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anonymous
 one year ago
Consider the graph of the cosine function shown below.
a. Find the period and amplitude of the cosine function.
b. At what values of θ for 0 ≤ θ ≤ 2pi do the maximum value(s), minimum value(s), and zeros occur?
anonymous
 one year ago
Consider the graph of the cosine function shown below. a. Find the period and amplitude of the cosine function. b. At what values of θ for 0 ≤ θ ≤ 2pi do the maximum value(s), minimum value(s), and zeros occur?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know the period is Pi and the amplitude is 2 but stuck on what to do next

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0period does not look like \(\pi\) to me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0by the time you get from 0 to pi your graph has gone down from 2 to 2, back up to 2 twice !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok can you help me on that part too?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you do it with your eyeballs it goes from 2 to 2 and back up to 2 that is one period it does it two times on the interval \([0,\pi]\) so the period is half of that, i.e. \(\frac{\pi}{2}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the max is evidently 2 and the min is 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is that it? this is a 7 point question so I am thinking there is more to it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this problem is all eyeballs

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you do have to find the zeros still

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so i was working on Max = n * Pi n=1, 2, 3, 4, 5............ Is this the right path?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh and also they do not ask for the max and the min, they ask where that occurs (the x values)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0min n*Pi/2 n = 1, 2, 3, 4, 5........

anonymous
 one year ago
Best ResponseYou've already chosen the best response.00 = n*Pi/2 n = 1, 2, 3, 4, ......

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no you are still thinking that the period is \(\pi\) but it is \(\frac{\pi}{2}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0as you can tell I am lost :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the max is occurring at \[0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi,...\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0don't get confused you see that it is 2 at \(x=0\) right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and also at \(\pi\) right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0BUT is s also 2 half way between yes?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, that is the climb back up on the graph?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so what is half way between 0 and pi?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol yeah or as we say in the math biz \[\frac{\pi}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is also your period since the curve does all its business on that interval goes down, and back up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so we have max. How do I proceed to get min & 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0min from your eyeballs

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but btw at this point we can write down exactlyl what the funciton is

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since the period is \(\frac{\pi}{2}\) we can solve \[\frac{2\pi}{b}=\frac{\pi}{2}\] and get \(b=4\) right away, making your function \[f(x)=2\cos(4x)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then if you want the zeros, \[\cos(\frac{\pi}{2})=0\]solve \[4x=\frac{\pi}{2}\] for \(x\) since

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you get \[x=\frac{\pi}{8}\] instantly

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0anther way it to break up the inteval \[[0,\frac{\pi}{2}]\] in to 4 equal parts

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I actually think I understand what you are saying. I am such a visual person and once given a formula I can usually see it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because that is the period, or at least one interval in 4 parts you have \[0,\frac{\pi}{8},\frac{\pi}{4},\frac{3\pi}{8},\frac{\pi}{2}\] and those are the points for which the function is \[2,0,2,0,2\] respectively

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0"I actually think I understand what you are saying." i will take that as a compliment

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0YES!! Compliment! :)
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