The maximum value of $\left(\frac{1}{x}\right)^x$ for $x > 0$ is

To find the maximum value of $\left(\frac{1}{x}\right)^x$ for $x > 0$, rewrite the expression in a simpler form:

$\left(\frac{1}{x}\right)^x = \frac{1^x}{x^x}$

$= \frac{1}{x^x}$

$= x^{-x}$

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Let $f(x) = x^{-x}$. To find the maximum, take the natural logarithm:

$\ln(f(x)) = \ln(x^{-x})$

$= -x \ln(x)$

Let $g(x) = -x \ln(x)$

Since $\ln$ is an increasing function, the maximum of $g(x)$ occurs at the same point as the maximum of $f(x)$.

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To find the maximum, take the derivative using the product rule:

$g'(x) = -\left[\ln(x) + x \cdot \frac{1}{x}\right]$

$= -[\ln(x) + 1]$

$= -\ln(x) - 1$

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Setting the derivative equal to zero:

$-\ln(x) - 1 = 0$

$\ln(x) = -1$

$x = e^{-1}$

$x = \frac{1}{e}$

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Substituting $x = \frac{1}{e}$ into the original expression:

$f\left(\frac{1}{e}\right) = \left(\frac{1}{\frac{1}{e}}\right)^{\frac{1}{e}}$

$= e^{\frac{1}{e}}$

Therefore, the maximum value is $e^{1/e}$.