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19 June, 22:25

Find the x-coordinates where f ' (x) = 0 for f (x) = 2x - sin (2x) in the interval [0, 2π].

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  1. 19 June, 22:42
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    The x-coordinates where f ' (x) = 0 for f (x) = 2x - sin (2x) in the interval [0, 2π] are: x=0, π, and 2π.

    Step-by-step 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.
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