Obviously on the RHS this graph is just a really steep exponential graph however problems arise on the LHS and I cannot find any graph sketching programs that can do. Some will give a graph but then simply say that the LHS is undefined which must be incorrect since negative values with odd powers must still work like $-3^{-3} = -1/27$ but then of course values like $(-1/2)^{-1/2}$ do not. I asked my maths teacher about this and my tutor and both didn't seem to have answers.
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1Do you known what is the definition of $x^x$? – Jun 04 '16 at 01:09
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You might want to see this. – J. M. ain't a mathematician Jun 04 '16 at 01:20
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You are failing to distinguish between $(-1/2)^{-1/2}$ and $-1/2^{-1/2}$. The latter means $-1/(2^{-1/2})$, which is the same as $-(1/2)^{-1/2}$, but different from $(-1/2)^{-1/2}$. $\qquad$ – Michael Hardy Jun 04 '16 at 01:45
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Note that on the interval $0<x<\text{a certain positive number}$, $x^x$ actually decreases as $x$ increases. $\qquad$ – Michael Hardy Jun 04 '16 at 01:48
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yeah thats intresting as well. At first I thought it would decrease between 0 < x < 1 but it seems to around x = 0.4. And this is my first time using the site sorry. Thanks for knowing what i mean. – PythoN00b Jun 05 '16 at 23:26
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When $x<0, x^x$ is undefined (in the real numbers) for most values of $x.$ It is defined for negative odd integers.
However in the complex numbers $x^x$ is defined for negative values of $x.$ It is essentially a spiral in the complex plane, touching the real axis when $x$ is a negative odd integer.
Doug M
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Yes, $(-3)^{-3}$ can be defined, but problems arise when you try to calculate rational powers of some negative value. For instance, think of $(-\frac{1}{2})^{-1/2}$. This can be written as $\sqrt{-2}$. As you can see, this is not a real value. This is why most graph sketching programs don't give us the graph for $x<0$.
zxcvber
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