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

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 $x>=0.$

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  • $\begingroup$ 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. $\endgroup$ – Mark McClure Oct 25 '18 at 14:10
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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:

enter image description here

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:

enter image description here

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.

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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.

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  • $\begingroup$ Wow. So is the graph on WolframAlpha incorrect? $\endgroup$ – user163862 Oct 24 '18 at 22:59
  • $\begingroup$ Yes. It is clear that there is also the part of the graph for which $x<0$. $\endgroup$ – José Carlos Santos Oct 24 '18 at 23:01
  • $\begingroup$ @user163862 No, the WolframAlpha graph is emphatically not incorrect. $\endgroup$ – Mark McClure Oct 24 '18 at 23:11
  • $\begingroup$ @MarkMcClure I disagree. The graphic is not the graphic of the function $f\colon\mathbb{R}\longrightarrow\mathbb R$ defined by $f(x)=2x+3x^{\frac23}$. $\endgroup$ – José Carlos Santos Oct 24 '18 at 23:17
  • $\begingroup$ @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. $\endgroup$ – Mark McClure Oct 24 '18 at 23:23

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