(11/16/2009, 04:53 AM)Amherstclane Wrote: I have recently taken great interest in studying the properties of the function f(x) = x^x , and I was wondering: is there any way to prove whether f(x) = x^x is an injective (i.e. one-to-one) function? I realize that if I can prove that if the inverse of f(x) = x^x is also a function, then f(x) = x^x is injective. The problem is: f^{-1}(x) is non-algebraic, so I can't figure out whether it's a function or not. Does anyone know another way to prove whether or not f(x) = x^x is injective?

NOTE: The domain of this function is real numbers.

Did you ever take a look at the graph of

?

First we have do termine the range of definition.

Surely

should be greater 0, because non-integer powers are not defined for

, e.g.

is not defined or

is not defined. At 0 we have sometimes the difficulty to compute

which is not defined in Analysis, so we also exclude this number. So our range of definition is the interval

.

Now you take some math software and plot

, say on

(because we can not show an infinitely long x-Axis):

There you see that the function is not injective on (0,1).

It is an interesting exercise to determine the minimum of the function. By solving

. This is also the way you can prove that the function is injective on

, i.e. by showing that the derivative is always greater 0.

PS: Can you please adapt your math skill level in your profile.