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Let $p (x)$ be an odd degree polynomial in one variable with coefficients from the set $\newcommand{\R}{\mathbb R} \R$ of real numbers. Let $g : \R → \R$ be a bounded continuous function. Prove that there exists an $x_0 ∈ \R$ such that $p (x_0) = g (x_0)$.

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If the degree of $p$ is odd, what can you say about the image of $p$? Then use the intermediate value theorem. – copper.hat Jul 23 '12 at 14:04
This is a good job for my students. +1 – Babak S. Jan 12 '13 at 9:32
up vote 3 down vote accepted

Let $f\colon x\mapsto p(x)-g(x)$ and $M$ such that $|g(x)|\leq M$ for all real number $x$. The degree of $p$ is odd, hence

  • first case: $\lim_{x\to +\infty}p(x)=+\infty$: in this case $f(x)\geq p(x)-M$ hence $\lim_{x\to +\infty}f(x)=+\infty$. We have $\lim_{x\to -\infty}p(x)=-\infty$ by the condition over the degree, and $f(x)\leq p(x)+M$ so $\lim_{x\to -\infty}f(x)=-\infty$. In particular, the continuous map $f$ takes negative and positives values, hence vanished (by the intermediate value theorem) at some point $x_0$.
  • second case $\lim_{x\to +\infty}p(x)=-\infty$: apply the previous case to $-p$ and $-g$ (which is continuous and bounded).
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