Calculus Question (Inverse Functions)

Question

Let $f(x)=x^3 +4x^2 +6x$ and $g(x)$ be its inverse. Then the value of $g'(-4)$ is?

My attempt at the solution: $$ g(x) = f^{-1}(x) \implies f(g(x))=x, $$ differentiating both sides, $$ f'(g(x)) \cdot g'(x)=1 \implies g'(x)=\frac1{f'(g(x))}. $$ I am stuck here as we do not know what $g(-4)$ is. As far as I know, there is no simple method to find the inverse of $f(x)$ as it is a cubic polynomial. Is the question missing some data? If not, how do I proceed from here?

Answer 1

$g(-4)$ is, by definition, the (unique) real solution to $f(x)=-4$.

Edit: to find it, for example use Gauss lemma (the solution is $x=-2$)

Answer 2

Hint: Interchange roles of f and g in what you wrote $$f'(x) = \frac{1}{g'(f(x))}$$ Find $x_0$ such that $f(x_0)=-4$ , then $$f'(x_0) = \frac{1}{g'(f(x_0))}$$ or $$f'(x_0) = \frac{1}{g'(-4)}$$ You know $f'(x)$ and you should have gotten $x_0$ so you're done.