anonymous
  • anonymous
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?
Algebra
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
I know the period is Pi and the amplitude is 2 but stuck on what to do next
anonymous
  • anonymous
period does not look like \(\pi\) to me

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anonymous
  • anonymous
by the time you get from 0 to pi your graph has gone down from 2 to -2, back up to 2 twice !
anonymous
  • anonymous
ok can you help me on that part too?
anonymous
  • anonymous
you 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
  • anonymous
the max is evidently 2 and the min is -2
anonymous
  • anonymous
is that it? this is a 7 point question so I am thinking there is more to it?
anonymous
  • anonymous
this problem is all eyeballs
anonymous
  • anonymous
you do have to find the zeros still
anonymous
  • anonymous
so i was working on Max = n * Pi n=1, 2, 3, 4, 5............ Is this the right path?
anonymous
  • anonymous
oh and also they do not ask for the max and the min, they ask where that occurs (the x values)
anonymous
  • anonymous
min n*Pi/2 n = 1, 2, 3, 4, 5........
anonymous
  • anonymous
0 = n*Pi/2 n = 1, 2, 3, 4, ......
anonymous
  • anonymous
no you are still thinking that the period is \(\pi\) but it is \(\frac{\pi}{2}\)
anonymous
  • anonymous
as you can tell I am lost :-)
anonymous
  • anonymous
the max is occurring at \[0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi,...\]
anonymous
  • anonymous
don't get confused you see that it is 2 at \(x=0\) right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
and also at \(\pi\) right?
anonymous
  • anonymous
yes :-)
anonymous
  • anonymous
BUT is s also 2 half way between yes?
anonymous
  • anonymous
yes, that is the climb back up on the graph?
anonymous
  • anonymous
so what is half way between 0 and pi?
anonymous
  • anonymous
.5pi ??
anonymous
  • anonymous
lol yeah or as we say in the math biz \[\frac{\pi}{2}\]
anonymous
  • anonymous
that is also your period since the curve does all its business on that interval goes down, and back up
anonymous
  • anonymous
period is pi/2
anonymous
  • anonymous
ok so we have max. How do I proceed to get min & 0
anonymous
  • anonymous
min from your eyeballs
anonymous
  • anonymous
but btw at this point we can write down exactlyl what the funciton is
anonymous
  • anonymous
since 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
  • anonymous
then if you want the zeros, \[\cos(\frac{\pi}{2})=0\]solve \[4x=\frac{\pi}{2}\] for \(x\) since
anonymous
  • anonymous
you get \[x=\frac{\pi}{8}\] instantly
anonymous
  • anonymous
anther way it to break up the inteval \[[0,\frac{\pi}{2}]\] in to 4 equal parts
anonymous
  • anonymous
I 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
  • anonymous
because 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
  • anonymous
"I actually think I understand what you are saying." i will take that as a compliment
anonymous
  • anonymous
YES!! Compliment! :-)

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