2. (2 points) Solve the equation $f^{-1}(g(x))=h(x)$, where $f^{-1}(x)$ is the inverse function of $f(x)$, if it is known that $f(x)=x^{3}+2 x^{2}+3 x+4, g(x)=x^{4}+2 x^{3}+1 x^{2}+11 x+11, h(x)=x+1$.
Answer: $\frac{-3 \pm \sqrt{5}}{2}$ and 1.

Solution: The original equation can be rewritten as $f(h(x))=g(x)$, that is, $(x+1)^{3}+2(x+1)^{2}+3(x+1)+4=x^{4}+2 x^{3}+x^{2}+11 x+11$.

Expanding the brackets and moving everything to the right side, we get: $x^{4}+x^{3}-4 x^{2}+x+1=0$. Notice that the polynomial on the left side is divisible by $x-1$, and more specifically by $(x-1)^{2}$. Therefore, we obtain the root 1 with multiplicity 2. Additionally, the quotient from dividing our fourth-degree polynomial by $(x-1)^{2}$ is the quadratic polynomial $x^{2}+3 x+1$, and its roots are $\frac{-3 \pm \sqrt{5}}{2}$.