Let p(x) be an odd-degree polynomial function. Prove that p(x)=0 has at least one real solution
Not sure if there is a formal proof of this, but one can see that as $\displaystyle x \to \infty$ then $\displaystyle p(x) \to \infty$ and as $\displaystyle x \to -\infty$ then $\displaystyle p(x) \to -\infty$. (assuming leading coefficient is positive, if negative its just the other way around)
By definition any polynomial is continuous on $\displaystyle \mathbb{R}$ and thus by the IVT we can say there exists at least one $\displaystyle x \in \mathbb{R}$ such that $\displaystyle p(x)=0$