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?
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?