Question

f(x) = − cos(x^2) + 2sin(x)

How to find critical points, absolute max, and absolute min of f(x) on the interval [1,3.5]?

Answer 1

well, we already did the critical points (where f'(x) = 0)
so now it's just a matter of checking the sign of f"(x) at the critical points to see whether they are max or min.
f'(x) = 2x sin(x^2) + 2cosx
f"(x) = 4x^2 cos(x^2) + 2sin(x^2) - 2sinx

Come back with your work if you get stuck.

Answer 2

I think oobleck did this earlier today, but I can't find it, so ....
f(x) = − cos(x^2) + 2sin(x)
f '(x) = 2x sin(x^2) + 2cosx
No easy way to solve 2x sin(x^2) + 2cosx = 0 , which we have to do for your problem

Wolfram says: https://www.wolframalpha.com/input/?i=solve+2x+sin%28x%5E2%29+%2B+2cosx+%3D+0

I see 3 solutions that fit within your given domain