f(x) = 3x^{3}- 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 = 3x^{3}- x

2. Switched the x and y variables in order to find the inverse.

x = 3y^{3}- y

3. Multiply both sides by natural log.

ln(x) = ln(3y^{3}- y)

4. Distributed the natural log on the other side.

ln(x) = ln(3y^{3}) - ln(y)

5. Use the log property, where subtracting two logs will mean dividing it.

ln(x) = ln(3y^{3})/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