# Understanding this graph

The function is given as $$f(x)=2x+3x^\left(2/3\right)$$. The enclosed graph is from wolfram..

My question is why doesn't $$f(-1)=1$$ show up to the left of the y axis. I don't see why the domain of this graph is $$x>=0.$$

• Please see my edited response, if you'd like to understand why this is not an error on WolframAlpha's part but, rather, a simple design decision. The answer also shows two simple ways to trigger the use of the real root, if that is what you desire. – Mark McClure Oct 25 '18 at 14:10

First off, the source of the confusion is that WolframAlpha uses the principal root so that $$(-1)^{2/3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i,$$ as you can see like so: (-1)^(2/3).

This is in no way an error on the part of WolframAlpha (as indicated in this comment on Jose's otherwise reasonable answer) but it is a simple design decision.

There are two simple ways to trigger the use of the real root, if you desire.

## Getting the real root with a link

If you make your screenshot just a little bit larger, you see the following:

Notice that there is a link that allows you to "use the real-valued root". When you press that, you generate the graph that you expect:

## Getting the real root with cbrt

If you know that you want to use the real root, you can use cbrt to denote that from the outset. Thus, cbrt(-1)^2 returns $$1$$, as you expect and 2x+3cbrt(x)^2 gives the graph that you'd expect.

Because when you write $$x^{\frac23}$$, what you have in mind is (I suppose) $$\sqrt[3]{x^2}$$. Yes, this is defined for every real number and it is another real number.

However, many computing systems “think” that$$x^{\frac23}=e^{\frac23\ln x}.$$ Since there are no (real) logarithms of negative real numbers, this ledas to a proble, which is reflected in the picture that you posted here.

• Wow. So is the graph on WolframAlpha incorrect? – user163862 Oct 24 '18 at 22:59
• Yes. It is clear that there is also the part of the graph for which $x<0$. – José Carlos Santos Oct 24 '18 at 23:01
• @user163862 No, the WolframAlpha graph is emphatically not incorrect. – Mark McClure Oct 24 '18 at 23:11
• @MarkMcClure I disagree. The graphic is not the graphic of the function $f\colon\mathbb{R}\longrightarrow\mathbb R$ defined by $f(x)=2x+3x^{\frac23}$. – José Carlos Santos Oct 24 '18 at 23:17
• @JoséCarlosSantos Right - but WolframAlpha doesn't assume that $f:\mathbb R \to \mathbb R$. I do know what I'm talking about here; I personally wrote the code that produces this graph. Here is my blog post describing how WolframAlpha deals with roots. – Mark McClure Oct 24 '18 at 23:23