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Thread: Where does sin(x^2+1)=0

  1. #1
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    Where does sin(x^2+1)=0

    I am given the function with the derivative f '(x)=sin(x^2+1). I need to find out how many relative extrema exist in the interval 2<x<4. To do this I know that I first need to find where the derivative is equal to 0. However I don't know where sin(x^2+1)=0 (and how prove it does at these points)
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  2. #2
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    $\displaystyle sin(n \pi) = 0$, so you require $\displaystyle x^2 + 1 = n \pi$. This occurs at $\displaystyle x = \sqrt{n \pi -1}$, in this case $\displaystyle n = 2,3,4,5$.
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  3. #3
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    Quote Originally Posted by rawkstar View Post
    I am given the function with the derivative f '(x)=sin(x^2+1). I need to find out how many relative extrema exist in the interval 2<x<4. To do this I know that I first need to find where the derivative is equal to 0. However I don't know where sin(x^2+1)=0 (and how prove it does at these points)
    When $\displaystyle sin(x^2+1)=0$, the angle must be some multiple of $\displaystyle \pi$. So $\displaystyle x^2+1=n\pi$ would be a solution.
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