# Finding the inverse of a function with two of the same variables

• Mar 14th 2013, 08:17 AM
Chaim
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 :)
• Mar 14th 2013, 08:29 AM
Plato
Re: Finding the inverse of a function with two of the same variables
Quote:

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

You want to find $\displaystyle h'(-2)=\frac{1}{f'(h(-2))}$.

It is clear that $\displaystyle f(-1)=-2$ so $\displaystyle h(-2)=-1.$
• Mar 14th 2013, 09:18 AM
Chaim
Re: Finding the inverse of a function with two of the same variables
Quote:

Originally Posted by Plato
You want to find $\displaystyle h'(-2)=\frac{1}{f'(h(-2))}$.

It is clear that $\displaystyle f(-1)=-2$ so $\displaystyle h(-2)=-1.$

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?
• Mar 14th 2013, 09:30 AM
Plato
Re: Finding the inverse of a function with two of the same variables
Quote:

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.
$\displaystyle f'(x)=9x^2-1$ so $\displaystyle f'(-1)=8.$
• Mar 14th 2013, 03:03 PM
HallsofIvy
Re: Finding the inverse of a function with two of the same variables
Quote:

Originally Posted by Plato
Well no.
$\displaystyle f'(x)=27x^2-1$ so $\displaystyle f'(-1)=26.$

??? In the original post, the function [tex]f(x)= 3x^3- x[/quote] which has derivative $\displaystyle f'(x)= 9x^2- 1$, as Chaim said, not $\displaystyle 27x^2- 1$.