(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 \( x^x \)?
First we have do termine the range of definition.
Surely \( x \) should be greater 0, because non-integer powers are not defined for \( x<0 \), e.g. \( (-1)^{0.5} \) is not defined or \( (-0.5)^{-0.5}=1/(-0.5)^{0.5} \) is not defined. At 0 we have sometimes the difficulty to compute \( 0^0 \) which is not defined in Analysis, so we also exclude this number. So our range of definition is the interval \( (0,\infty) \).
Now you take some math software and plot \( x^x \), say on \( (0,2) \) (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 \( \frac{\partial x^x}{\partial x}=0 \). This is also the way you can prove that the function is injective on \( (1,\infty) \), i.e. by showing that the derivative is always greater 0.
PS: Can you please adapt your math skill level in your profile.
