f(x)=-cos(x^2)+2sin(x) [1,3.5]
Find the three roots of f'(x) on the given interval
f'(x) = 2x sin(x^2) + 2cosx so you want to find where
x sin(x^2) + cosx = 0
Good luck doing that by hand. A graphical solution yields roots at
x = 1.744, 2.572, 3.015
See
https://www.wolframalpha.com/input/?i=2x+sin%28x%5E2%29+%2B+2cosx%3D0%2C+1%3C%3Dx%3C%3D3.5