`f(x) = sin(x)/(1 + cos^2(x))` Consider the function on the interval (0, 2pi). Find the open intervals on which the function is increasing or decreasing, apply first derivative test to...

`f(x) = sin(x)/(1 + cos^2(x))` Consider the function on the interval (0, 2pi). Find the open intervals on which the function is increasing or decreasing, apply first derivative test to identify all relative extrema.

Expert Answers
sciencesolve eNotes educator| Certified Educator

You need to determine the relative extrema of the function, hence, you need to find the solutions to the equation f'(x) = 0.

You need to evaluate the derivative of the function, using the quotient rule:

`f'(x) = (sin ' x * (1 + cos^2 x) - sin x*(1 + cos^2 x)')/((1 + cos^2 x)^2)`

f'`(x) = (cos x* (1 + cos^2 x) + sin x*(2sin x*cos x))/((1 + cos^2 x)^2)`

`f'(x) = (cos x* (1 + cos^2 x) + 2sin^2 x*cos x)/((1 + cos^2 x)^2)`

Replace 1` - cos^2 x` for `sin^2 x,` such that:

`f'(x) = (cos x* (1 + cos^2 x)+ 2(1 - cos^2 x)*cos x)/((1 + cos^2 x)^2)`

`f'(x) = (cos x + cos^3 x + 2cos x- 2cos^3 x)/((1 + cos^2 x)^2)`

`f'(x) = (3cos x- cos^3 x)/((1 + cos^2 x)^2)`

You need to solve for x the equation f'(x) = 0:

`(3cos x- cos^3 x)/((1 + cos^2 x)^2) = 0`

`3cos x- cos^3 x = 0`

Factoring out `cos x` :

`cos x*(3 - cos^2 x) = 0 => cos x = 0 or 3 - cos^2 x = 0`

The function cosine cancels on `(0,2pi)` at `x = pi/2` and` x = 3pi/2.`

Solving for x the equation `3 - cos^2 x = 0` , yields:

`3 - cos^2 x = 0 => 3 = cos^2 x => cos x = +-sqrt 3` impossible since` cos x in [-1,1].`

The function f'(x) is negative on intervals `(pi/2, pi)` and `(pi,3pi/2)` , and positive on `(0,pi/2), (3pi/2,2pi)` , hence, the function increases on `(0,pi/2) U(3pi/2,2pi)` , and it decreases on `(pi/2, pi)U (pi,3pi/2). ` The function has relative extrema at `(pi/2,f(pi/2)) ` and `(3pi/2,f(3pi/2)).`