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 points

y = x^{x}, x > 0

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}})

f (x) = x^{3} - 4 x^{2} - 3 x + 7

f (x) = x^{2} + y^{2} - 2 y - 9

f (x) = - 0.4 x^{2} + 90 x - 2000

f (x) = x^{3} + 3 x^{2} + 4, - 3 \leq x \leq 2

f (x) = \frac{( x - 5 ) ^{2}}{x + 9}

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