# Thread: Increasing/Decreasing and Abs Max and Min

1. ## Increasing/Decreasing and Abs Max and Min

i'd like your thoughts on this one. thank you!

Find the intervals on which f is increasing/decreasing and state any relative extrema
1)$\displaystyle f(\theta) = 2cos\theta - cos2\theta\,,\,[0,2\pi]$
So I find the derivative to find the critical numbers:

$\displaystyle f\,'\,(\theta) = 2(\,-\,sin\theta) - (\,-\,sin\,2\theta)(2)$

$\displaystyle = -2sin\theta + 2sin\,2\theta$

So I set that equal to zero: $\displaystyle -2sin\,\theta + 2sin\,2\theta = 0$

Divide by -2: $\displaystyle sin\theta - sin2\theta = 0$

Add $\displaystyle sin 2\theta$: $\displaystyle sin\theta = sin 2\theta$

Sub $\displaystyle sin2\theta = 2sin\theta cos\theta$: $\displaystyle sin\theta = 2sin\theta cos\theta$

Divide by $\displaystyle sin \theta$: $\displaystyle 1 = 2cos\theta$

So: $\displaystyle cos\theta = \frac{1}{2}$

So I went on the unit circle and found all the places where the x coordinate is one half and included the negative of those angles angles since $\displaystyle cos (\,-\,x) = cosx$

Based on that I got $\displaystyle \theta = \pm\frac{\pi}{3}\,,\,\pm\frac{5\pi}{3}$

Is this right so far?

2. you missed solutions when you divided by $\displaystyle \sin{\theta}$ , note that the correct procedure is to factor ...

$\displaystyle \sin{\theta} - 2\sin{\theta}\cos{\theta} = 0$

$\displaystyle \sin{\theta}(1 - 2\cos{\theta}) = 0$

$\displaystyle \sin{\theta} = 0$

$\displaystyle \theta = 0 , \pi , 2\pi$

$\displaystyle \cos{\theta} = \frac{1}{2}$

$\displaystyle \theta = \frac{\pi}{3} , \frac{5\pi}{3}$

also, note that the solution interval is $\displaystyle 0$ to $\displaystyle 2\pi$ , so the negative solutions are not needed.

3. Hi

You should not divide by $\displaystyle \sin \theta$ but rather factor

$\displaystyle \sin \theta - \sin 2\theta = 0$

$\displaystyle \sin \theta \1 - 2 \cos \theta) = 0$ which leads to

$\displaystyle \sin \theta = 0$ or $\displaystyle \cos \theta = \frac12$

And since the interval is $\displaystyle [0,2\pi]$

$\displaystyle \theta=0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}, 2\pi$

EDIT : too late !

4. thank you both!