Question

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

Answer 1

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.

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

Answer 2

Answer:

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

Step-by-step 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 $[0,2\pi]$

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