$x^x$ graph freaks out

Question

So I was goofing around in desmos when I plugged the function
$f(x)=x^x$
And noticed that desmos shows only the positive part of the graph. Why does this happen. The function has real values. Can you explain why it doesn't work?

Answer 1

Desmos doesn't work with imaginary numbers, and (for example) the expression $(-1)^{0.5}$ is imaginary. the exponent must be an integer when x is negative, and that is not continuous, so perhaps Desmos "decides" not to show it.

Answer 2

I suppose it is a convention not to show the negative part of the graph (though the function is real valued at integral points on the negative side of $x$) to avoid the disastrous case of fractional negative indices as $x$ is also in the exponent.

For example,

At $x=-2.3$, $f(x)=(-2.3)^{-2.3}$ which can be evaluated as the reciprocal of $-2.3$ raised to $2.3$ or the tenth root of $\frac{-1}{2.3}$ raised to $23$ i.e. $\sqrt[10]{(\frac{-1}{2.3})^{23}}$ which is an eventh root under a negative expression and hence is complex.

The thing gets interesting here as the expression can also be written as the hundredth root of $\frac{-1}{2.3}$ raised to $230$ i.e. $\sqrt[100]{(\frac{-1}{2.3})^{230}}$ which is a real valued expression indeed. The same expression cannot be complex and real at the same time.

Due to this apparent contradiction, we conventionally don't allow negative exponents to muddle up with negative bases together.

i.e. You won't get the graph of $a^x$ (where, $a$ is -ve) in conventional graphing calculators. Try graphing $(-2)^x$ in desmos; you won't get it.

Thus the graph is discontinuous for the negative inputs and hence we don't get the negative $x$-axis part of it, though it is perfectly defined at negative integral inputs.

I think this is the same reason for which we don't allow negative bases in logarithms so as to avoid the cases in which the arguments are not integral powers of the base.