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ERoseM

  • 3 years ago

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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  1. anonymous
    • 3 years ago
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    hmm what a weird looking interval you have

  2. ERoseM
    • 3 years ago
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    Scratch that make it [pi/2,-pi/2]

  3. anonymous
    • 3 years ago
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    no such interval

  4. ERoseM
    • 3 years ago
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    [-pi/2,pi/2]

  5. anonymous
    • 3 years ago
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    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 \(f(\frac{\pi}{2})=\frac{\pi}{2}-\cos(\frac{\pi}{2})=\frac{\pi}{2}-0=\frac{\pi}{2}\)

  6. anonymous
    • 3 years ago
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    and \[f(-\frac{\pi}{2})=-\frac{\pi}{2}\] similar to above therefore \[\frac{f(\frac{\pi}{2})-f(-\frac{\pi}{2})}{\frac{\pi}{2}+\frac{\pi}{2}}\] \[=\frac{(\frac{\pi}{2}+\frac{\pi}{2})}{\pi}=1\]

  7. anonymous
    • 3 years ago
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    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

  8. ERoseM
    • 3 years ago
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    Right so I take the arcsin of 0 and get the answer?

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