# Math Help - Where does sin(x^2+1)=0

1. ## 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)

2. $sin(n \pi) = 0$, so you require $x^2 + 1 = n \pi$. This occurs at $x = \sqrt{n \pi -1}$, in this case $n = 2,3,4,5$.

3. 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.