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Interesting Derivative Proof Question

  1. Feb 11, 2015 #1
    1. The problem statement, all variables and given/known data
    I recently searched around SE, and found:



    http://math.stackexchange.com/questions/1142546/how-to-solve-this-derivative-of-f-proof


    2. Relevant equations

    Below

    3. The attempt at a solution

    The answer is interesting.

    "A function given that $$f(x)=f''(x)+f'(x)g(x)$$ could be an exponential function, sine, cosine , quadratic polynomial or $$f\equiv0$$. So we can say that the function is a continuous function $\in C^2$.

    The right negation is that $$f(x)\ge0 , \in (a, b)$$ and exist a point c | $$f(c)>0$$.


    You have that $$f''(x_1)\ge0$$ (the function in that point is convex) so in that point you have a minima so there are two case

    1. $$f(x_1)<0$$ (obviously contradiction)
    2. $$f(x_1)=0$$ (it's impossible because this imply that f(x) = 0)

    Analog for the other case"

    But how does $$f''(x_1) > 0$$ show that $$f(x_1) < 0$$? I dont understand the contradiction for #1??

    Thanks!
     
  2. jcsd
  3. Feb 11, 2015 #2

    PeroK

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    The proof is saying that if f is not identically 0, then it must attain its maximum (or minimum) at x_1 in (a, b). And f''(x_1) = f(x_1) . But, using the second derivative test, if f(x_1) > 0, then we have a local minimum and if f(x_1) < 0, we have a local maximum. Contradiction.
     
  4. Feb 11, 2015 #3

    epenguin

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    Finding this strange. This should generate an infinite number of zeroes in the interval?
     
  5. Feb 12, 2015 #4
    But how is that a contradiction and to what?
     
  6. Feb 12, 2015 #5

    PeroK

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    It's basic calculus. If f(x_1) > 0, how can that be a local minimum?
     
  7. Feb 12, 2015 #6

    $$f(x) = x^2 + 1$$?

    It has a minimum f(x_1) > 0?
     
  8. Feb 12, 2015 #7

    PeroK

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    We're not talking here about an arbitrary function. We're talking about one that meets the key criteria stated in the problem. The first of these is that f(x) = 0 at two distinct points. The second criterion is that f(x) = f''(x) whenever f'(x) = 0.

    If such an f has a minimum it must be < 0. And f'' must be < 0 at that point. But, f'' < 0 implies it's a local maximum. That's the contradiction.
     
  9. Feb 12, 2015 #8
    Ok, this is starting to make sense.


    The only part I do not understand is.

    "If such an f has a minimum it must be < 0"

    Why does this have to be true? Thanks!
     
  10. Feb 12, 2015 #9

    PeroK

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    To be more precise, the proof was:

    If f(x) < 0 for some x in (a, b), then f attains its absolute minimum on (a, b).

    And, if f(x) > 0 for some x in (a, b), then f attains its absolute maximum on (a, b).
     
  11. Feb 12, 2015 #10

    ???

    But I am not asking what the question asked? I was asking,

    How if f(x) < 0 it is a contradiction to have f"(x) > 0 ??
     
  12. Feb 12, 2015 #11

    PeroK

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    It isn't! Lots of functions have that property. But NOT the one in the question which meets certain criteria.

    Do you not understand that a proof may apply to a function with certain properties, but not to all functions?
     
  13. Feb 12, 2015 #12

    I do understand it.

    But can you show that for our function it is a contradiction that,

    if f(x) < 0 tthen the contradiction is to have f''(x) > 0??
     
  14. Feb 12, 2015 #13

    PeroK

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    That's not correct either. We have f(x) < 0, f'(x) = 0 and f''(x) = f(x) < 0.
     
  15. Feb 12, 2015 #14
    Yes but how will it be a contradiction?
     
  16. Feb 13, 2015 #15

    epenguin

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    Is the misunderstanding here that Amad is trying to prove a statement true by contradiction, while PeroK has proved it false by contradiction?
    Perhaps the problem has been wrongly copied or misunderstood, for which there are other indications.

    Also I saw earlier without even looking at the posts that if a function is such that between any two zeroes it has another zero, then it will have an infinite number of zeroes. Which is not impossible, but it would be a funny function to have an excercise on at this level.
     
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