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I have misgivings about the graphs of power functions below.

  1. Are they correct?
  2. Even if correct, while the slope near ∞ for n +even and near 0 for n -even are indeed infinite, the cusps are a bit hard to swallow and I am not up to doing the computations. Is there some informal argument that would make it easier to accept them?

enter image description here

schremmer
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  • Adding details to your post would help your question get more attention. Exactly which functions are those the graphs of? – Mike Pierce Aug 18 '17 at 15:49
  • I have added the words "of power functions" but which were in the title. – schremmer Aug 18 '17 at 15:52
  • You are asking if the graphs in the image are correct. What functions are they supposed to be the graphs of? I would imagine they are the functions given by $x$ being sent to $x^2$, $x^3$, $1/x^2$, and $1/x$ respectively. I assume you knew what they are supposed to be, and providing that information takes just burden of guessing off the reader. – Mike Pierce Aug 18 '17 at 16:06
  • Given that this is a site for mathematicians and that I had titled power functions, I thought anything more might be offending. However, no contest. – schremmer Aug 18 '17 at 16:10
  • That's a reasonable thought to have for not including details. :) But yeah, this site is for mathematicians "of any level," so it's definitely best to include as many details as possible.

    Also, I'm not quite sure what you are trying to say with, "... n +even and near 0 for n -even ... ".

     – Mike Pierce Aug 18 '17 at 16:46
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    I had forgotten that stackexchange uses MathJax. But, at least, I should have written something like when "n is +even". To return a bit: for whatever reason, precalculus students strongly expect the graphs of $x^{+2m}$ to be like a pancake draped on the sphere. Pointing out that the draping of the plane onto the sphere treats both the $x$ and the $f(x)$ directions the same way while the graph goes "faster" in the $f(x)$ direction than in the $x$ direction does not seem to help them overcome their disbelief in having to go "forth and back". – schremmer Aug 18 '17 at 20:39
  • The point of all this is to let the students consider infinity by giving them a way to visualize it. And, after all, real space is curved while planes are only a local approximation, therefore abstract. – schremmer Aug 18 '17 at 20:44

1 Answers1

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I think a nice way to accept those cusps at $\infty$ would be to compare the limits of the derivatives of those functions as $x \to \infty$ in each direction in this clever way. I imagine $f$ will have no cusp at $\infty$ if

$$\lim_{x\to \infty} \left(f'(x) - f'(-x)\right) = 0\,.$$

We can see in the case of $f(x) = x^2$ that this limit is nonzero, so there should be a cusp, but in the case $f(x) = x^3$ the limit is zero.

And then in the case of functions like $f(x) = 1/x^2$ or $f(x) = 1/x$ that approach infinity as $x \to 0$, you can shift the discontinuity to infinity and use the same trick. So if your function $f$ has a discontinuity caused by an asymptote at $x=a$, then you could look at

$$\lim_{x\to \infty} \;f'\left(\frac{1}{x-a}\right) - f'\left(-\frac{1}{x-a}\right) = 0\,.$$

Mike Pierce
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  • Of course but that was not really my question. Students in Pre calculus seem to like the idea of 1-pt compactification but are puzzled by the cusps. My own diffident response has been that $x^2$ goes faster than $x$ but how to explain the effect of the curvature of the sphere. – schremmer Aug 18 '17 at 15:59
  • @schremmer That's exactly what I meant about your question needing more details: you didn't mention precalculus students at all, and when you use the word "compactification" I assumed the target audience of this question/answer would be people who had some background in topology (and so derivatives would be appropriate). – Mike Pierce Aug 18 '17 at 16:02
  • I acknowledge. Sorry and I apologize. By the way, compactifying on a torus is nice too. – schremmer Aug 18 '17 at 16:04