Prove an injective polynomial $f: \Bbb{R}\to \Bbb{R}$ is surjective over $\Bbb{R}$

Question

Prove an injective polynomial $f: \Bbb{R}\to \Bbb{R}$ is surjective over $\Bbb{R}$

How can I prove that using basic calculus methods?

Answer 1

If $\operatorname{deg}(P)$ is even, then $f$ cannot be injective, because it either has a local maximum or a local minimum. Therefeore $\operatorname{deg}(P)$ is odd and every polynomial with odd degree is surjective.

Answer 2

If you have a polynomial $$f(x)=a_nx^n+\cdots+a_0.$$ If it is injective this implies that $n$ is odd because if $n$ is even you have that $$\lim_{x\to \infty}f(x)=\lim_{x \to -\infty}f(x)=\pm \infty.$$ So $n$ is odd and this imply that $$\lim_{x\to \infty}f(x)= \pm \infty \qquad \mbox{and} \qquad \lim_{x \to -\infty}f(x)=\mp \infty,$$ and such that a polynomial is continuos it is surjective.

Answer 3

First, as $f$ is injective it cannot be a polynomial of even degree as $f$ tends to $+\infty$ in both $+\infty$ and $-\infty$, and it can't be constant. Now $f$ is a continuous function that goes to $-\infty$ in $-\infty$ and to $+\infty$ in $+\infty$. From the theorem of intermediate values it is easy to conclude that $f$ is surjective.

Answer 4

Hints:

You only need the Intermediate value theorem: if a polynomial function $p(x)$ (which is different from a polynomial) is injective, it is monotonic, hence its derivative $p'(x)$ has a constant sign. As the limit of a polynomial function at $\pm\infty$ is the limit of its leading term, this implies this leading term has an even degree, so $p(x)$ has an odd degree. Supposing w.l.o.g. that the leading coefficient of $p(x)$ is positive, we deduce that $$\lim_{x\to+\infty} p(x)=+\infty,\qquad\lim_{x\to-\infty} p(x)=-\infty.$$ Can you end the proof?