does anyone know how to demonstrate this proposition?
“Closed half-lines” are subsets of $\Bbb{R}$ of the form ${[x ∈ \Bbb{R}: x ≤b}]$ or ${[x ∈ \Bbb{R}: x ≥b]}$ for real numbers b. A polynomial of degree n on R is a function $x → a_nx^n + · · · + a_1x + a_0$ with $a_n= 0$. Show that the range of any polynomial of degree n ≥ 1 is $\Bbb{R}$ for n odd and a closed half-line for n even.
Now, I know already the difference between $x^2$ and $x^3$ (since + * +=+ and - * -=+, the former will be always positive, while the latter has range $\Bbb{R}$), although I wouldn't know how to start in order to demonstrate it mathematically...
Polynomials are continuous functions, so their range is an interval. Consider $a_0+a_1x+...+a_nx^{n}$ where $a_n \neq 0$ and $n$ is odd. We can write this as $x^{n}(a_n+\frac {a_{n-1}} x+...+\frac {a_0} {x^{n}})$. From this it follows that the polynomial tends to $\pm \infty$ as $ x \to \pm \infty$ (not necessarily in that order) and so the range is the entire real line. Now suppose $n$ is even and $a_n>0$. Then the polynomial tends to $ \infty$ as $ x \to \pm \infty$. So the right end point of the range is $\infty$. By continuity it has a finite minimum so the range is $[a,\infty)$ for some $a$. Similarly when $a_n <0$ the range is $(-\infty, b]$ for some $b$.