# Derivatives of Inverse Functions

## Homework Statement

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

## 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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- 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 $$y=f^{-1}(x)$$, what you actually have is $$(f^{-1})'(x) = \frac{1}{3(f^{-1}(x))^2 + 1}$$.

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Okay, and we want x to be 10.

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

Okay, I think that makes sense now. Thank you!

HallsofIvy
Another way: f(x)= $x^3+ x$ so f'(x)= $3x^2+ 1$ and f'(2)= 13. Since f(2)= 10, f-1(10)= 2 and f'(10)= 1/13.