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Analysis: Inverse Function Theorem

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  • #2
Dick
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The function is monotone increasing, it does have an inverse even if the derivative happens to be zero at point. And you don't have to evaluate f'(pi/2) to solve the problem, you have to find f'(f^(-1)(-1)). What's f^(-1)(-1)?
 
  • #3
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The function is monotone increasing, it does have an inverse even if the derivative happens to be zero at point. And you don't have to evaluate f'(pi/2) to solve the problem, you have to find f'(f^(-1)(-1)). What's f^(-1)(-1)?
I know I'm not evaluating at that point. To me, it seems that the theorem states that for every x in the interval the derivative of the function cannot be zero. Pi/2 is in that interval.

The answer to the problem is (c).
 
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Dick
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I know I'm not evaluating at that point. To me, it seems that the theorem states that for every x in the interval the derivative of the function cannot be zero. Pi/2 is in that interval.
That's not what it says. It's says f' has to be continuously differentiable and nonzero NEAR the point f^(-1)(-1). Not everywhere.
 
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Dick
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That's not what it says. It's says f' has to be continuously differentiable and nonzero NEAR the point f^(-1)(-1). Not everywhere.
What are you talking about? It says it right here.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20111109_161232.jpg [Broken]

I'm interpreting I = (0, 2pi).
 
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  • #7
Dick
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What are you talking about? It says it right here.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20111109_161232.jpg [Broken]

I'm interpreting I = (0, 2pi).
You don't need to take I=(0,2pi). f^(-1)(-1)=pi, so take I=(pi-1/4,pi+1/4). Or any other small interval around pi that doesn't include pi/2. Then it fits your statement, doesn't it? That's what I mean by NEAR pi.
 
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  • #8
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You don't need to take I=(0,2pi). f^(-1)(-1)=pi, so take I=(pi-1/4,pi+1/4). Or any other small interval around pi that doesn't include pi/2. Then it fits your statement, doesn't it? That's what I mean by NEAR pi.
Yeah, I just realized that maybe setting that given interval equal to I is my problem. Thanks again.
 

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