Showing a function $f$ has a zero I have been working on this problem and was wondering if anyone could give me some hints as to how to answer this. So far as $p$ is a polynomial it is continuous and $g$ is given as being continuous, and so by algebra of continuity $f$ is continuous, now does the solution include the use of the Intermediate Value Theorem?
Any help would be appreciated.

Let $f = p + g$ , where $p$ is an odd degree polynomial and $g:$ $\mathbb{R}$ $\rightarrow$ $\mathbb{R} $ is a bounded, continuous function. Show that f must have a zero.


Suppose that $g$ is continuous and bounded by $M>0$ and $p(x)= x^n+\ldots+a_0$ where $n$ is odd. Then $f(x)=x^n+ a_{n-1}x^{n-1}\ldots+a_0+g(x)$ which may be written as $x^n\big(1+\ldots\frac{a_0}{x^n}+\frac{g(x)}{x^n}\big)$ when $x\not=0$, the basic idea is that when $|x|$ is big $x^n$ dominate and $f$ contain the sign of this.
More formally we need to estimate the quantity inside the  parenthesis. 
\begin{align}\bigg|\frac{a_{n-1}}{x}+\ldots+\frac{a_0}{x^n}+\frac{g(x)}{x^n}\bigg|\le\frac{|a_{n-1}|}{|x|}+\ldots+\frac{|a_0|}{|x^n|}+\frac{|g(x)|}{|x^n|}\\
\le\frac{|a_{n-1}|}{|x|}+\ldots+\frac{|a_0|}{|x^n|}+\frac{M}{|x^n|}\end{align}
Let choose $|x|\ge N=\max\{1,2^2|a_{n-1}|\ldots,2^{n+1}|a_0|,2^{n+2}M\}$
Thus 
\begin{align}\frac{|a_{n-1}|}{|x|}+\ldots+\frac{|a_0|}{|x^n|}+\frac{M}{|x^n|}\le\frac{|a_{n-1}|}{|x|}+\ldots+\frac{|a_0|}{|x|}+\frac{M}{|x|}\\
\le\frac{1}{2^2}+\ldots+\frac{1}{2^{n+1}}+\frac{1}{2^{n+2}}\\
<\sum_{n=2}^\infty \frac{1}{2^n}=\frac{1}{2}\end{align}
Hence $\bigg|\frac{a_{n-1}}{x}+\ldots+\frac{a_0}{x^n}+\frac{g(x)}{x^n}\bigg|< \frac{1}{2}$ and so 
$$\frac{1}{2}< 1+\frac{a_{n-1}}{x}+\ldots+\frac{a_0}{x^n}+\frac{g(x)}{x^n}$$
Then for $x\ge N$ we therefore have 
$$0<\frac{x^n}{2}< x^n \bigg(1+\frac{a_{n-1}}{x}+\ldots+\frac{a_0}{x^n}+\frac{g(x)}{x^n}\bigg)=f(x)$$
and for $y\le -N$ we thus have 
$$0>\frac{y^n}{2}>y^n \bigg(1+\frac{a_{n-1}}{y}+\ldots+\frac{a_0}{y^n}+\frac{g(y)}{y^n}\bigg)=f(y)$$
Since $f$ is continuous then is continuous on $[y,x]$ and since $f(y)<0<f(x)$. Hence by the IVT we have that there is a $c\in (y,x)$ such that $f(c)=0$.
