Differing graphs for simple inverse exponential problem

In class, we are learning exponential functions.

The following inverse exponential problem is bothering me: $y=x^{-\frac{1}{9}}$.

When graphed, I feel that it should look like it does on Desmos:

But other software is giving me different results. Look at, for example, Google:

and WolframAlpha:

and my trusty TI-84 calculator (which may only look incorrect because of its low pixel density):

There appears to be unreal (imaginary) answers involved, as is shown in the WolframAlpha screenshot. But, as we've learned in class, with any odd root there shouldn't be unreal answers, even for negatives (for example, $(-64)^{\frac{1}{2}} = 8i$, but $(-64)^{\frac{1}{3}} = -4$).

Can you please explain to me which one is right? (Or, if all are right in their own respectable ways?)

• The meaning of the -1/9-th power of a negative number is up for debate. However, I think Google is just plain wrong here :-( Commented Mar 29, 2016 at 23:45
• @DanPiponi How so? How is it possible that there exists other, non-real answers to the -1/9th power with negative (real) inputs? We've learning nothing of such answers. Commented Mar 29, 2016 at 23:48
• It's a result of something known as complex analysis that the nth root of any number has n answers even if some are complex. One such phenomenon which occurs often are the nth roots of unity (the number 1) Commented Mar 29, 2016 at 23:57
• The problem is that there are 9 9th roots of any non-zero number. For positive numbers there's an obvious one that stands out. For negative numbers it's not clear which one you should pick. The real one (like Desmos did) or the first one you meet as you walk anticlockwise around the circle starting at a positive real number (as Mathematica does). Commented Mar 29, 2016 at 23:58

When you have something like $f(x) = x^{\frac{-1}{9}}$ you have many choices as to what to plot. For example, let us consider the argument $x = -1$. There are 9 complex numbers that are of the form $(a + bi)^{-9} = -1$.

You may think think all we have is $f(-1) = -1$, however, we also have many imaginary roots.

Soon enough you will learn Euler's formula and you will figure out these other roots.

$$e^{i \theta} = \cos \theta + i \sin \theta$$.

For example, another possible root is $e^{\frac{-i \pi}{9}}$

See what happens when you raise that to the $-9$th power.

As for the graphing software, it's really up to them what to graph. (you should notice that you have clicked on the complex valued plot for Wolfram)

Your example of $(-64)^\frac{1}{2}$ has two roots, them being $8i$ and $-8i$.

Additionally, $(-64)^\frac{1}{3}$ has 3 roots. We have $4e^{\frac{i \pi}{3}}, \>-4, \>4e^{\frac{-i \pi}{3}}$.

• Although I am unsure what this means right now, I will be sure to ask my teacher about this. Thanks! Commented Mar 30, 2016 at 0:03
• And the ninth power of that e expression is very interesting. I never would've thought it equals -1... Commented Mar 30, 2016 at 0:05
• @JonathanLam Desmo is plotting the real solution, but wolfram alpha is plotting what it's called the principal root, which contains an imaginary number
– lEm
Commented Mar 30, 2016 at 0:06
• @JonathanLam If you know the fundamental theorem of algebra, then something of the form $x^n+\dots=0$ has $n$ different answers for $x$, which is what is being said here. Commented Mar 30, 2016 at 23:30