Hey guys.

I've got the following problem:

Let $\displaystyle U = (0, \infty ) \times R \subset R^2 $ og let $\displaystyle f : U \rightarrow R^2$ be defined by : $\displaystyle f(x,y) = (x, y^3+xy)$

(a) Show that $\displaystyle f(U) = U$Done that already.

(b) Show that $\displaystyle f$ is injective

(Hint: When is $\displaystyle g: R \rightarrow R$ given by $\displaystyle g(y)= y^3 + ay + b$ monotonic?)

(c)... some more questions.

I've already found the solution to (a), but I cant quite figure out how to solve (b). This is pretty much what I've tried so far:

For $\displaystyle f : U \rightarrow R$ to be injective, the following has to be true : $\displaystyle q \neq p \Rightarrow f(q) \neq f(p) \forall q, p \in U$.

So, for proof by contradiction, we assume that for a pair $\displaystyle q \neq p \in U$ we have that $\displaystyle f(p) = f(q)$.

Since $\displaystyle f(q) = f(p)$ we have that $\displaystyle x_q = x_p \wedge y_q^3+x_q y_q = y_p^3+x_p y_p$ (1).

But since $\displaystyle q \neq p$ we have that $\displaystyle x_q \neq x_p \vee y_q \neq y_p$.

According to (1) we know that $\displaystyle x_q = x_p$ so $\displaystyle y_q \neq y_p$ must be the case.

From here I dont know exacly where to go. I can see that I'm not really using the hint.

I know that all monotonic functions are injective, but how can I prove that f is monotonic?

I can see by visualization that f will be monotonic, because x is defined to be positive, but how do I prove that rigorously?

Thanks for your help guys.

Morten