Given $f(x) = x^3 - 3x^2 -1, x\ge2$, find $f^{-1}{'}(-1)$
I know I should show work I have done on this question when I ask on stackexchange but I literally get stucked at the beginning. I cannot find the inverse of this function and needless to say the derivative of it. Any help would be appreciated!
You do not need to find the algebraic expression of $f^{-1}(x)$
You just need the derivative of the inverse at $x=-1$
So you calculate where $f(x)=-1;\;x\ge 2$ solving
$$x^3-3 x^2-1=-1;\;x\ge 2\to x=3$$ And apply the rule of the derivative of the inverse $$f^{{-1}'}(-1)=1/f'(3)$$ As $f'(x)=3x^2-6x$ we have $f'(3)=9$ and finally $$f^{{-1}'}(-1)=1/9$$