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
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 9x^{2} - 1
Then I plug in the -1, which makes the denominator = -10 right?
Then it would be -1/10?