Analysis: Inverse Function Theorem

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The discussion centers on the application of the Inverse Function Theorem, particularly regarding the derivative being zero at x = pi/2. One participant argues that the theorem cannot be applied because the derivative is zero within the interval. However, another clarifies that the theorem requires the derivative to be nonzero only near the point of interest, not throughout the entire interval. The conclusion drawn is that by selecting a smaller interval around pi that excludes pi/2, the theorem can still be applied effectively. The participants ultimately agree on the importance of understanding the conditions of the theorem in relation to the specific problem.
Shackleford
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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)?
 
Dick said:
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).
 
Last edited:
Shackleford said:
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.
 
Shackleford said:
The answer to the problem is (c).

Sure it is.
 
Dick said:
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

I'm interpreting I = (0, 2pi).
 
Last edited by a moderator:
Shackleford said:
What are you talking about? It says it right here.

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

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.
 
Last edited by a moderator:
Dick said:
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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