Find the right end point for f(x)=-2-cos(2x-pi).
Could also be written as -cos(2x-pi)-2
So our starting point will be when our angle is zero.\[\Large\rm 2x-\pi=0\]Solving for x gives us:\[\Large\rm x=\frac{\pi}{2}\]That's probably what you got for your starting point, yes? Our ending point will be when we've completed one revolution, in other words, when our angle is 2pi,\[\Large\rm 2x-\pi=2\pi\]
Are you listing your start and end points as ordered pairs? Or just the x-coordinates?
Just like in the other problems. I did pi/2 for my phase shift (starting/left end point).
Right end point = c/b(phase shift) + 2pi/b(period).
Mmmm ok yes, if we're going to stick with that plan, yes yes yes :) Looks good.
\[\Large\rm \text{right end point}=\frac{\pi}{2}+\frac{2\pi}{2}\]So something like that, yah? :d
Right. Then I went a step further by getting the sum. But technically what I got was an improper fractions, which isn't unheard of, but have not come across in this situation.
Oh ok. Yah that's one of those things to get used to in big boy math: `We don't like mixed numbers, those are bad`. Improper fractions are good :)
You could say that it's `1 and a half` pi's. But we don't like mixed numbers :o no no no! If you're more comfortable with decimals, it's ok to call it 1.5pi
Because 3 divided by 2 = 1.5
I guess I'm making an assumption. I'm hoping that you ended up with 3pi/2, and not 3pi/4 :)
3pi/2 is what I ended up with and I think she doesn't want us using decimals in this case. So, yay, my math was right!
yay! :)
:D
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