How do I find the inverse of a derivative at $x=-1$?

Question

Let $f(x)=x^3-3x^2-1$, $x\geq2$. Find the value of $\left(df/dx\right)^{-1}$ at $x=-1$.

This is the work that I've done so far:

1. Found $f'(x)=3x^2-3x$

2. Found the inverse of $f'(x)$ ►$3y^2-6y-x=0$

3. Found the roots of $y$ ►$[6±(36+12x)^{1/2}]/6$

4. Plugged $x=-1$ into $[6±(36+12x)^{1/2}]/6]$ and got $[3±(6)^{1/2}]/3$

Is $[3±(6)^{1/2}]/3$ the correct answer?

Answer 1

$f(x)=x^3-3x^2-1$.

• $f'(x)=3x^2-6x$
• For $(f'(x))^{-1} $ complete the squares. $y=3x^2-6x+3-3=3(x-1)^2-3\implies x=1+\sqrt{\dfrac {y+3}{3}},\:y\in[-3,\infty)$