- #1
Moogie
- 168
- 1
Hi
I'm doing some work on finding the derivative of an inverse function and getting a bit lost, probably because I am confusing myself with the inverses. So i think i need to clear up the inverse part before I move on to the calculus bit.
You have the function
f(x) = 1/3x3 - x2 + 5x -11
This function has an inverse f-1(x). Set this equal to y
y = f-1(x)
You can then say f(y) = x
Is this because f(y) = f(f-1(x)) = x
I'm now getting confused, due to notation I think. The f by definition is
1/3(some input)3 - (some input)2 + 5(some input) -11
So f(y) = 1/3y3 - y2 + 5y -11 = x
Is this right?
I'm doing some work on finding the derivative of an inverse function and getting a bit lost, probably because I am confusing myself with the inverses. So i think i need to clear up the inverse part before I move on to the calculus bit.
You have the function
f(x) = 1/3x3 - x2 + 5x -11
This function has an inverse f-1(x). Set this equal to y
y = f-1(x)
You can then say f(y) = x
Is this because f(y) = f(f-1(x)) = x
I'm now getting confused, due to notation I think. The f by definition is
1/3(some input)3 - (some input)2 + 5(some input) -11
So f(y) = 1/3y3 - y2 + 5y -11 = x
Is this right?