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Theorem on continuous function crossing x-axis

  1. Aug 16, 2014 #1
    I think this is a theorem, and I'm telling myself that I've proved it. Just a shout out for any possible counter-examples:
    If a function f(x) is continuous on some interval and has non-zero derivatives at its root(s) (where f(x')=0 ) then there is some interval around the roots where there are no other roots, and f(x)><0 for x><x' as f'(x')><0.

    It just says a function crossing the x-axis comes from below and goes above or comes from above and goes below when the derivative at the point is not zero. Any counter-examples?
     
  2. jcsd
  3. Aug 16, 2014 #2
    There is a more general theorem that can be proven:

    If ##f## is differentiable at ##a## and ##f'(a)\neq 0##, then there is an open interval ##I## with ##a\in I## such that for all ##x\in I\setminus\{a\}##, ##f(x)\neq f(a)##.
     
  4. Aug 16, 2014 #3

    jbunniii

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    Here's a simple proof for the result mentioned by gopher_p.

    Suppose that ##f'(a) \neq 0##. By definition, this means that
    $$\lim_{h \rightarrow 0}\frac{f(a+h) - f(a)}{h} = f'(a) \neq 0$$
    Therefore if we set ##\epsilon = |f'(a)|/2## then there is a ##\delta > 0## such that
    $$\left|\frac{f(a+h) - f(a)}{h}\right| > \frac{|f'(a)|}{2} > 0$$
    for all ##h## satisfying ##0 < |h| < \delta##. Therefore, for all such ##h##, we have
    $$|f(a+h) - f(a)| > |h|\frac{|f'(a)|}{2} > 0$$
    and as a result, ##f(a+h) \neq f(a)## for all ##h \in (-\delta, \delta) \setminus \{0\}##.
     
    Last edited: Aug 17, 2014
  5. Aug 16, 2014 #4

    WWGD

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    if you don't mind the heavy machinery; using a tank to kill a fly and f' is continuous, just use the inverse function theorem.
     
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