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

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- anonymous

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- anonymous

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

- anonymous

period does not look like \(\pi\) to me

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## More answers

- 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

ok can you help me on that part too?

- 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

the max is evidently 2
and the min is -2

- anonymous

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

- anonymous

this problem is all eyeballs

- anonymous

you do have to find the zeros still

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

as you can tell I am lost :-)

- anonymous

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

- anonymous

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

- anonymous

yes

- anonymous

and also at \(\pi\) right?

- anonymous

yes :-)

- anonymous

BUT is s also 2 half way between yes?

- anonymous

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

- anonymous

so what is half way between 0 and pi?

- anonymous

.5pi ??

- anonymous

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

- anonymous

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

- anonymous

period is pi/2

- anonymous

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

- anonymous

min from your eyeballs

- anonymous

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

- 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

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

- anonymous

you get \[x=\frac{\pi}{8}\] instantly

- anonymous

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

- 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

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

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

- anonymous

YES!! Compliment! :-)

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