Y=X^X Graph Problem?

Question

I plotted a graph on my calculator, which was Y=X^X. It drew a graph that was correct on the positive side of the X axis, but the negative side was blank. It was as if that the graph had gone straight back where it came from. Is this correct, because surely -1^-1=-1, so there should be at least another point on the negative axis.

I've taken this picture from the graph program. And thanks to your answers, I have found there are two small dots on the negative X axis!

See it at http://www.kiano.net/maths.bmp

Answer 1

But your graphical calculator probably doesn't plot -1 unless the x-value of a pixel is exactly -1.

Answer 2

This is interesting. If your calculator could show them there would be an infinite number of points corresponding to negative x values. This is because when x is an integer x^x certainly gives a real result. For instance (-5)^(-5) = 1/((-5)^5) and this is the real number
-0.00032 but the result is so small that the calculator effectively plots it on the negative x axis. More interestingly x^x also has a real result for negative rational fractions with an odd denominator. Consider (-a/b)^(-a/b) = 1/((-a/b)^(a/b) = (-b/a)^(a/b)
= bth root (-b^a)/bth root(a^a). Now if b is odd then the top of the fraction is real and of course the bottom always is so the result is real. I wonder why your calculator did not plot for example
(-2/3)^(-2/3) = -1.311 approx. Perhaps it's because calculators don't work in fractions.
Therefore the graph on the negative x side should look like a curve underneath the x axis with an infinite number of holes (where x is not rational with even denominator) infinitely close together but you would never see them! Apart from this, the line also has occasional points above the axis where x is a negative even integer. Isn't it a weird function? I really enjoyed working on it.
By the way (-1)^(-1) DOES = -1 because (-1)^(-1) = 1/(-1) = -1

Later edit. I now realise that there is one small error in the above explanation but it is EMPHATICALLY not in the last line.

The error concerns the sign of the final result when using rational fractions. In the case of (-a/b)^(-a/b) I had forgotten that in the final part, the bth root of (-b^a), the sign will depend on whether a is odd or even. Thus (-1/3)^(-1/3) = -1.442 approx, i.e. negative but
(-2/3)^(-2/3) = +1.311 approx. positive. Thus the points above the axis are not isolated and the "curve with holes" has two parts, one above the axis and one below.

Final comment to the person below and some above. We are not just doing simple multiplication. The situation is much more complicated than that. Please do not make dogmatic assertions unless you are really sure of yourself, particularly when you are contradicting someone who appears to know what they are talking about. I do make mistakes but not ones that simple.

Answer 3

theres nothing wrong with your calculator and you are correct about -1^-1 = -1
as you get more negative, your function is decreasing. And since your graphing window is set so that you see the increase on the positive end, you will not see the negative end. kinda get what i mean? only applies to integers
there is also the fact that there are complex numbers. any fractional part, u get a root of some kind

Answer 4

I would not think the other side was empty.
-1^-1 = -1
-2^-2 = 0.25
-5^-5 = 0.00032
So you should have a number of values plotted, however, they are all INTEGER values. So you need to turn off grid lines if you have them on.
Any negative number raised to a negative non-integer will not result in a real number.

Answer 5

x^x=1 when x = -2 and is complex when x = -0.5 for example. So I get the impression that it alternates between 1 and -1 for x a negative integer and is complex elsewhere.

Perhaps the calculator objects to complex numbers, after all how could it draw a graph of them?

Answer 6

You can only have positive as -1 times -1 = 1
so any negative number multiplied by negative number becomes positive.

Answer 7

there should be no negative x-axis. this is simply because, when you have said that y=x^2 , when x is a negative number and you squared it, it will become a positive number. so there's no way you can have a postive one. cos all squares of a negative number are positive.
or that maybe your calculator dont have -1
use this website below

Answer 8

The problem is fractional powers. Numbers like (-0.5)^(-0.5) are imaginary, and in general (-x)^(-x) is complex, so the graph doesn't actual exist on real axes for -ve x. The results of (-x)^(-x) is only real for x integer.

Answer 9

The function x^x is only defined if the base x >0

Answer 10

No, -1^-1=1, It is NOT correct
Please give me best answer thanks!