Consider the function on the interval (0, 2π). f(x) = x − 2 sin x (a) Find the open intervals on which the function is increasing or decreasing. (Enter your answers using interval notation.)

For continuous functions, we can safely assume between a given critical point and the next one, the function keeps its behavior, i.e. it's increasing or decreasing in the interval formed by both points.

So, we find the critital points of

[tex]f(x)=x-2sinx[/tex]

Taking the derivative

[tex]f'(x)=1-2cosx[/tex]

Equating to 0

[tex]1-2cosx=0[/tex]

Solving

[tex]\displaystyle cosx=\frac{1}{2}[/tex]

There are two solutions in the interval [tex](0,2\pi)[/tex]

[tex]\displaystyle x=\frac{\pi}{3},\ x=\frac{5\pi}{3}[/tex]

Now we compute the second derivative

[tex]f''(x)=2sinx[/tex]

Evaluating for both critical points

[tex]\displaystyle f''(\frac{\pi}{3})=2sin\frac{\pi}{3}=\sqrt{3}[/tex]

Since it's positive, the point is a minimum

[tex]\displaystyle f''(\frac{5\pi}{3})=2sin\frac{5\pi}{3}=-\sqrt{3}[/tex]

Since it's negative, the point is a maximum

In the interval

[tex](0,\frac{\pi}{3})[/tex]

the function is decreasing

In the interval

[tex](\frac{\pi}{3},\frac{5\pi}{3})[/tex]

the function is increasing

In the interval

[tex](\frac{5\pi}{3},2\pi)[/tex]

the function is decreasing