# `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.

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### 1 Answer

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)).` **