The problem says that if $f(x)=x^3+x^2+2x$ find $(f^{-1} )' (-2)$.
Now I solved similar problems however there was given another hint, like $f(-2)=b$ or something. I don't know how to approach this problem, I just know that if I denote $g(x)$ to be the inverse of $f(x)$ then $g'((x)f(x))\cdot f'(x)=1$ Could you help me
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According to the inverse function theorem, we have $$(f^{-1})'(a)=\frac{1}{f'(f^{-1}(a))}$$ with $a=-2.$ With $f'(x)=3x^2+2x+2$ and $f^{-1}(-2)=\frac{1}{(-2)^3+(-2)^2+2\cdot(-2)}=-\frac{1}{8}$ we get $$(f^{-1})'(-2)=\frac{1}{3(-\frac{1}{8})^2+2(-\frac{1}{8})+2}=\frac{64}{115}.$$
Edit: I have $\color{red}{\textrm{misinterpreted}}$ the notation $f^{-1}$ as being the multiplicative inverse of $f$ whereas it's the inverse function of $f$. I may or may not correct this in near future - in the meantime it may still serve as an example for a popular pitfall. So, again and explicitly: What I have written above may be wrong.
With $a=-2$, find $f^{-1}(a)$ by the following
$$f(b)=a$$
$$b^3+b^2+2b+2=0$$
$$b^2(b+1)+2(b+1)=(b^2+2)(b+1)=0$$
Real solutions give that $b=-1$
$$\therefore-1=f^{-1}(a)$$
Now
$$f^{-1}{'}(a)=\frac{1}{f'(f^{-1}(a))}$$
$$f'(-1)=\left.\left(3x^2+2x+2\right)\right|_{-1}=3$$
Therefore we conclude that
$$f^{-1}{'}(a)=\frac{1}{3}$$