What is the extreme y=x^x,x>0 ?

The extreme y=x^x,x>0 is Minimum(1/e ,(1/e)^{1/e})

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extreme y=x^x,x>0

extreme

Minimum (\frac{1}{e}, (\frac{1}{e})^{\frac{1}{e}})

Solution steps

Find the critical points:

x = \frac{1}{e}

Domain of

x^{x}, x > 0 : x > 0

Find intervals:

Decreasing

: 0 < x < \frac{1}{e},

Increasing

: \frac{1}{e} < x < \infty

Plug the extreme point

x = \frac{1}{e}

into

x^{x} \Rightarrow y = (\frac{1}{e})^{\frac{1}{e}}

Minimum (\frac{1}{e}, (\frac{1}{e})^{\frac{1}{e}})

What is the extreme y=x^x,x>0 ?
The extreme y=x^x,x>0 is Minimum(1/e ,(1/e)^{1/e})