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ERoseM
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Determine whether the Mean Value Theorem for Derivatives applies for F(x)= xcosx, on [x/2,x/2] and find the coordinates of the points whose existence is guaranteed by the theorem
 2 years ago
 2 years ago
ERoseM Group Title
Determine whether the Mean Value Theorem for Derivatives applies for F(x)= xcosx, on [x/2,x/2] and find the coordinates of the points whose existence is guaranteed by the theorem
 2 years ago
 2 years ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.1
hmm what a weird looking interval you have
 2 years ago

ERoseM Group TitleBest ResponseYou've already chosen the best response.0
Scratch that make it [pi/2,pi/2]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
no such interval
 2 years ago

ERoseM Group TitleBest ResponseYou've already chosen the best response.0
[pi/2,pi/2]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
that makes sense yes, the mvt applies because your function is continuous how you are going to solve i have no idea, but \[f(x)=x\cos(x)\] so \(f(\frac{\pi}{2})=\frac{\pi}{2}\cos(\frac{\pi}{2})=\frac{\pi}{2}0=\frac{\pi}{2}\)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
and \[f(\frac{\pi}{2})=\frac{\pi}{2}\] similar to above therefore \[\frac{f(\frac{\pi}{2})f(\frac{\pi}{2})}{\frac{\pi}{2}+\frac{\pi}{2}}\] \[=\frac{(\frac{\pi}{2}+\frac{\pi}{2})}{\pi}=1\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
and \[f'(x)=1+\sin(x)\] so you need so solve \[1+\sin(x)=1\] for \(x\) in the interval should not be too hard actually
 2 years ago

ERoseM Group TitleBest ResponseYou've already chosen the best response.0
Right so I take the arcsin of 0 and get the answer?
 2 years ago
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