Thread: Finding the inverse of a function with two of the same variables

f(x) = 3x³ - x. If h(x) is the inverse of f(x) then h'(-2) =

Ok this is what I did:
1. Switched f(x) and replaced it with y, even though they're the same thing, but it's easier to see.
y = 3x³ - x

2. Switched the x and y variables in order to find the inverse.
x = 3y³ - y

3. Multiply both sides by natural log.
ln(x) = ln(3y³ - y)

4. Distributed the natural log on the other side.
ln(x) = ln(3y³) - ln(y)

5. Use the log property, where subtracting two logs will mean dividing it.
ln(x) = ln(3y³)/ln(y)

6. Brought the exponent into the front.
ln(x) = 3ln(3y)/ln(y)

But then... I think I'm just experimenting and moving stuff around...
I don't think I'm really sure if I'm getting anywhere with this.
At first I stopped at step #2, and tried using natural log to see if it'll work.

But so far, I still have all 'y' on one side.
Anyone think they can help me out please?
Thanks

Originally Posted by Chaim

f(x) = 3x³ - x. If h(x) is the inverse of f(x) then h'(-2) =

You want to find .

It is clear that so

Originally Posted by Plato

You want to find .

It is clear that so

Oh wow, that makes more sense now.
So basically to find the inverse, you could just place under the denominator of 1 right?
Then after that you find out the derivative of f, which is 9x² - 1
Then I plug in the -1, which makes the denominator = -10 right?

Then it would be -1/10?

Originally Posted by Chaim

Then after that you find out the derivative of f, which is 9x² - 1
Then I plug in the -1, which makes the denominator = -10 right?
Then it would be -1/10?

Well no.
so

Originally Posted by Plato

Well no.
so

??? In the original post, the function [tex]f(x)= 3x^3- x[/quote] which has derivative , as Chaim said, not .