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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
 one year ago
 one year ago
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
 one year ago
 one year ago

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satellite73Best ResponseYou've already chosen the best response.1
hmm what a weird looking interval you have
 one year ago

ERoseMBest ResponseYou've already chosen the best response.0
Scratch that make it [pi/2,pi/2]
 one year ago

satellite73Best 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 \(f(\frac{\pi}{2})=\frac{\pi}{2}\cos(\frac{\pi}{2})=\frac{\pi}{2}0=\frac{\pi}{2}\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
and \[f(\frac{\pi}{2})=\frac{\pi}{2}\] similar to above therefore \[\frac{f(\frac{\pi}{2})f(\frac{\pi}{2})}{\frac{\pi}{2}+\frac{\pi}{2}}\] \[=\frac{(\frac{\pi}{2}+\frac{\pi}{2})}{\pi}=1\]
 one year ago

satellite73Best 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
 one year ago

ERoseMBest ResponseYou've already chosen the best response.0
Right so I take the arcsin of 0 and get the answer?
 one year ago
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