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I have a question about the graph of $f(x) = x^x$. How come the graph doesn't extend into the negative domain? Because, it is not as if the graph is undefined when $x=-5$. But according to the graph, that seems to be the case. Can someone please explain this?


enter image description here

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Very related post – amWhy May 21 '13 at 15:26
@amWhy I was actually inspired by that post but I didn't think they were related for some reason. Maybe both posts can complement each other? – Jeel Shah May 21 '13 at 15:27
Negative numbers cannot be raised to arbitrary real powers and produce real outputs. Just consider what $(-1/2)^{-1/2}$ would be. However you can extend this function in a "nice" way by looking at $|x|^x$. – alex.jordan May 21 '13 at 15:28
@alex.jordan I put $(-1/2)^{-1/2}$ and I get imaginary but when I put $(-5)^{-5}$ I get answer that is $-0.00032$. Can you provide more detail please? – Jeel Shah May 21 '13 at 15:50
@gekkostate Raising a negative number to a whole number is not a problem. $(-5)^{-5}$ means $1/((-5)(-5)(-5)(-5)(-5))$. But $x^{-1/2}$ should be a square root of $x$,and if $x$ is negative, no square roots are real. And then there are things like $(-2)^{\pi}$. What would that possibly mean to you? At least with $2^{\pi}$ we could approximate $\pi$ with a rational number $m/n$ and $2^{\pi}$ would be something close to $\sqrt[n]{2^m}$. But for negative bases we'll have trouble executing that $\sqrt[n]{}$. – alex.jordan May 21 '13 at 15:56

1 Answer 1

Well, for a START, what would happen if your exponents were to be -0.5, -1.5, -2.5, -3.5 etc? Convert that into radical notation and try to evaluate for those negative x-values. (And there is lots more going on for negative x-values)

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