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Thread: 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 $\displaystyle f'(a)=0$, was that supposed to be $\displaystyle f'(c)=0$

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

    It is just different to see $\displaystyle f'(a)=0$ 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 $\displaystyle y=f'(c)x-cf'(c)+f(c)$.
    To pass through (0,0) we must have $\displaystyle f(c)-cf'(c)=0$.
    Consider the function $\displaystyle \frac{f(x)}{x}$ 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 $\displaystyle g(x)=\frac{f(x)}{x}$ then $\displaystyle g(a)=g(b)=0$.
    Moreover, on $\displaystyle (a,b)$ $\displaystyle g'(x) = \frac{{xf'(x) - f(x)}}{{x^2 }}$ so if $\displaystyle g'(c)=0 $ then $\displaystyle cf'(c) - f(c)=0 $.
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