`f(x) = cos^2 (x) - 2sin(x), 0<=x<=2pi` (a) Find the intervals on which `f` is increasing or decreasing. (b) Find the local maximum and minimum values of `f`.
a) You need to determine the intervals where the function increases and decreases, hence, you need to determine where f'(x)>0 or f'(x)<0.
You need to determine the first derivative, using the chain rule:
`f'(x) = (cos^2 x - 2sin x)`
`f'(x) = 2cos x*(-sin x) - 2cos x`
You need to solve for x the equation f'(x) = 0:
`2cos x*(-sin x) - 2cos x = 0`
You need to factor out 2cos x:
`2cos x*(-sin x - 1) = 0`
`2cos x = 0 => cos x = 0` for `x = pi/2` and `x = 3pi/2`
`-sin x - 1 = 0 => sin x = -` 1 for `x = 3pi/2`
Since f'(x) >0 for `x in (pi/2,3pi/2)` , the function increases. Since f'(x)<0 for `x in (0,pi/2)U(3pi/2,2pi)` yields that the function decreases.
b) The function reaches the extrema at the values of x for f'(x) = 0.
From point a) yields that the function has a maximum point at `x = 3pi/2` and a minimum point at `x = pi/2` .