document.write( "Question 1197013: If f(x) = 9^x/(3 + 9^x), prove that:
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Algebra.Com's Answer #830080 by ikleyn(52803)\"\" \"About 
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\n" ); document.write( "If f(x) = 9^x/(3 + 9^x), prove that:
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document.write( "As first step, let's prove that f(x) + f(1-x) = 1  for any value of x.\r\n" );
document.write( "We have \r\n" );
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document.write( "    f(1-x) = by the definition of function f(x) = \"9%5E%281-x%29%2F%283%2B9%5E%281-x%29%29\" = \r\n" );
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document.write( "           = \"9%2F%289%5Ex%2A%283%2B9%2F9%5Ex%29%29%29\" = \"%289%2A9%5Ex%29%2F%289%5Ex%2A%283%2A9%5Ex%2B9%29%29\" = \"9%2F%283%2A9%5Ex%2B9%29\" = \"3%2F%289%5Ex%2B3%29\" = \"3%2F%283%2B9%5Ex%29\".\r\n" );
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document.write( "    THEREFORE,  f(x) + f(1-x) = \"9%5Ex%2F%283+%2B+9%5Ex%29\" + \"3%2F%283%2B9%5Ex%29\" = \"%289%5Ex%2B3%29%2F%283%2B9%5Ex%29\" = 1,\r\n" );
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document.write( "    and the statement is proved.\r\n" );
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document.write( "As the next step, let's write two identical sums in direct and inverse order\r\n" );
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document.write( "    f(1/2016)     + f(2/2016)    + f(3/2016)    + . . . + f(2015/2016)\r\n" );
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document.write( "    f(20125/2016) + f(2014/2016) + f(2013/2016) + . . . + f(1/2016)\r\n" );
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document.write( "and add them. Pairing the addends vertically, we have 2015 pairs of the form  \"f%28i%2F2016%29\" + \"f%281-i%2F2016%29\", \r\n" );
document.write( "and the sum in each such a pair equals 1.\r\n" );
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document.write( "So, the doubled sum equals 2015 and the sum itself is  \"2015%2F2\",  exactly as the problem states.\r\n" );
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