lf $x > 0$ then the minimum value of $x^{x}$ is

e^{- 1 / e}

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e^{1 / e}

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e^{e}

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e

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Solution

The correct option is A $e^{- 1 / e}$
$f^{'} (x) = x^{x} (l o g x + 1)$
$f^{'} (x) = 0$ when $x = \frac{1}{e}$
$f^{'} (x) < 0$ If $0 < x < \frac{1}{e}$
$f^{'} (x) > 0$ If $x > \frac{1}{e}$
So $f (x)$ is minimum at $x = \frac{1}{e}$
$f (x) = (\frac{1}{e})^{\frac{1}{e}}$

Suggest Corrections

Similar questions

f (x) = \frac{x (e^{1 / x} - e^{- 1 / x})}{e^{1 / x} + e^{- 1 / x}} x \neq 0

x = 0

f (0) =

{(\frac{1}{x})}^{x}

e^{1 / e}

{(\frac{1}{x})}^{x}

e^{1 / e}

lim x \to 0 \frac{e^{\frac{1}{x}} - e^{\frac{- 1}{x}}}{e^{\frac{1}{x}} + e^{\frac{- 1}{x}}}

f (x) = x^{x}, x > 0

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