$$h(x) = x^2 + 2x^{-1} $$

a)Identify the natural domain of $h$, asymptotes and the interval(s) on which $h(x)\lt0$

Based on the given function, it is undefined at $x = 0$, so the natural domain would be any number except $0$. And based on the function, the function has a vertical asymptote ie at $x=0$, which is basically the $y$ axis.

I'm a bit confused with the last question although the only rational way would be to set the equation to be $\le0$ and try to solve for $x$ which didn't give me a precise interval. Could someone explain intuitively or arithmetically as to how to go about it?

  • 1
    $\begingroup$ Is it $<$ or $\le$ ? $\endgroup$ – Yves Daoust Feb 16 '17 at 10:43
  • $\begingroup$ What do you mean it doesn't give you a precise interval? $\endgroup$ – Jack M Feb 16 '17 at 10:44
  • $\begingroup$ Sorry, it's <0. I mean computing that would give me the cube root of negative 2 which would be approximately -1.26, so would x<-1.26 be my interval for the same? $\endgroup$ – Gary Andrews30 Feb 16 '17 at 10:45
  • $\begingroup$ $(-\sqrt[3]2, 0)$ sounds like a pretty precise interval to me. $\endgroup$ – Jack M Feb 16 '17 at 12:34

You are asked to solve the inequation


Assuming $x>0$, you can multiply by $x$ and get

$$x^3+2<0,$$ which is not possible.

Then assuming $x<0$,

$$x^3+2>0,\\x^3>-2,\\ x>-\sqrt[3]2.$$



This curve is known as the "Trident of Newton", or the "Parabola of Descartes". http://www.mathcurve.com/courbes2d/trident/trident.shtml

  • $\begingroup$ +1 For the additional information about the name of the function:). $\endgroup$ – MrYouMath Feb 16 '17 at 11:17
  • $\begingroup$ @MrYouMath: the site linked to is a sheer wonder. $\endgroup$ – Yves Daoust Feb 16 '17 at 11:19
  • $\begingroup$ Already bookmarked the site ;-). $\endgroup$ – MrYouMath Feb 16 '17 at 11:21

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