Or
$\frac{dy}{dx}=y\left(ln\right(x)+1)$
$=x^x\left(ln\right(x)+1)$
Now for the critical points $\frac{dy}{dx}=0$
Or
$x^x\left(ln\right(x)+1)=)$
However
$x^x$ cannot be equal to 0.
Thus we are left with
$ln\left(x\right)+1=0$
$x=e^{-1}$
$f"\left(x\right)$
$=x^{x-1}+x^x\left(ln\right(x)+1{)}^2$
$=x^x[\frac{1}{x}+(ln\left(x\right)+1{)}^2]$
Now
$f"\left(e^{-1}\right)$
$=(\frac{1}{e}{)}^{\frac{1}{e}}.e$
Thus
$f"\left(e^{-1}\right)>0$ ...(minima).
Hence $f\left(x\right)$ attains a minimum value at $x=\frac{1}{e}$.