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A continuous function having an inverse <=> conditions on a derivative?

  1. Jan 16, 2013 #1
    Sorry for the poorly-worded title.

    I help tutor kids with pre-calculus, and they're working inverse functions now. They use the "horizontal line test" to see if a function will have an inverse or not by seeing visually if it's one-to-one.

    I was thinking about what that might imply. If a function passes the horizontal line test (let's assume it's domain is ℝ and that it's continuous everywhere), then it must not have any extremum, since then it'd fail the test. So this would imply that:
    [tex]f'(x) \le 0 \hspace{3pt}\mathrm{or }\hspace{3pt} f'(x) \ge 0 \hspace{3pt} \forall x[/tex]
    I'm wondering if this is always true, and so if it's possible to prove that a bijective function must have a derivative that's either zero or positive/negative everywhere, and I'm wondering if the converse is also true.

    I'm more of a physics-guy than a math-guy, but I do find math interesting and these are the types of questions that I like to think about. Any thoughts?
  2. jcsd
  3. Jan 16, 2013 #2

    Simon Bridge

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    I would imagine that it is always true for the kinds of functions and inverses you are considering and proving the horizontal line test would be sufficient proof of this too. I think the converse logically follows by completeness - to have an inverse, the derivative of the bilinear form must be definite.
  4. Jan 17, 2013 #3
    Consider the function
    [itex]f(x)=\begin{cases}(x+2)^3,&x<-2,\\0,&-2\leq x\leq 2,\\(x-2)^3,&x>2.\end{cases}[/itex]
    You can see that [itex]f(x)[/itex] is continuous on [itex]\mathbb{R}[/itex] and that [itex]f'(x)\geq0\forall x\in\mathbb{R}.[/itex] However, this function fails the horizontal line test on [itex][-2,2][/itex] and therefore does not have a continuous inverse. A sufficient condition for the converse to be true would be a strict inequality.
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