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If $f(x) = x^x$ for $x > 0$, then $f$ has a minimum at $x = $:
A1
B$1/e$ (minimum value = $e^{-1/e}$)
C0
D$e$
Answer & 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: $f(1/e) = (1/e)^{1/e} = e^{-1/e}$.
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