# 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?

Sep 13, 2016

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

#### Explanation:

We have, in this case $b = - 9$. We need ${f}^{-} 1 \left(- 9\right)$.

That is: we need $x$ in the domain of $f$ such that $f \left(x\right) = - 9$.

We must solve $f \left(x\right) = - 9$ for solutions in the domain of $f$.

Solve
${x}^{3} - 5 {x}^{2} + 2 x - 1 = - 9$.

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!

$f \left(- x\right) = - 9$ 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.)

$f ' \left(x\right) = 3 {x}^{2} - 10 x + 2$,

and $f \left(- 1\right) = 15$.

$\frac{d}{\mathrm{dx}} \left({f}^{-} 1 \left(- 9\right)\right) = \frac{1}{f ' \left({f}^{-} 1 \left(- 9\right)\right)} = \frac{1}{f ' \left(- 1\right)} = \frac{1}{15}$.