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Determine whether the Mean Value Theorem for Derivatives applies for F(x)= x-cosx, on [x/2,-x/2] and find the coordinates of the points whose existence is guaranteed by the theorem

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hmm what a weird looking interval you have
Scratch that make it [pi/2,-pi/2]
no such interval

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Other answers:

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 \(f(\frac{\pi}{2})=\frac{\pi}{2}-\cos(\frac{\pi}{2})=\frac{\pi}{2}-0=\frac{\pi}{2}\)
and \[f(-\frac{\pi}{2})=-\frac{\pi}{2}\] similar to above therefore \[\frac{f(\frac{\pi}{2})-f(-\frac{\pi}{2})}{\frac{\pi}{2}+\frac{\pi}{2}}\] \[=\frac{(\frac{\pi}{2}+\frac{\pi}{2})}{\pi}=1\]
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
Right so I take the arcsin of 0 and get the answer?

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