# `f(theta) = 2cos(theta) + cos^2 (theta), 0<=theta<=2pi` (a) FInd the intervals of increase or decrease. (b) Find the local maximum and minimum values. (c) Find the inflection points.

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### 1 Answer

You need to determine the monotony of the function, hence, you need to find where the derivative is positive or negative.

You need to evaluate the first derivative, using the chain rule:

`f'(theta) = -2sin theta - 2cos theta*sin theta`

You need to solve for `theta` `f'(theta) = 0.`

`-2sin theta - 2cos theta*sin theta = 0`

Factoring out `-2 sin theta` , yields:

`-2sin theta(1 + cos theta) = 0 => -2sin theta = 0 => sin theta = 0` for `theta = 0, theta = pi` and `theta = 2pi`

`1 + cos theta = 0 => cos theta = -1` for `theta = pi`

**You need to notice that `f'(theta)<0` , hence the function decreases, for `theta in (0,pi)` and `f'(theta)>0` , hence, the function increases, for `theta in (pi,2pi).` **

b) **The function has a minimum point at theta = pi and it has maximum points at `theta = 0` and `theta = 2pi.` **

c) You need to evaluate the inflection points of the function, hence, you need to solve for theta the equation `f''(theta) = 0.`

`f''(theta) =(- 2sin theta - sin 2theta)'`

`f''(theta) = -2cos theta - 2cos 2theta`

`f''(theta) = -2cos theta - 2(2cos^2 theta - 1)`

`f''(theta) = -2cos theta - 4cos^2 theta + 2`

You need to solve for theta, `f''(theta) = 0` , such that:

`-2cos theta - 4cos^2 theta + 2 = 0`

`2cos^2theta + cos theta - 1 = 0`

`cos theta = (-1+-sqrt(1+8))/4`

`cos theta = 1/2` for `theta = pi/3` and `theta = 5pi/3`

or `cos theta = -1` for `theta = pi`

**Hence, the function has inflection points at `theta = pi/3, theta = pi` and `theta = 5pi/3.` **