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Math Help - Rolle's Theorem

  1. #1
    No one in Particular VonNemo19's Avatar
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    Rolle's Theorem

    Determine whether Rolle's theorem can be applied to f on the interval and, if so, find all values of c in the open interval (a,b) such that f'(c)=0.

    f(x)=4x-\tan{\pi{x}}, [-\frac{1}{4},\frac{1}{4}]

    Here's what I've done:

    4x is everywhere cont./diff. and \tan{\pi{x}} is everywhere cont. on [-\frac{1}{4},\frac{1}{4}] and everywhere diff. on (-\frac{1}{4},\frac{1}{4}), therefore f(x) is everywhere cont./diff. on (-\frac{1}{4},\frac{1}{4}). Furthermore, f(-\frac{1}{4})=0=f(\frac{1}{4}), so Rolle's theorem applies.

    f'(x)=4-\pi\sec^2\pi{x}=0
    \pi\sec^2\pi{x}=4

    \sec{\pi{x}}=\pm\frac{2}{\sqrt{\pi}}

    {x}=\frac{1}{\pi}cos^{-1}\left(\pm\frac{\sqrt{\pi}}{2}\right)\Rightarrow{  c}=\frac{1}{\pi}cos^{-1}\left(\pm\frac{\sqrt{\pi}}{2}\right)

    Could someone please verify my result?
    Last edited by VonNemo19; June 28th 2009 at 11:21 AM.
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  2. #2
    MHF Contributor Amer's Avatar
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    I think it is correct I did not see any mistakes but you should take the solution which is in the interval [\frac{-1}{4},\frac{1}{4}]

    c=\frac{1}{\pi} cos^{-1}\left(\frac{\sqrt{\pi}}{2}\right)\approx 0.1533172924205\in [\frac{-1}{4},\frac{1}{4}]


    c=\frac{1}{\pi}cos^{-1}\left(-\frac{\sqrt{\pi}}{2}\right)\approx 0.8466827075795 \ne [\frac{-1}{4},\frac{1}{4}]
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