The problem statement, all variables and given/known data
Show that if ##f:\mathbb{R}\to \mathbb{R}## is a polynomial function of odd degree, then ##f(\mathbb{R}) = \mathbb{R}##.
The attempt at a solution
How can I rigorously prove this? What is the most direct and concise way to prove this?
What I have is the following:
Any polynomial in ##x_1, x_2, \ ... \ ,x_n## with coefficients in ##\mathbb{R}## is continuous. Since ##\mathbb{R}## is connected, ##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 thus 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}##. Therefore, ##f(\mathbb{R})\in \mathbb{R}.##
Show that if ##f:\mathbb{R}\to \mathbb{R}## is a polynomial function of odd degree, then ##f(\mathbb{R}) = \mathbb{R}##.
The attempt at a solution
How can I rigorously prove this? What is the most direct and concise way to prove this?
What I have is the following:
Any polynomial in ##x_1, x_2, \ ... \ ,x_n## with coefficients in ##\mathbb{R}## is continuous. Since ##\mathbb{R}## is connected, ##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 thus 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}##. Therefore, ##f(\mathbb{R})\in \mathbb{R}.##
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