Let f(x)=x1/x. Take log: lnf=xlnx.
Differentiating: f(x)f′(x)=x21−lnx.
Setting f′(x)=0 gives lnx=1, so x=e. Second-derivative test confirms maximum. So f(e)=e1/e is the maximum value.
Consider the function f(x)=xx1. Its
Held on 30 Aug 2022 · Verified 13 Jul 2026.
minimum value is ee1
maximum value is ee1
minimum value is ee
maximum value is (e1)e
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