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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?

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    I know the period is Pi and the amplitude is 2 but stuck on what to do next

  3. anonymous
    • one year ago
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    period does not look like \(\pi\) to me

  4. anonymous
    • one year ago
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    by the time you get from 0 to pi your graph has gone down from 2 to -2, back up to 2 twice !

  5. anonymous
    • one year ago
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    ok can you help me on that part too?

  6. anonymous
    • one year ago
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    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. \(\frac{\pi}{2}\)

  7. anonymous
    • one year ago
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    the max is evidently 2 and the min is -2

  8. anonymous
    • one year ago
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    is that it? this is a 7 point question so I am thinking there is more to it?

  9. anonymous
    • one year ago
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    this problem is all eyeballs

  10. anonymous
    • one year ago
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    you do have to find the zeros still

  11. anonymous
    • one year ago
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    so i was working on Max = n * Pi n=1, 2, 3, 4, 5............ Is this the right path?

  12. anonymous
    • one year ago
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    oh and also they do not ask for the max and the min, they ask where that occurs (the x values)

  13. anonymous
    • one year ago
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    min n*Pi/2 n = 1, 2, 3, 4, 5........

  14. anonymous
    • one year ago
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    0 = n*Pi/2 n = 1, 2, 3, 4, ......

  15. anonymous
    • one year ago
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    no you are still thinking that the period is \(\pi\) but it is \(\frac{\pi}{2}\)

  16. anonymous
    • one year ago
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    as you can tell I am lost :-)

  17. anonymous
    • one year ago
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    the max is occurring at \[0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi,...\]

  18. anonymous
    • one year ago
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    don't get confused you see that it is 2 at \(x=0\) right?

  19. anonymous
    • one year ago
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    yes

  20. anonymous
    • one year ago
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    and also at \(\pi\) right?

  21. anonymous
    • one year ago
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    yes :-)

  22. anonymous
    • one year ago
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    BUT is s also 2 half way between yes?

  23. anonymous
    • one year ago
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    yes, that is the climb back up on the graph?

  24. anonymous
    • one year ago
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    so what is half way between 0 and pi?

  25. anonymous
    • one year ago
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    .5pi ??

  26. anonymous
    • one year ago
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    lol yeah or as we say in the math biz \[\frac{\pi}{2}\]

  27. anonymous
    • one year ago
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    that is also your period since the curve does all its business on that interval goes down, and back up

  28. anonymous
    • one year ago
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    period is pi/2

  29. anonymous
    • one year ago
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    ok so we have max. How do I proceed to get min & 0

  30. anonymous
    • one year ago
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    min from your eyeballs

  31. anonymous
    • one year ago
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    but btw at this point we can write down exactlyl what the funciton is

  32. anonymous
    • one year ago
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    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 \[f(x)=2\cos(4x)\]

  33. anonymous
    • one year ago
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    then if you want the zeros, \[\cos(\frac{\pi}{2})=0\]solve \[4x=\frac{\pi}{2}\] for \(x\) since

  34. anonymous
    • one year ago
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    you get \[x=\frac{\pi}{8}\] instantly

  35. anonymous
    • one year ago
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    anther way it to break up the inteval \[[0,\frac{\pi}{2}]\] in to 4 equal parts

  36. anonymous
    • one year ago
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    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.

  37. anonymous
    • one year ago
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    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

  38. anonymous
    • one year ago
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    "I actually think I understand what you are saying." i will take that as a compliment

  39. anonymous
    • one year ago
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    YES!! Compliment! :-)

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