How can I find the inverse of $h(x)=-x(x^3 +1)$ How can I find the inverse of $h(x)=-x(x^3+1)$? it's asked also to find $h^{-1}(2)$ and $h^{-1}(-2)$. I think it's easy to find a domain where this function is bijective. I've already find $h^{-1}(-2)=1$. My problem is to find $h^{-1}(2)$ and the inverse itself.
Thanks
 A: It's complicate to find the inverse of a non bijective function... In particular, your function is neither one to one, nor onto. Indeed, $$h(-1)=h(0)=0$$ and $$h^{-1}(2)=\emptyset,$$
that's the reason that you can't find $h^{-1}(2)$ ;-)
Restriction of the domain and codomain that gives $h$ bijective
$$h'(x)=-(x^3+1)-3x^3=-4x^3-1$$
Therefore it's one to one on $\left]-\infty ,-\frac{1}{\sqrt[4]3}\right]$ or $\left[\frac{-1}{\sqrt[3]4},+\infty \right[$. By intermediate value theorem, since $h$ has a maximum at $x=-\frac{1}{\sqrt[3]4}$ and that $$\lim_{x\to-\infty }h(x)=\lim_{x\to \infty }h(x)=-\infty, $$ you have the surjectivity on $\mathbb R\to \left]-\infty ,h\left(\frac{-1}{\sqrt[3]4}\right)\right]$. To conclude, 
$$h_1:\left]-\infty ,\frac{-1}{\sqrt[3]4}\right]\longrightarrow \left]-\infty ,h\left(\frac{-1}{\sqrt[3]4}\right)\right]$$
define by $h_1(x)=h(x)$ and
and
$$h_2:\left[\frac{-1}{\sqrt[3]4},+\infty \right[\longrightarrow \left]-\infty ,h\left(\frac{-1}{\sqrt[3]4}\right)\right]$$
define by $h_2(x)=h(x)$ are bijective.
