Derivatives of Inverse Functions

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Homework Help Overview

The problem involves finding the derivative of the inverse function of f(x) = x³ + x at the point where f(2) = 10. Participants are exploring the relationship between a function and its inverse, particularly focusing on the application of implicit differentiation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss switching the variables in the function and differentiating implicitly to find the derivative of the inverse function. There is a focus on the correct application of the derivative formula for inverse functions, with some questioning the next steps after deriving the expression for (f-1)'.

Discussion Status

Some participants have provided guidance on the differentiation process and the evaluation of the derivative at a specific point. There are multiple interpretations being explored regarding the steps to take after finding the derivative expression, but productive discussion is ongoing.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is a focus on ensuring clarity in the notation used for derivatives and inverse functions.

jumbogala
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Homework Statement


f(x) = x3+ x. Note that f(2) = 10. Find (f-1)`(10).


Homework Equations





The Attempt at a Solution


Note that where I have written ` it denotes prime (as in the derivative of).

- Switch the x and y variables. x= y3 + y

- Differentiate implicitly 1= (3y2 + 1)y`

- Solve for y`. y`= 1 / (3y2 + 1). Since y = f-1, then y` = (f-1)`

Therefore (f-1)` = 1 / (3y2 + 1)

Since f(2) = 10, f-1(10)=2. But my equation is(f-1)`, not just
f-1! I don't know what to do after this point!
 
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jumbogala said:
- Solve for y`. y`= 1 / (3y2 + 1). Since y = f-1, then y` = (f-1)`

Therefore (f-1)` = 1 / (3y2 + 1)

Since f(2) = 10, f-1(10)=2. But my equation is(f-1)`, not just
f-1! I don't know what to do after this point!
Since you let [tex]y=f^{-1}(x)[/tex], what you actually have is [tex](f^{-1})'(x) = \frac{1}{3(f^{-1}(x))^2 + 1}[/tex].
 
Last edited:
Okay, and we want x to be 10.

So (f-1)`(10)= 1/13

Okay, I think that makes sense now. Thank you!
 
Another way: f(x)= [itex]x^3+ x[/itex] so f'(x)= [itex]3x^2+ 1[/itex] and f'(2)= 13. Since f(2)= 10, f-1(10)= 2 and f'(10)= 1/13.
 

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