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Minimum Value Muse 2 (Posted on 2014-11-20) Difficulty: 3 of 5
The function F(x) is defined for all positive real values of x and is such that:

F(3x) = 3F(x), and:
F(x) = 1 - abs(x-2) for 1 ≤ x ≤ 3.

Determine the minimum value of x for which F(x) = F(2001).

*** abs(x) refers to the absolute value of x.

No Solution Yet Submitted by K Sengupta    
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Solution sollution Comment 2 of 2 |

F(2001)=3*F(2001/3)= (3^2)*F(2001/(3^2))=…. (3^5)* F(2001/(3^5))=729*F(2.744856).

F(2.7444856)=1-.744856)=.255144

F(2001)=729*.255144  =185.999976


rem: I've used more digits after 2.7448590... and

the answer  came out exactly  186.


Edited on November 20, 2014, 10:02 am
  Posted by Ady TZIDON on 2014-11-20 09:58:57

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