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

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• Nov 25th 2009, 02:03 PM
rawkstar
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)
• Nov 25th 2009, 02:13 PM
GeoH2102
$sin(n \pi) = 0$, so you require $x^2 + 1 = n \pi$. This occurs at $x = \sqrt{n \pi -1}$, in this case $n = 2,3,4,5$.
• Nov 25th 2009, 02:14 PM
adkinsjr
Quote:

Originally Posted by rawkstar
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 $sin(x^2+1)=0$, the angle must be some multiple of $\pi$. So $x^2+1=n\pi$ would be a solution.