`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`.

1 Answer

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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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` .