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Proving that a certain f(x) can't be zero by differentiating

  1. Jul 1, 2009 #1
    Hello. I have the following query. Consider a continuous function f(x). We want to know whether this function ever gets zero or not. So we make the assumption that it does. If (e.g) we have [itex]f(x)=x^2 + 4x + 2 - cosx[/itex]

    we'll assume that [itex]f(x)=0 (=) x^2 + 4x + 2 -cosx =0 [/itex]
    now we differentiate with respect to x
    [itex]2x + 4 +sinx =0[/itex] and again
    [itex]2 + cosx = 0 (=) cosx = -2 [/itex] which is impossible.

    Right now we have proven that f ''(x) can't be zero. Is there any theorem that, given some prerequisites for f(x), can show that f(x) can't be zero too, using the above demonstrated way?

    Thanks in advance, I hope you understand the core of my question. (I am not looking for the Bolzano Theorem, btw)

    Edit1: Another way of asking this is : Is there any theorem that proves, given prerequisites, that if [itex]f^n (x)=0 then f(x)=0[/itex], I won't specify how many (n) times differentiated.
  2. jcsd
  3. Jul 1, 2009 #2
    Certainly not in general, as you can well verify that in the example you gave f(x) does take on both positive and negative values even though its second derivative is always strictly positive. It's up to you to come up with the necessary prerequisites of a function to be able to ascertain something on the lines you are proposing. If you succeed let us know:P We could have a theorem named after you :)
    Last edited: Jul 1, 2009
  4. Jul 1, 2009 #3


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    The method that you've shown does nothing to prove or show that f(x) = 0. In fact, the function that you name is zero at 2 points.

    There are a couple of problems with what you posted. First, if f(x) = 0, this implies that given any x that is part of the domain of f(x), f(x) = 0. This is not true of your example function (f(0) = 1 if f(x) = x^2 + 4x + 2 - cos(x)).

    Second, what you really mean to say is suppose for some value x = a, f(a) = 0. Then we have, f(a) = a^2 + 4(a) + 2 - cos(a) = 0. However, since this is a constant if we differentiate f(a) we find that (f(a))' = 0. Note, this isn't the same as f'(a)! Ultimately, this result should not be too surprising.

    We can use derivatives to determine if a function is increasing or decreasing on an interval and this information may help us determine whether or not a function is zero and we can use derivatives to help us find zeroes (Newton's Method for example). However, the fact that the nth derivative of a function is zero alone does not suggest much about whether or not a function is zero.
    Last edited: Jul 1, 2009
  5. Jul 1, 2009 #4
    Okay, first of all @Bobbybear: That was very flattering :)

    @jgens: I think you misunderstood my post. I was not presenting a way to prove that f(x)=0 via its f'(x),f ''(x)... I was merely asking if there is a theorem, that given a derivative of the function f(x), be it f ' (x), can't be zero { f '(x) ≠ 0 } can tell us whether the initial function f(x) ≠0 too. That happening of course, given several restrictions for f(x).

    If I am still being unclear , let me know :)
  6. Jul 1, 2009 #5


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    Sorry for misunderstanding your post.

    To my very limited knowledge, no such theorem exists. However, given the nature of the theorem you propose, I very much doubt that one exists (at least that is known to the mathematical community) since such a theorem would probably be fairly significant - i.e. you'd probably be learning about it in your calculus class.
  7. Jul 1, 2009 #6
    Yes, I think so too. That's partly a reason why I'm asking :)
  8. Jul 8, 2009 #7
    You cannot "differentiate both sides" of [itex] f(x) = 0 [/itex] unless this equation holds for all x in the domain of f.

    [itex] f(x) = 2x [/itex] is continuous on R.

    Suppose [itex] 2x = 0 [/itex].

    "Differentiating both sides" :

    [itex] 2 = 0 [/itex]. (is not a valid equation)

    However we know that [itex]f(0) = 2(0) = 0[/itex].
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