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Math Help - show that there is a number c such that f(c) = 1000

  1. #1
    s3a
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    show that there is a number c such that f(c) = 1000

    Question: If f(x) = x^2 + 10 sinx, show that there is a number c such that f(c) = 1000.

    I'm not too sure I understand the question but I plugged in 1000 as the y value in that rule and don't really no how to solve for x.

    Any help would be greatly appreciated!
    Thanks in advance!
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    Quote Originally Posted by s3a View Post
    Question: If f(x) = x^2 + 10 sinx, show that there is a number c such that f(c) = 1000.

    I'm not too sure I understand the question but I plugged in 1000 as the y value in that rule and don't really no how to solve for x.

    Any help would be greatly appreciated!
    Thanks in advance!
    f(31) \approx 957

    f(32) \approx 1030

    what does the Intermediate Value Theorem tell you?
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  3. #3
    s3a
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    That there is a point (c,N) in between (31, 957) and (32, 1030)?
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    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by s3a View Post
    Question: If f(x) = x^2 + 10 sinx, show that there is a number c such that f(c) = 1000.

    I'm not too sure I understand the question but I plugged in 1000 as the y value in that rule and don't really no how to solve for x.

    Any help would be greatly appreciated!
    Thanks in advance!
    f(0)=0, f(\infty)=\lim\nolimits_{x\to\infty}f(x)=\infty, because of the continuity of f and the intermediate value theorem, there exists x_{0}\geq0 such that f(x_{0})=y_{0} for any fixed y_{0}\geq0.
    By the way, you dont need to solve the value to show its existence.
    http://en.wikipedia.org/wiki/Intermediate_value_theorem
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