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Mathematics 4 Online
OpenStudy (anonymous):

determine whether Rolle's theorem can be applied, if it can be applied, determine all of the values of c. f(x)=cos2x [-π,π]

OpenStudy (anonymous):

if someone could help me out by solving this problem step by step that'd be great, thanks!

OpenStudy (anonymous):

it can certainly be applied because cosine is a continuous and differentiable function

OpenStudy (anonymous):

so how exactly what i go with this function, would I find the derivative of cos2x and plug in π and -π?

OpenStudy (anonymous):

into the derivative equation

OpenStudy (anonymous):

oh it is \(\cos(2x)\) sorry no matter it is still true that \(\cos(2\pi)=\cos(-2\pi)\)

OpenStudy (anonymous):

since the function is equal at the endpoints, rolle's theorem says that the derivative is equal to zero somewhere between \(-\pi\) and \(\pi\)

OpenStudy (anonymous):

the derivative is \(-2\sin(2x)\) set it equal to zero and solve for \(x\)

OpenStudy (anonymous):

so then the interval is only there so that I can make sure the equation applies to rolle's theorem? and I just have to solve the derivative for x then?

OpenStudy (anonymous):

your last job is to solve \[-2\sin(2x)=0\] for \(x\)

OpenStudy (anonymous):

don't think too hard for this one, since \(\sin(0)=0\) you get \(2x=0,x=0\)

OpenStudy (anonymous):

oh ok, I just didn't know what the intervals where there for, thanks for the help I appreciate it.

OpenStudy (anonymous):

np you see that 0 is right there in the middle of your interval

OpenStudy (anonymous):

yeah I definitely understand that since it's in the middle of π and -π

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