f(x)=-cos(x^2)+2sin(x) [1,3.5]

Find the three roots of f'(x) on the given interval

1 answer

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
Similar Questions
  1. I need to find the exact solutions on the interval [0,2pi) for:2sin^2(x/2) - 3sin(x/2) + 1 = 0 I would start:
    1. answers icon 0 answers
  2. y= 2sin^2 xy=1- sinx find values of x inthe interval 0<x<360 if 2sin^x = 1-sinx this can be arranged into the quadratic. 2sin^2
    1. answers icon 0 answers
  3. Find the solutions to the given equation in the interval [0,2π)?2sin^3(x)−sin(x)cos(x)=2sin(x) check my answers π 4π)/3
    1. answers icon 3 answers
  4. 2sin(x)cos(x)+cos(x)=0I'm looking for exact value solutions in [0, 3π] So I need to find general solutions to solve the
    1. answers icon 6 answers
more similar questions