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Math Help - Rolle Theorem and MVT?

  1. #1
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    Rolle Theorem and MVT?

    In the interval [0,3] f(x)= {1 , x=0
    ax+b , 0 less than x less than or equal to 1
    x2+4x+c , 1 less than x less than or equal to 3
    (if it isnt clear, f(x) is a piecewise function)

    How do i solve for a, b and c?
    thanks!

    Anyone?? Please??
    Last edited by painterchica16; October 23rd 2008 at 07:14 PM.
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  2. #2
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    you didn't say so, so let me guess ... you need to find the values of a, b, and c so that the MVT will apply to f(x) on the interval [0,3] ?

    If the MVT applies, then f(x) must satisfy what two conditions?
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  3. #3
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    Sorry for not making it clear!

    I have to determine the values of a,b and c such that the function f satisfies the hypotheses of the Mean Value Theorem on the interval [0,3]
    Does that make more sense?
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  4. #4
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    f(x) needs to be continuous on [0,3] and differentiable on (0,3).

    \lim_{x \to 0^+} ax + b = 1

    so , that means b = 1


    to be continuous at x = 1 ...

    f(1) = a+1

    \lim_{x \to 1^-} ax+1 = \lim_{x \to 1^+} x^2 + 4x + c

    a+1 = 5+c


    to be differentiable at x = 1 ...

    \lim_{x \to 1^-} \frac{(ax+1) -  (a+1)}{x - 1}

    \lim_{x \to 1^-} \frac{a(x-1)}{x - 1} = a


    \lim_{x \to 1^+} \frac{(x^2+4x+c) -  (5+c)}{x - 1}

    \lim_{x \to 1^+} \frac{x^2+4x-5}{x - 1}

    \lim_{x \to 1^+} \frac{(x+5)(x-1)}{x - 1} = 6

     a = 6

     c = 2
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  5. #5
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    Thanks so much. I appreciate it!
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