Let $f(x)$ be a monic polynomial of odd degree. Prove that $\exists A\in \mathbb{R}$ s.t. $f(A)<0$ and there exists $B \in \mathbb{R}$ such that $f(B)>0$.
Deduce that every polynomial of odd degree has a real root.
There are questions that answer the final part, but they do not do so by proving the first part. I am fairly sure that this involves the intermediate value theorem, but not sure how to implement it in this case.
Hint: consider the limit of $f(x)$ as $x\to\infty$ and separately the limit as $x\to-\infty$.
For the final part. If you have $f(A)< 0$ and $f(B)>0$ then by the IVT every value in $[f(A),f(B)]$ is attained by $f(x)$ for some $x$ between $A$ and $B$, and this includes $0$.
To show the existence of the $A$ and $B$ show that for $x $ large one has that the sign of $f(x)$ is the sign of the leading coefficient. And, if the degree is odd for small $x$ one has that the sign of $f(x)$ is the opposite sign of the leading coefficient.
Let $f(x)=x^n+a_{n-1}x^{n-1}+\dotsm+a_0$ be a monic polynomial of odd degree. Let $f(x)=x^ng(x)$ for $x\neq 0$, where $g(x)=1+\frac{a_{n-1}}{x}+\dotsm+\frac{a_0}{x^n}$. For $|x|>1$, observe that $$|g(x)-1|\leq \frac{|a_{n-1}|}{|x|}+\frac{|a_{n-2}|}{|x|^2}+\dotsm+\frac{|a_{0}|}{|x|^n}\leq \frac{|a_{n-1}|}{|x|}+\frac{|a_{n-2}|}{|x|}+\dotsm+\frac{|a_{0}|}{|x|}\leq\frac{a}{x},$$ where $a=\sum\limits_{i=0}^{n-1}{|a_i|}$. So if $|x|>\max\{1,2a\}$ then $|g(x)-1|<\frac{1}{2}$ and hence $\frac{1}{2}<g(x)<\frac{3}{2}$. Hence $g(x)>0$ for all $x$ satisfying $|x|>\max\{1,2a\}$. If we choose $b>\max\{1,2a\}$ then both $g(b)$ and $g(-b)$ are positive. But then $f(b)>0$ and $f(-b)<0$. So by the intermediate value theorem, there exists a $c\in[-b,b]$ such that $f(c)=0$.
