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Proving if f'=g', then f=g+k for all k constant

  1. Oct 27, 2012 #1
    1. The problem statement, all variables and given/known data
    Let f and g be two continuous functions that are defined on [a,b] and differentiable on (a,b).

    Claim: If f'=g' on (a,b), then there exists a constant k such that f(x)=g(x)+k for all x in [a,b]


    2. Relevant equations



    3. The attempt at a solution

    Let h:[a,b]->R be defined by h(x)=g(x)+k-f(x)

    h is continuous and differentiable on [a,b] by the assumptions of f and g already made.

    Since f'=g' => g'-f'=0

    So h'(x) = g'(x) - f'(x) = 0

    By the mean value theorem, there exists a c in (a,b) so that

    0=(h(b)-h(a))/(b-a) => h(a)=h(b)

    I think I showed rolle's theorem. I am trying to use my assumption that f'=g' to show the claim. Any hints on this?
     
  2. jcsd
  3. Oct 27, 2012 #2

    jgens

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    Consider the function defined by [itex]h(x) = f(x)-g(x)[/itex] and apply the mean value theorem to show that this is a constant function.
     
  4. Oct 27, 2012 #3
    If h(x)=f(x)-g(x), then h'(x)=f'(x)-g'(x) and, since f'=g' by assumption, f'(x)-g'(x)=0 for all x in (a,b). By MVT, there is a c in (a,b) so that h'(c)=h(b)-h(a) = 0 or

    [f(b)-g(b)-f(a)+g(a)]/(b-a) = 0 => f(b)-f(a)=g(b)-g(a) ?

    If h'(x) = 0 for all x in (a,b), then h is constant.

    Is this what you meant?
     
  5. Oct 27, 2012 #4

    jgens

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    Not exactly. The idea is to suppose [itex]x,y \in (a,b)[/itex] and compute [itex]h(x)-h(y)[/itex] using the mean value theorem. This will tell you that the function is constant and complete the proof.
     
  6. Oct 27, 2012 #5
    So choose x,y in (a,b) arbitrary so that h(x)-h(y) = 0?
     
  7. Oct 27, 2012 #6

    Dick

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    Do a proof by contradiction. Define h(x)=f(x)-g(x) and suppose it's not constant. What does the MVT tell you?
     
  8. Oct 27, 2012 #7

    Ray Vickson

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    Yes, but you need to prove that, not just write it down.

    RGV
     
  9. Oct 27, 2012 #8
    MVT tells me that there is a point in the interval (a,b) such that h'(c) = [h(b)-h(a)]/(b-a)

    So f'(c)-g'(c) = [h(b)-h(a)]/(b-a) and f'(x) = g'(x) for all x in [a,b]

    So 0 = [h(b)-h(a)]/(b-a) => h(b)=h(a) making it a constant function.
     
  10. Oct 27, 2012 #9

    Dick

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    You aren't really stating that it in "proof by contradiction" form. Suppose h(a) is NOT equal to h(b) what would bother you about that?
     
  11. Oct 27, 2012 #10
    What would bother me that is, when f'(x)=g'(x), is that h(a) is h(b). And when h'(x) = 0 for all x in (a,b), then h(x) is constant.
     
  12. Oct 27, 2012 #11

    Dick

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    You are just stating what you already think you know without proving anything. If h(a)-h(b) is not zero then (h(a)-h(b))/(a-b) is not zero. The MVT should give you a contradiction that would prove that can't happen.
     
  13. Oct 28, 2012 #12
    Let me try this again, then...

    I want to show that if f'=g' on (a,b), then there exists a constant k such that f(x)=g(x)+k for all x in [a,b]

    Let f and g be two continuous functions that are defined on [a,b] and differentiable on (a,b).
    Assume f'(x)=g'(x) for all x in [a,b]

    Let h:[a,b]->ℝ be defined by h(x)=f(x)-g(x)

    Suppose h(a)≠h(b)

    Then there is no point c in (a,b) so that h'(c)=0

    By the MVT, there is a point d in (x,y)[itex]\subset[/itex][a,b] such that h'(d) = [h(y)-h(x)]/(y-x)

    Since h is differentiable on all of [a,b], h'(x)=f'(x)-g'(x)

    And f'-g'=0 by assumption.

    So h'(d)=0=[h(y)-h(x)]/(y-x) => h(y) = h(x) for all x,y in [a,b]

    This means h'(x)=0 for all x in [a,b] and by the monotonicity theorem, h is constant on [a,b].

    So f and g must be the same.
     
  14. Oct 28, 2012 #13

    Dick

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    Things start going wrong about here. That is just plain not true. You are negating statements wrong, I think. What does h(a)≠h(b) really imply?
     
  15. Oct 28, 2012 #14
    That the function has different values at the end points of the interval.
     
  16. Oct 28, 2012 #15

    Dick

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    What does h(a)≠h(b) imply if you use the mean value theorem?
     
  17. Oct 28, 2012 #16
    h'(c) = [h(b)-h(a)]/(b-a) ≠ 0
     
  18. Oct 28, 2012 #17

    Dick

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    Yes! There is c in (a,b) such that h'(c)≠0. But you know h'(x)=0 for all points c in (a,b). So?
     
  19. Oct 28, 2012 #18
    h'(c) must be 0 too since c is in (a,b)
     
  20. Oct 28, 2012 #19

    Dick

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    Yes. So h'(c) must be both zero and nonzero. That's impossible, so your initial assumption h(a)≠h(b) must be wrong. So?
     
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