Integration with inverse functions

In summary, the conversation discusses finding the derivative of the inverse of a given function. The formula for the derivative of inverse functions is mentioned and it is suggested to start with f(f^(-1)(x))=x. The conversation also clarifies the purpose of inverse functions.
  • #1
Shannabel
74
0

Homework Statement


let f(x)=(4t^3+4t)dt(between 2 and x)
if g(x) = f^(-1)(x), then g'(0)=?


Homework Equations





The Attempt at a Solution


f'(x) = 4x^3+4x
annd i already don't know where to go from here.. help?
 
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  • #2
There's a formula for the derivative of inverses
http://en.wikipedia.org/wiki/Inverse_functions_and_differentiation"

If you start with f(f-1(x)) = x, differentiate both sides and rearrange and you'll get something like
b80fffc5e854e0c44c3e6accbfadf7a2.png
 
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  • #3
Bohrok said:
There's a formula for the derivative of inverses
http://en.wikipedia.org/wiki/Inverse_functions_and_differentiation"

If you start with f(f-1(x)) = x, differentiate both sides and rearrange and you'll get something like
b80fffc5e854e0c44c3e6accbfadf7a2.png

so
[f^(-1)(0)]' = 1/[f'(f^(-1)(0))]
but where do i go from here?
because i don't know what f^(-1)(0) is...
 
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  • #4
Shannabel said:
so
[f^(-1)(0)]' = 1/[f'(f^(-1)(0))]
but where do i go from here?
because i don't know what f^(-1)(0) is...

If f(b) = 0, then taking the inverse of both sides gives you f-1(0) = b. Then you apply this to the original function you were given to find f-1(0)
 
  • #5
Bohrok said:
If f(b) = 0, then taking the inverse of both sides gives you f-1(0) = b. Then you apply this to the original function you were given to find f-1(0)

got it :)
one other thing, at the beginning you started with f(f^(-1)(x))=x
... where did that come from?
 
  • #6
Shannabel said:
at the beginning you started with f(f^(-1)(x))=x
... where did that come from?

That's the purpose of the inverse functions: the compositions of inverse functions return the input x, f(f-1(x)) = f-1(f(x)) = x. http://en.wikipedia.org/wiki/Inverse_function" has a pretty good article with examples.
 
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  • #7
Bohrok said:
That's the purpose of the inverse functions: the compositions of inverse functions return the input x, f(f-1(x)) = f-1(f(x)) = x. http://en.wikipedia.org/wiki/Inverse_function" has a pretty good article with examples.

thanks!
 
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Related to Integration with inverse functions

1. What is integration with inverse functions?

Integration with inverse functions is the process of finding the original function when given its derivative. It involves using inverse functions to solve for the original function and then integrating it to find the general solution.

2. How is integration with inverse functions different from regular integration?

Integration with inverse functions involves finding the inverse of the derivative function, while regular integration involves finding the anti-derivative of a function. This means that in integration with inverse functions, we are finding the original function, while in regular integration, we are finding a function whose derivative is the given function.

3. Why is integration with inverse functions important?

Integration with inverse functions is important because it allows us to solve for the original function when given its derivative. This is useful in many applications, such as in physics, engineering, and economics, where we often have information about the derivative of a function but need to find the original function.

4. What is the process for integrating with inverse functions?

The process for integrating with inverse functions involves using inverse functions to solve for the original function and then integrating it to find the general solution. This usually involves using algebraic manipulation and inverse function rules to solve for the original function.

5. Are there any special techniques for integrating with inverse functions?

Yes, there are special techniques for integrating with inverse functions, such as substitution and integration by parts. These techniques can make the process of solving for the original function easier and more efficient.

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