Question

If g is the inverse of f and if `f(x)=x^3-5x^2+2x-1`, how do you calculate g'(-9) if the domain of f(x) is the set of integers less than 0? Thanks in advance for taking time to help me out. Steve?

Answer 1

Use `d/dx(f^-1(b)) = 1/(f'(f^-1(b))`

Explanation:

We have, in this case `b=-9`. We need `f^-1(-9)`.

That is: we need `x` in the domain of `f` such that `f(x) = -9`.

We must solve `f(x) = -9` for solutions in the domain of `f`.

Solve
`x^3-5x^2+2x-1= -9`.

If you know the rational zeros theorem, use it. Otherwise, you'll need to solve "by inspection". (That's the fancy name for "guess and check".)

Try `x=1`, `x=2`, `x=-1` STOP!

`f(-x)=-9` and `-1` is in the domain of `f`, so if the problem is well written that should be the only solution in the domain. (If there are multiple solutions, then `f` is not invertible.)

`f'(x) = 3x^2-10x+2`,

and `f(-1) = 15`.

`d/dx(f^-1(-9)) = 1/(f'(f^-1(-9))) = 1/(f'(-1)) = 1/15`.