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Mathematics 7 Online
OpenStudy (anonymous):

Could someone help me find the absolute maximum and minimum?

OpenStudy (anonymous):

zepdrix (zepdrix):

Find s' ok? :)

hero (hero):

\[s^{(1)}t\]

OpenStudy (anonymous):

is it -t sin(t)?

zepdrix (zepdrix):

Hmm yah that sounds right :D

zepdrix (zepdrix):

So we want to find critical points within the interval, plug them into the function to get a value. Then we'll check the end points, plugging them into the function to get a value. Then we'll compare and see which gives us the largest and smallest values! :D So to get critical points what do we do? c:

OpenStudy (anonymous):

Do we substitute the [0, 2pi] into the original equation?

zepdrix (zepdrix):

Ummm yes, we'll do that eventually, those will be our values for the end points. But for now, we also want to find critical points within the interval. We've found s', now we'll set s' equal to zero, and solve for t.

OpenStudy (anonymous):

For t I got 0. Is that correct?

zepdrix (zepdrix):

Sint=0 Hmm that's when t=0 and the other t gave us 0 also. Hmm I guess this problem will be a little easier than we though :O since we only need to check 0 and 2pi.

OpenStudy (anonymous):

So since we on need to check 0 and 2 pi, do we skip this step?

zepdrix (zepdrix):

We found a critical point, t=0. That is also one of our end points though, so we don't need to worry about it, since we were planning on checking t=0 anyway (because it's an end point). So yes, we can skip right to checking the end points now since we didn't have any important critical points to find. s(0)=? s(2pi)=? The smaller value is your min, larger your max :D

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