# 2009 Indonesia MO Problems/Problem 6

## Problem

Find the lowest possible values from the function $$f(x) = x^{2008} - 2x^{2007} + 3x^{2006} - 4x^{2005} + 5x^{2004} - \cdots - 2006x^3 + 2007x^2 - 2008x + 2009$$ for any real numbers $x$.

## Solution

We start the search for potential minimums by taking the derivative of $f(x)$ and setting it to equal 0. $$f'(x) = 2008x^{2007} - 4014x^{2006} + 6018x^{2005} - \cdots - 6018x^2 + 4014x - 2008 = 0$$ Notice from the symmetry of the function that $f'(1) = 0$. This makes $f(1)$ a possible minimum of the function.

To check, we find that \begin{align*} f(x) &= (x-1)(x^{2007} - x^{2006} + 2x^{2005} - 2x^{2004} + \cdots + 1004x - 1004) + 1005 \\ &= (x-1)^2(x^{2006} + 2x^{2004} + 3x^{2002} + \cdots + 1004) + 1005 \end{align*} From the Trivial Inequality, $(x-1)^2 \ge 0$ and $x^{2006} + 2x^{2004} + 3x^{2002} + \cdots + 1004 > 0$, so $(x-1)^2(x^{2006} + 2x^{2004} + 3x^{2002} + \cdots + 1004) \ge 0$. Thus, the minimum of $f(x)$ is $\boxed{1005}$, obtained when $x = 1$.