Question

x^x
find local min and max
help please

Answer 1

local max?

Answer 2

yea

Answer 3

first write what this really is:
\[x^x=e^{x\ln(x)}\]

Answer 4

i did find the first and second derivative..

Answer 5

first derivative is
\[e^{x\ln(x)}(\ln(x)+1)\] if i am not mistaken
set that equal to zero and solve

Answer 6

the first factor is never zero, so you only need to solve
\[\ln(x)+1=0\]

Answer 7

yea thats right i guess i didnt convert it in ln form.

Answer 8

if not you got
\[x^x(\ln(x)+1)\] same thing

Answer 9

and since \(x^x\) is not zero, you still only need solve \(\ln(x)+1=0\) you get
\[x=\frac{1}{e}\]

Answer 10

which is the \(x\) value of your local min
the min itself is \(f(\frac{1}{e})\)

Answer 11

yea there is no local max. its only local min