# Analysis: Inverse Function Theorem

• Shackleford
In summary, the inverse function theorem cannot be applied to this problem because the derivative is zero at x = pi/2, which is in the given interval. However, the theorem states that the derivative must be continuously differentiable and nonzero near the point f^(-1)(-1), not everywhere. By choosing a smaller interval that does not include pi/2, the theorem can be applied and the answer to the problem is (c).

#### Shackleford

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).

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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 [Broken]

I'm interpreting I = (0, 2pi).

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Shackleford said:
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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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.