Question

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

Answer 1

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

Answer 2

`x=0,x=\pi,x=2\pi` on
`[0,2\pi]`

Explanation:

The given function is
`f(x)=2x-\sin(2x)`.

We differentiate the given function with respect to x, to obtain;

`f'(x)=2-2\cos(2x)`

We find the x-coordinates where
`f'(x)=0`.

`2-2\cos(2x)=0`

`\Rightarrow \cos(2x)=1`

In the first quadrant,

`2x=cos^(-1)(1)`

`2x=0`

`\Rightarrow x=0`

In the fourth quadrant,

`2x=2\pi-0`

`\Rightarrow 2x=2\pi`

`\Rightarrow x=\pi`

Since this function has a period of
`\pi` another will occur at

`\Rightarrow x=\pi+\pi=2\pi`

Therefore
`f(x)=2x-\sin(2x)` has horizontal tangents at
`x=0,x=\pi,x=2\pi` on
`[0,2\pi]`

see attachment