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Math Help - One Mean Value Thm Problem

  1. #1
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    One Mean Value Thm Problem

    I'm new to M.V.T. and Rolle's Thm, so I guess I'm not quite comphrehending the concepts... Here's the problem:

    Suppose that f is continuous on [0,4] and differentiable on (0,4). Also, suppose f(0)=1 and 3<f'(x)<6 for all x in (0,4). Which of the following must be true?

    A) 13<f(4)<25

    B) 3<f(4)<6

    C) 12<f(4)<24

    D) f(4) is undefined


    Any help/explanation that would aid in my understanding of the concepts would be very appreciated!
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  2. #2
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    Quote Originally Posted by obsmith08 View Post
    I'm new to M.V.T. and Rolle's Thm, so I guess I'm not quite comphrehending the concepts... Here's the problem:

    Suppose that f is continuous on [0,4] and differentiable on (0,4). Also, suppose f(0)=1 and 3<f'(x)<6 for all x in (0,4). Which of the following must be true?

    A) 13<f(4)<25

    B) 3<f(4)<6

    C) 12<f(4)<24

    D) f(4) is undefined


    Any help/explanation that would aid in my understanding of the concepts would be very appreciated!
    the MVT says that for at least one value of x in the given interval ...

    f'(x) = \frac{f(4) - f(0)}{4 - 0}

    f'(x) = \frac{f(4) - 1}{4}

    since 3 < f'(x) < 6 ...

    3 < \frac{f(4) - 1}{4} < 6

    find the solution interval for f(4)
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  3. #3
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    Quote Originally Posted by skeeter View Post
    the MVT says that for at least one value of x in the given interval ...

    f'(x) = \frac{f(4) - f(0)}{4 - 0}

    f'(x) = \frac{f(4) - 1}{4}

    since 3 < f'(x) < 6 ...

    3 < \frac{f(4) - 1}{4} < 6

    find the solution interval for f(4)

    Thank you so much! I can see now that I was making this problem out to be much more difficult than it actually is! (Math--especially Calculus--isn't really my forte though)
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