(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.
^{0.5})
is not defined or
^{-0.5}=1/(-0.5)^{0.5})
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.
