Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Find the answer

Example + Given that $f(x)$ is an even function defined on $\mathbf{R}$ , if $g(x)$ is an odd function, and $g(x)=f(x-1)$ , $g(1)=2003$ , find the value of $f(2004)$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

$g(x)$ is an odd function, so, $g(-x)=-g(x)$ , and $g(x)=f(x-1)$ , therefore, $g(-x)=$ $-f(-x-1)$ . Since $f(x)$ is an even function, then $f(x-1)=-f(x+1)$
Thus,
$f(x-1)=-f(x+1)=-[-f(x+1+2)]=f(x+3)$
$f(x)$ is a periodic function with a period of 4,
$f(0)=g(1)=2003, f(2004)=f(0)=2003$

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