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Hi there, the question and the provided answer to it is actually a little bit broken.

Indeed, if \(n=\dfrac{1}{3m}\), then \(h^{-1}\) will exist, but the converse doesn't hold. That is, it's not necessarily true that \(n=\dfrac{1}{3m}\) if \(h^{-1}\) exists.

We can still learn from this question however.

Recall that: the inverse of a function will exist

(it may only be a partial function) if and only if it is

**one-to-one**.

So, in the case of the cubic function \(h(x)=mx^3+x^2+nx+p\), where \(m,n,p\in\mathbb{R}\setminus\{0\}\), you just need to ensure that there are either no stationary points, or that if there is one that it's a point of inflection.

We have \[h'(x)=3mx^2+2x+n=0\implies x=\frac{-1\pm\sqrt{1-3mn}}{3m},\ \ \text{for suitable }m\ \text{and }n,\] and so for \(h^{-1}\) to exist, we require \[1-3mn\leq 0\implies \boxed{n\geq\frac{1}{3m}\ \text{and}\ m>0\quad \mathbf{or}\quad n\leq\frac{1}{3m}\ \text{and}\ m<0.\ }\] The answer you gave is only a small subset of all the possible relationships between \(m\) and \(n\). For example, it's clear that \(m=n=-2\) does not satisfy \(n=\dfrac{1}{3m}\), yet if you sketch the graph of \(h\), one would find that its inverse exists.

Hope this helps