Analysis: Inverse Function Theorem

[Shackleford]

My contention is that you cannot apply the Inverse Function Theorem to this problem because there is a point in the interval at which the derivative is zero.

At x = pi/2, f ' = 0.

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled.png

 

[Dick]

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)?

 

[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)?

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).

 

[Dick]

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.

 

[Dick]

The answer to the problem is (c).

Sure it is.

 

[Shackleford]

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).

 

[Dick]

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.

 

[Shackleford]

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.