Given- fx-x1-x-x-0 has the maximum value at x-e- then

The correct option is A e^π>π^e Since, f(x) has a maximum at x=e ⇒ f(e) gt;f(x) ⇒ e^1/e gt;x^1/x ...

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Conduction Law

Given: fx=x...

Question

Given: $f (x) = x^{\frac{1}{x}}, (x > 0)$ has the maximum value at $x = e$ , then

e^{π} > π^{e}

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e^{π} > π^{π}

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e^{π} = π^{e}

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e^{π} \leq π^{e}

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Solution

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Similar questions

\frac{π^{e}}{x - e} + \frac{e^{π}}{x - π} + \frac{π^{π} e^{e}}{x - π - e} = 0

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

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