Differentiating Inverse Functions

[BuBbLeS01]

1. The problem statement, all variables and given/known data
f(x) = x^3 + 2x - 1 when a=2


2. The attempt at a solution

I thought you did...
1/(f '(f-1(x)))
but I am not sure how to solve for x?

0=x^3 + 2x - 1
1=x^3 + 2x -1
I tried factoring but that did not work either.

[frasifrasi]

what are you looking for?

[frasifrasi]

if it is f-1(a) when a = 2...

set the first equation equal to 2, which will happen when x = 1.

So, f-1(2) = 1.

Now if you want f-1'(x), you have:

= 1 / f'(f-1(x))

so 1/ f'(1)

find the derivative of f(X):

3x^2 + 2

so,
answer = 1/(3(1) +2) = 1/5.

[BuBbLeS01]

thank you so very much!

[BuBbLeS01]

How do you know it is 1? Because thats the only number without an X term?

[BuBbLeS01]

How do you know f-1(2) = 1?

[EnumaElish]

frasifrasi was not supposed to give away the answer, if that's the answer. That's not how this forum's supposed to work.

f -1(2) is the answer to question, "at what value of x does x^3 + 2x -1 = 2"?

frasifrasi assumed f(x) = a, which may or may not be justified. Your statement of the problem does not indicate what a is. If that assumption is right, then f -1(f(x)) = f -1(a), and by the definition of an inverse function, f -1(f(x)) = x. So x = f -1(a). You can verify that when x = 1, f(1) = 2. Therefore 1 = f -1(2).

[BuBbLeS01]

Yea I thought it was weird that he just gave the answer. But I am trying to figure out how you figured out that x^3 + 2x - 1 = 2 when x= 1?

[HallsofIvy]

He solved the equation of course! Cubics can be difficult to solve so I suspect he did what I would: try some easy numbers for x and hope one works. In "real life" that is seldom true but in "made up" exercises it often is.