Let f be the function with derivative given by f'(x) = sin(x^2 + 1). How many relative extrema does f have on the interval 2

Question

Let f be the function with derivative given by f'(x) = sin(x^2 + 1). How many relative extrema does f have on the interval 2 < x < 4 ? How would I solve this without a calculator?

anonymous · Answer

i guess you have to find the zeros of $f'$

anonymous · Answer

Yes

anonymous · Answer

not that hard since you know that $\sin(0)=0$

anonymous · Answer

But x^2 cannot be negative.

anonymous · Answer

oh good point, i thought it said $\sin(x^2-1)$

anonymous · Answer

That would be alot easier!

anonymous · Answer

might try setting 
$$x^2+1=\pi$$ hten

anonymous · Answer

see if get something in the interval

anonymous · Answer

No, it's not on the interval.

anonymous · Answer

yes i think it is

anonymous · Answer

looks like there are two of them in fact http://www.wolframalpha.com/input/?i=sin%28x^2%2B1%29

anonymous · Answer

$$x^2+1=\pi$$
$$x^2=\pi-1$$
$$x=\pm\sqrt{\pi-1}$$

anonymous · Answer

That is 1.46 though.

anonymous · Answer

U WILL get one maxima and other minima 
x=3.4,x=3.8

anonymous · Answer

Yes, I know, there is a variable in front of the Root(pi-1), but just root(pi-1) did not fall in the interval. I have the answer, I needed an explination.

anonymous · Answer

$$\sin(x^2+1)=0 for \sin (npi)$$
n=0,1,2,3,4 .........   g