If $n$ is even, prove $P(x)$ cannot have an inverse function

Question

If $P(x)$ is a polynomial of degree $n$, how can I prove that if $n$ is even, $P(x)$ cannot have an inverse function.

Answer 1

I'm assuming that the domain is $\mathbb{R}. $If $p$ is injective, then it should be increasing or decreasing, but its derivative is a polynomial of odd degree, hence its image contains both positive and negative values.

Remark: it doesn' work for arbitraty rings. In $\mathbb{F}_{2}[x]$, for example, the polynomial $x^2$ is a bijective function.

Answer 2

^{This answer assumes that you are talking about polynomials over the real numbers.}

Hint.

• Consider the degree of $P'(x)$.
• Then note that every polynomial of odd degree must have one real zero.
• Suppose $P$ has an inverse $Q$ so that $P(Q(x))=x$ on $\mathbb{R}$. Take the derivatives on both sides to get a contradiction.

Answer 3

This may be an overly complicated solution, but:

Without loss of generality $P$ is monic, so $P(x)\to\infty$ as $x\to\pm\infty$. Fix $M$ greater than $|P|$'s value at any of its turning points, as there are finitely many of these. Applying the intermediate value theorem to $1/P$ either side of the smallest inteval containing $P$'s real roots, $P(x)=M$ has roots both side of this integral.

Answer 4

$P(x)$ can be written as $P(x) = a_0 + a_1x + ... + a_nx^n,\ a_n \neq 0,\ $ for some even integer $\ n.$

By writing $P(x) = x^n\left( a_n + \frac{a_{n-1}}{x}+ ... + \frac{a_2}{x^{n-2}}+ \frac{a_1}{x^{n-1}}+ \frac{a_0}{x^n} \right),\ $ we see that $\displaystyle\lim_{x\to+\infty}P(x) = \displaystyle\lim_{x\to-\infty}P(x) = (\pm\infty)^n \times a_n = \infty \times a_n\ $ because $\ n\ $ is even, so both limits equal either $\ +\infty\ $ or $\ -\infty,\ $ depending on the sign of $\ a_n.\ $ Now suppose WLOG that $\ a_n>0,\ $ so that $\displaystyle\lim_{x\to+\infty}P(x) = \displaystyle\lim_{x\to-\infty}P(x)= +\infty\\ $ and let $x_1 \in \mathbb{R}.\ $ Then due to the continuity of $P(x)\ $ and the definition of $\displaystyle\lim_{x\to-\infty}P(x)= +\infty\ $ and $\displaystyle\lim_{x\to+\infty}P(x)= +\infty,\ \exists\ x'< x_1\ $ such that $P(x') = P(x_1) + 1,\ $ and $\ \exists\ x''> x_1\ $ such that $P(x'') = P(x_1) + 1,\ $ i.e. $P(x')=P(x''),\ $ showing that $P$ is not injective. Therefore $P(x)$ doesn't have an inverse function.