Verify $\displaystyle f(x) = 1 - 2x - x^3$ has an inverse, and find $\displaystyle (f^{-1})'(4)$.

So I take the derivative and get $\displaystyle f'(x) = -2 -3x^2$.

It's negative for all x, so it's always decreasing, so it's 1-to-1 and has an inverse.

Now, how do I find $\displaystyle (f^{-1})'(4)$? Is this the same as $\displaystyle \frac{1}{f'(4)} = \frac{1}{-2-3(4)^2} = -\frac{1}{50} $?