Analysis: Inverse Function Theorem

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).
  • #1
Shackleford
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  • #2
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
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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  • #4
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.
 
  • #5
Shackleford said:
The answer to the problem is (c).

Sure it is.
 
  • #6
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).
 
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  • #7
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.
 
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  • #8
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.
 

1. What is the inverse function theorem?

The inverse function theorem is a mathematical theorem that states that if a function is differentiable at a point and has a non-zero derivative at that point, then it has a local inverse function that is also differentiable at that point.

2. Why is the inverse function theorem important?

The inverse function theorem is important because it allows us to find the inverse of a function that is differentiable and has a non-zero derivative at a given point. This is useful in many areas of mathematics, including optimization and differential equations.

3. How does the inverse function theorem relate to the chain rule?

The inverse function theorem is closely related to the chain rule in calculus. The chain rule states that the derivative of a composite function is equal to the product of the derivatives of its individual functions. The inverse function theorem can be used to find the derivative of the inverse function, which is necessary for applying the chain rule.

4. Can the inverse function theorem be applied to any function?

No, the inverse function theorem can only be applied to functions that are differentiable and have a non-zero derivative at a given point. It also requires that the function be one-to-one, meaning that each input has a unique output. Functions that do not meet these criteria may not have a well-defined inverse function.

5. Are there any limitations to the inverse function theorem?

Yes, the inverse function theorem has some limitations. It only guarantees the existence of a local inverse function, meaning that it may not be valid for the entire domain of the original function. Additionally, it does not provide a method for finding the inverse function, only the guarantee of its existence. Finally, the theorem may not hold for functions with more than one variable.

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