Difference between revisions of "2001 AIME II Problems/Problem 8"

 
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== Problem ==
 
== Problem ==
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A certain function <math>f</math> has the properties that <math>f(3x) = 3f(x)</math> for all positive real values of <math>x</math>, and that <math>f(x) = 1 - \mid x - 2 \mid</math> for <math>1\leq x \leq 3</math>. Find the smallest <math>x</math> for which <math>f(x) = f(2001)</math>.
  
 
== Solution ==
 
== Solution ==
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{{solution}}
  
 
== See also ==
 
== See also ==
* [[2001 AIME II Problems]]
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{{AIME box|year=2001|n=II|num-b=7|num-a=9}}

Revision as of 00:43, 20 November 2007

Problem

A certain function $f$ has the properties that $f(3x) = 3f(x)$ for all positive real values of $x$, and that $f(x) = 1 - \mid x - 2 \mid$ for $1\leq x \leq 3$. Find the smallest $x$ for which $f(x) = f(2001)$.

Solution

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See also

2001 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions