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Math Help - Find c in (a, b) so tangent passes through 0

  1. #1
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    Find c in (a, b) so tangent passes through 0

    Let a; b be real numbers such that 0 < a < b. Let f be a function continuous on [a, b]
    and dif erentiable on (a, b). Assume that f satis es the property
    f(a) = f(b) = 0 and f'(a) = 0:
    Show that there exist c in (a; b) such that the tangent to f at c passes through 0.

    Any help? I am totaly lost.
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  2. #2
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    Sounds like the Mean Value Theorem
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  3. #3
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    hmm or could it be Rolles Theorem...
    Anyon else?
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  4. #4
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    When you wrote f'(a)=0, was that supposed to be f'(c)=0

    If you meant f'(c)=0, then it is Mean Value.

    It is just different to see f'(a)=0 since that doesn't come up in either theorem.
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  5. #5
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    No it is actually f'(a)...
    That is why I am getting confused. I have never seen that before
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  6. #6
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    The tangent line is y=f'(c)x-cf'(c)+f(c).
    To pass through (0,0) we must have f(c)-cf'(c)=0.
    Consider the function \frac{f(x)}{x} apply Rolle's Theorem.
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  7. #7
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    is x = a?
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  8. #8
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    Suppose that g(x)=\frac{f(x)}{x} then g(a)=g(b)=0.
    Moreover, on (a,b) g'(x) = \frac{{xf'(x) - f(x)}}{{x^2 }} so if  g'(c)=0 then  cf'(c) - f(c)=0 .
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