Find the open intervals on which the function is increasing or decreasing. Apply the First Derivative Test to identify the relative extrema. f(x)=x-2sin(x)

Answer 1

The given function is f(x) = x - 2sin(x), based on the question, we have to take the first derivative of this function. Therefore, we will have the following.

f (x) = x - 2sin(x) ; original function

f '(x) = 1 - 2cos(x) ; derivative of original function

To obtain the derivative, we have to know the constant multiple rule for 'x', and the trigonometric derivatives, in this case for '2sin(x)'

Our next step will be to set our derivative equal to zero and then solve for x. Therefore,

1 - 2cos(x) = 0 ;derivative equal to zero

1 = 2cos(x) ; add 2cos(x) to both sides

1/2 = cos(x) ; divide both sides by 2 to isolate the cosine function.

arccos(1/2) = x ; arccos is the inverse of cos. It could also be denoted by cos^-1(1/2)

At this point, we are supposed to know the values at which cosine will give us 1/2.

If not, we can go to the unit circle and look for the angles of cosine.

These points will be called our Critical Points (CP).

cos will only be positive on QI and QIV. Therefore the angles should be:

Radians: π/3 and 5π/3

Degrees: 60º and 300º

We will use radians from this point on.

Now, these critical points will divide our intervals as follow:

(0,π/3) , (π/3,5π/3) , (5π/3,2π)

We have to test these intervals by plotting random points that go within them in the derivative such as:

0 < x < π/3 ; we could test π/6 ;

π/3 < x < 5π/3 ; we could test π ;

5π/3 < x < 2π ; we could test 11π/6 ;

We have to plug these testing points into the derivative to see if the end result is positive (increasing) or negative (decreasing)

f '(x) = 1 - 2cos(x) ; derivative

such as:

f '(π/6) = 1 - 2cos(π/6) = 1 - √3 => negative

f '(π) = 1 - 2cos(π) = 1 + 2 = 3 => positive

f '(11π/6) = 1 - 2cos(11π/6) = 1 - √3 => negative

We can conclude that the function is...

Increasing from (π/3, 5π/3)

Decreasing from (0,π/3) U (5π/3,2π)