Define a polynomial
![Rendered by QuickLaTeX.com \[ f(x) = \sum_{k=0}^n c_k x^k, \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-2918a1a2d2d093d82d980cf1a78730d2_l3.png)
such that
and
have opposite signs. Prove there is some
such that
.
Proof. Since
![Rendered by QuickLaTeX.com \[ f(0) = \sum_{k=0}^n c_k 0^k = c_0 \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-bcec8a5f46cc7f65950bcc244a426ac6_l3.png)
we see that
has the same sign as
.
Next,
![Rendered by QuickLaTeX.com \[ f(x) = c_n x^n + c_{n-1}x^{n-1} + \cdots + c_1 x + c_0 = c_n x^n \left( 1+ \frac{c_{n-1}}{c_n} \cdot \frac{1}{x} + \cdots + \frac{c_0}{c_n} \cdot \frac{1}{x^n} \right). \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-04a1f375090af2f673dfa587004c84b8_l3.png)
Then we claim that for sufficiently large
,
![Rendered by QuickLaTeX.com \[ 1 + \frac{c_{n-1}}{c_n} \frac{1}{x} + \cdots + \frac{c_0}{c_n} \frac{1}{x^n} > 0; \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a2e662a662646ae7d104038c588cacff_l3.png)
hence,
will have the same sign as
for sufficiently large
, since
and so
![Rendered by QuickLaTeX.com \[ x^n \left( 1 + \frac{c_{n-1}}{c_n} \frac{1}{x} + \cdots + \frac{c_0}{c_n} \frac{1}{x^n} \right) > 0. \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-99164fd5fad4af066e6178e9dc78c761_l3.png)
(So, when we multiply a positive number by
the result will have the same sign as
.)
So, we need to show that the claimed term is indeed positive. First,
![Rendered by QuickLaTeX.com \[ \left( 1 + \frac{c_{n-1}}{c_n} \cdot \frac{1}{x} + \cdots + \frac{c_0}{c_n} \cdot \frac{1}{x^n} \right) \geq 1 - \left( \left| \frac{c_{n-1}}{c_n} \cdot \frac{1}{x} \right| + \cdots + \left| \frac{c_0}{c_n} \cdot \frac{1}{x^n} \right| \right). \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-73d33d689cdaaa8463b7b271efc1a5f8_l3.png)
This is true since for each term in the sum
![Rendered by QuickLaTeX.com \[ \left| \frac{c_i}{c_n} \cdot \frac{1}{x^{n-i}} \right| \geq \frac{c_i}{c_n} \cdot \frac{1}{x^{n-i}}, \qquad \text{for all } 1 \leq i \leq n-1. \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-1ce539a976e0c275424db409608e3555_l3.png)
Now, since we are showing there is sufficiently large
such that our claim is true, we let
![Rendered by QuickLaTeX.com \[ x > \max \left\{ 1, \left| \frac{c_n}{n \cdot c_a} \right| \right\}, \qquad \text{where} \qquad |c_a| = \max\{ |c_0|, |c_1|, \ldots, |c_n|\}. \]](http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5aaa35b5b72060fa5a919ab9dc49ac0d_l3.png)
Since
, we know
for all
, so

This proves our claim, and so
has the same sign as
for sufficiently large
. Hence,
and
have different signs (since
and
had different signs by assumption); thus, there is some
such that