1. ## Ciritcal Numbers help!

Find and classify all critical numbers of f(0)= 2cos(0) - cos(2(0)) on the interval [0, 2pi]

0 represents theta.

2. Originally Posted by Hibijibi
Find and classify all critical numbers of f(0)= 2cos(0) - cos(2(0)) on the interval [0, 2pi]

0 represents theta.

$f(\theta) = 2\cos(\theta) - \cos(2\theta)$

Critical numbers occur where there is a break in the slope, or the slope is horizontal.

So first we need to find the equation for the slope, which we can do by finding the derivative:
$\frac d{d\theta}f(\theta) = -2\sin(\theta) + 2\sin(2\theta)$

Now we know that sine is a continuous function, so there are no places where it will break (ie, where there is no graph such as at zero on 1/x)

So we set it equal to zero and solve
$0 = -2\sin(\theta) + 2\sin(2\theta)$

$\sin(\theta)=\sin(2\theta)$

For this step, if you visualize it in your mind (the unit circle, with theta moving along it's perimeter, and 2theta moving along twice as fast, ie visualize their values in your head by moving theta along the circle, and seeing how 2theta moves twice as fast, this will help you find the correct values), you can see that starting at zero, both of these will return zero. As you move the one around (I'm doing this mentally), you can see that as you go through pi/6, the one that is twice as far will be on pi/3, which has a larger sine, then up to pi/4, and the 2x has a value of pi/2, then pi/3, and the other has a value of 2pi/3, which both have the same sine, so theta = pi/3 is another answer. Then keep moving it, you hit (pi/2, pi), (2pi/3, 4pi/3), (3pi/4, 3pi/2), (5pi/6, 5pi/3), (pi,2pi) and these two both have the same height on the unit circle, so they both return zero as well.

Keep moving the value of theta in your mind like this, and you will see that there should be one last value, between 3pi/2 and 2pi, it turns out that value is 5pi/3, so your solutions are theta = 0, pi/3, pi, 5pi/3

These are then your critical numbers.

I don't think you need to include the endpoints, but I don't remember.

The reason these values are critical, is because when the derivative is equal to zero, the slope of your function is horizontal. Any time the slope changes directions, it must level out first, so these critical points are potentially where maximum and minimum values are.

3. Thank you very much! My derivative looked different, simply because I converted cos(2(0)) into something familiar to me, but I did end up with the same answers. Appreciate the help and explanation.