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Math Help - A Bit Lost...

  1. #1
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    A Bit Lost...

    Hello, and thanks for the help in advance.

    I'm not sure where to begin for solving this problem, so I'll simply state it below and then mention how I might solve it.

    Suppose that f'(x) is greater than or equal to 3 and less than or equal to 5. Show that the quantity (f(8) - f(2)) is greater than or equal to 18 and less than or equal to 30.

    This chapter has only had two sections consisting of finding absolute minimums and maximums, critical points, etc. and this section, consisting of Rolle's Theorem and the Mean Value Theorem. I'm not sure how I could apply any of these concepts for a solution of this problem.

    Thanks again for the help; feel free to go beyond the scope of this chapter to solve it, I'd just like to understand!
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  2. #2
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    Well if you assume that f'(x) is minimum (constant f'(x)=3) then it is just a straight line
    with a slope 3. So if f(2)=A then f(8)=A+(8-2)*3.

    Do the same to find the upper bound.
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  3. #3
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    Quote Originally Posted by NBrunk View Post
    Hello, and thanks for the help in advance.

    I'm not sure where to begin for solving this problem, so I'll simply state it below and then mention how I might solve it.

    Suppose that f'(x) is greater than or equal to 3 and less than or equal to 5. Show that the quantity (f(8) - f(2)) is greater than or equal to 18 and less than or equal to 30.

    This chapter has only had two sections consisting of finding absolute minimums and maximums, critical points, etc. and this section, consisting of Rolle's Theorem and the Mean Value Theorem. I'm not sure how I could apply any of these concepts for a solution of this problem.

    Thanks again for the help; feel free to go beyond the scope of this chapter to solve it, I'd just like to understand!
    assuming f(x) is continuous and differentiable ...

    MVT says there exists a value x = c such that f'(c) = \frac{f(b)-f(a)}{b-a}

    further, for at least one value x = c , 2 < c < 8 ...

    f'(c) = \frac{f(8)-f(2)}{8-2}

    since 3 \le f'(c) \le 5 ...

    3 \le \frac{f(8)-f(2)}{8-2} \le 5

    3 \le \frac{f(8)-f(2)}{6} \le 5

    18 \le f(8) - f(2) \le 30
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