Polynomial functions of odd degree are surjective

Prove if the function $f: \mathbb{R} \to \mathbb{R}$ is a polynomial function of odd degree, then $f(\mathbb{R}) = \mathbb{R}.$

We know a polynomial, $f(x)=a_nx^n +a_{n−1}x^{n−1} ...a_1x+a_0$ with real coefficients is continuous. Also, $\mathbb{R}$ is connected now since $\mathbb{R}$ is connected then $f(\mathbb{R})$ is connected, thus we can apply the intermediate value theorem. Now any polynomial of odd degree has at least one real root since every polynomial, with real coefficients, has as many roots as it has degrees. Also, any complex roots are paired with their complex conjugates thus they take up an even number of roots. So, an odd degree polynomial has only an even number of roots that can be complex therefore at least one root must be real. Therefore, there exists $p,q\in \mathbb{R}$, $p<q$ such that $f(p)<0$ and $f(q)>0$. Thus we can then choose $p\in \mathbb{R}$ such that $f(p) \ge 0,$ $f(p)\le 0$ and obtain any $f(\mathbb{R})\in \mathbb{R}$.

Similarly, suppose that the leading coefficient of $f$ is positive then, $\lim_{x\to \infty} f(x) = \infty$, and $\lim_{x\to -\infty} f(x) = -\infty$. So, any arbitrarily large or small value can be attained. Due to the intermediate value theorem and since $f$ is continuous, any value between such a large value $M$ and a small value $m$ is attained. Hence, $f(\mathbb{R}) = \mathbb{R}$. And a similar argument follows if the leading coefficient of $f$ is negative. Therefore, $f(\mathbb{R}) = \mathbb{R}$.

Is this correct?

Solution 1:

That is correct, but all you really need is the last part : Assume without loss of generality that the leading coefficient is positive (else apply this result to $-f$), and note that $$ \lim_{x\to +\infty} f(x) = +\infty, \text{ and } \lim_{x\to -\infty} f(x) = -\infty $$ Then, for any $y\in \mathbb{R}$, there is $M > 0$ such that $$ x > M \Rightarrow f(x) > y, \text{ and } x < -M \Rightarrow f(x) < y $$ Hence, by intermediate value theorem, there is $x_0 \in [-M,M]$ such that $f(x_0) = y$.

This is true for all $y \in \mathbb{R}$, and hence $f(\mathbb{R}) = \mathbb{R}$