Answer:
The x-coordinates where f '(x) = 0 for f(x) = 2x - sin(2x) in the interval [0, 2π] are: x=0, π, and 2π.
Explanation:
f(x)=2x-sin(2x)
f'(x)=[f(x)]'
f'(x)=[2x-sin(2x)]'
f'(x)=(2x)'-[sin(2x)]'
f'(x)=2-cos(2x) (2x)'
f'(x)=2-cos(2x) (2)
f'(x)=2-2 cos(2x)
f'(x)=0, then:
2-2 cos(2x)=0
Solving for x. First solving for cos(2x): Subtracting 2 from both sides of the equation:
2-2 cos(2x)-2=0-2
-2 cos(2x)=-2
Dividing both sides of the equation by -2:
-2 cos(2x) / (-2)=-2 / (-2)
cos(2x)=1
If cos(2x)=1, then:
2x=2nπ
Dividing both sides of the equation by 2:
2x/2=2nπ/2
x=nπ
If n=-1→x=-1π→x=-π, and -π is not in the interval [0,2π], then x=-π is not in the solution.
If n=0→x=0π→x=0, and 0 is in the interval [0,2π], then x=0 is in the solution.
If n=1→x=1π→x=π, and π is in the interval [0,2π], then x=π is in the solution.
If n=2→x=2π, and 2π is in the interval [0,2π], then x=2π is in the solution.
If n=3→x=3π, and 3π is not in the interval [0,2π], then x=2π is not in the solution.
Solution: x=0, π, and 2π.