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) $f(\theta) = 2cos\theta - cos2\theta\,,\,[0,2\pi]$
So I find the derivative to find the critical numbers:

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

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

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

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

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

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

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

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

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 $cos (\,-\,x) = cosx$

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

Is this right so far?

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

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

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

$\sin{\theta} = 0$

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

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

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

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

3. Hi

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

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

$\sin \theta \1 - 2 \cos \theta) = 0" alt="\sin \theta \1 - 2 \cos \theta) = 0" /> which leads to

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

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

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

EDIT : too late !

4. thank you both!