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Math Help - [SOLVED] Derivatives of real valued functions

  1. #1
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    [SOLVED] Derivatives of real valued functions

    Let f and g be twice differentiable real-valued functions defined on \mathbb{R}. If f'(x)>g'(x) \forall x>0, which of the following must be true for all x>0?
    (a) f(x)>g(x)
    (b) f''(x)>g''(x)
    (c) f(x)-f(0)>g(x)-g(0)
    (d) f'(x)-f'(0)>g'(x)-g'(0)
    (e) f''(x)-f''(0)>g''(x)-g''(0)

    The answer is c but I thought it was d. Can someone show me how to prove it is c?
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  2. #2
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    Integrate f'>g' from 0 to x.
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  3. #3
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    Quote Originally Posted by maddas View Post
    Integrate f'>g' from 0 to x.
    That is just f(x)-f(0) and same for g
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  4. #4
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    f(x)-f(0) = \int_0^x f' > \int_0^x g' = g(x)-g(0)...

    edit: to see its not (d), take f(x)=2x and g(x)=x.
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  5. #5
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    Then why wouldn't the derivative of f'(x)>g'(x) also be an answer?
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  6. #6
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    (a) is not true, take f(x) = 2x and g(x) = x+1.

    (b), (d), and (e) are not true, take f(x)=2x, g(x) = x.
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  7. #7
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    Quote Originally Posted by dwsmith View Post
    Let f and g be twice differentiable real-valued functions defined on \mathbb{R}. If f'(x)>g'(x) \forall x>0, which of the following must be true for all x>0?
    (c) f(x)-f(0)>g(x)-g(0)
    Can someone show me how to prove it is c?
    We know the derivative of f-g is f '-g '>0.
    Therefore f-g is increasing or f(x)-g(x)>f(0)-g(0).
    Thus answer (c).
    Last edited by Plato; May 9th 2010 at 12:54 PM.
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