Answer
$$\left.\frac{df^{-1}}{dx}\right|_{x=f(3)}=\frac{1}{9}$$
Work Step by Step
$$f(x)=x^3-3x^2-1\hspace{1cm}x\ge2$$
To find $df^{-1}/dx$ at $x=-1=f(3)$, we apply Theorem 3, the Derivative Rule for Inverses. In detail,
1) Find $\left.\frac{df}{dx}\right|_{x=3}$:
We have $$\frac{df}{dx}=3x^2-6x$$
So, $$\left.\frac{df}{dx}\right|_{x=3}=3\times3^2-6\times3=27-18=9$$
2) According to Theorem 3:
$$\left.\frac{df^{-1}}{dx}\right|_{x=f(3)}=\frac{1}{\left.\frac{df}{dx}\right|_{x=3}}=\frac{1}{9}$$
