If $f(x) = x^x$ for $x > 0$, then $f$ has a minimum at $x = $:
- 1
- $1/e$ (minimum value = $e^{-1/e}$)
- 0
- $e$
Show answer and solution
Correct answer: B. $1/e$ (minimum value = $e^{-1/e}$)
Take $\ln f = x \ln x$.
**Differentiate:** $f'/f = \ln x + 1 = 0$ ⇒ $\ln x = -1$ ⇒ $x = 1/e$.
$f''$ check confirms min. Value: $f(1/e) = (1/e)^{1/e} = e^{-1/e}$.