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Thread: 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!
    $\displaystyle f(31) \approx 957$

    $\displaystyle f(32) \approx 1030$

    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!
    $\displaystyle f(0)=0$, $\displaystyle f(\infty)=\lim\nolimits_{x\to\infty}f(x)=\infty$, because of the continuity of $\displaystyle f$ and the intermediate value theorem, there exists $\displaystyle x_{0}\geq0$ such that $\displaystyle f(x_{0})=y_{0}$ for any fixed $\displaystyle y_{0}\geq0$.
    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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