## 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

2. anonymous

I know the period is Pi and the amplitude is 2 but stuck on what to do next

3. anonymous

period does not look like $$\pi$$ to me

4. anonymous

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

ok can you help me on that part too?

6. 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. $$\frac{\pi}{2}$$

7. anonymous

the max is evidently 2 and the min is -2

8. anonymous

is that it? this is a 7 point question so I am thinking there is more to it?

9. anonymous

this problem is all eyeballs

10. anonymous

you do have to find the zeros still

11. anonymous

so i was working on Max = n * Pi n=1, 2, 3, 4, 5............ Is this the right path?

12. anonymous

oh and also they do not ask for the max and the min, they ask where that occurs (the x values)

13. anonymous

min n*Pi/2 n = 1, 2, 3, 4, 5........

14. anonymous

0 = n*Pi/2 n = 1, 2, 3, 4, ......

15. anonymous

no you are still thinking that the period is $$\pi$$ but it is $$\frac{\pi}{2}$$

16. anonymous

as you can tell I am lost :-)

17. anonymous

the max is occurring at $0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi,...$

18. anonymous

don't get confused you see that it is 2 at $$x=0$$ right?

19. anonymous

yes

20. anonymous

and also at $$\pi$$ right?

21. anonymous

yes :-)

22. anonymous

BUT is s also 2 half way between yes?

23. anonymous

yes, that is the climb back up on the graph?

24. anonymous

so what is half way between 0 and pi?

25. anonymous

.5pi ??

26. anonymous

lol yeah or as we say in the math biz $\frac{\pi}{2}$

27. anonymous

that is also your period since the curve does all its business on that interval goes down, and back up

28. anonymous

period is pi/2

29. anonymous

ok so we have max. How do I proceed to get min & 0

30. anonymous

31. anonymous

but btw at this point we can write down exactlyl what the funciton is

32. 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 $f(x)=2\cos(4x)$

33. anonymous

then if you want the zeros, $\cos(\frac{\pi}{2})=0$solve $4x=\frac{\pi}{2}$ for $$x$$ since

34. anonymous

you get $x=\frac{\pi}{8}$ instantly

35. anonymous

anther way it to break up the inteval $[0,\frac{\pi}{2}]$ in to 4 equal parts

36. 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.

37. 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

38. anonymous

"I actually think I understand what you are saying." i will take that as a compliment

39. anonymous

YES!! Compliment! :-)