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Math Help - Prove using (ONLY) Rolle's Theorem (??)

  1. #1
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    Prove using (ONLY) Rolle's Theorem (??)

    How do I prove the following theorem, using ONLY Rolle's theorem?

    Suppose that for some real number a,
    f(x) is continuous on [a, inf)
    and f '(x) exists on (a,inf)
    with f '(x)>0 for all x>a
    Then f(x)>f(a) for all x>a

    I think I'm right in saying that Rolle's theorem is;

    if a function f of real value is continuous on [a,b], differentiable on (a,b) and f(a)=f(b) then there exists a C in (a,b) such that f'(C)=0

    but how do i PROVE the above theorem from that?

    Please help
    Thank you
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by DBsHo View Post
    How do I prove the following theorem, using ONLY Rolle's theorem?

    Suppose that for some real number a,
    f(x) is continuous on [a, inf)
    and f '(x) exists on (a,inf)
    with f '(x)>0 for all x>a
    Then f(x)>f(a) for all x>a

    I think I'm right in saying that Rolle's theorem is;

    if a function f of real value is continuous on [a,b], differentiable on (a,b) and f(a)=f(b) then there exists a C in (a,b) such that f'(C)=0

    but how do i PROVE the above theorem from that?

    Please help
    Thank you
    Clearly, we must have that either f(x)<f(a) or f(x)>f(a) for all x\in[a,\infty]. To see this suppose that there were some points where f(x)<f(a) and some where f(x)>f(a). Then, since f is continuous the IVT furnishes us with a \xi\in[a,\infty] such that f\left(\xi\right)=f(a) at which point Rolle's theorem tells us that f'(x)=0 for some point in (a,\infty), this is a contradiction.
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  3. #3
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    Thank you for your reply, I find these hard to grasp, I over complicate and confuse myself. Could you possibly elaborate futher?
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  4. #4
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    Basically he has shown that you must have that either f(x)>f(a) or f(x)<f(a) for all x\in (a,\infty). Otherwise you would contradict f'(x)>0  \forall x\in (a,\infty). Then since f'(x)> you know which of the choices must be correct.
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