The maximum value of x1/x, x 0 is __________ .

Question

The maximum value of x1/x, x > 0 is __________ .

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e^1/e 	ext { Given }: f\left( x ight) = x^1/x 	ext { Taking log on both sides, we get } \log f\left( x ight) = 1/x\log x 	ext { Differentiating w . r . t . x, we get } \frac{1}{f\left( x ight)}f'\left( x ight) = - 1/x^2\log x + 1/x^2 \Rightarrow f'\left( x ight) = f\left( x ight)1/x^2\left( 1 - \log x ight) \Rightarrow f'\left( x ight) = x^1/x \left( 1/x^2 - 1/x^2\log x ight) . . . \left( 1 ight) \Rightarrow f'\left( x ight) = x^1/x - 2 \left( 1 - \log x ight) 	ext { For a local maxima or a local minima, we must have } f'\left( x ight) = 0 \Rightarrow x^1/x - 2 \left( 1 - \log x ight) = 0 \Rightarrow \log x = 1 \Rightarrow x = e 	ext { Now,} f''\left( x ight) = x^1/x \left( 1/x^2 - 1/x^2\log x ight)^2 + x^1/x \left( - 2/x^3 + 2/x^3\log x - 1/x^3 ight) = x^1/x \left( 1/x^2 - 1/x^2\log x ight)^2 + x^1/x \left( - 3/x^3 + 2/x^3\log x ight) 	ext { At }x = e: f''\left( e ight) = e^1/e \left( 1/e^2 - 1/e^2\log e ight)^2 + e^1/e \left( - 3/e^3 + 2/e^3\log e ight) = - e^1/e \left( 1/e^3 ight) < 0 	ext { So, x = e is a point of local maxima }. 	herefore 	ext { Maximum value } = f\left( e ight) = e^1/e

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