Algebra.Com's Answer #846422 by ikleyn(52799)![\"\"](\"/images/stars/stars-13.gif\")  
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document.write( "If f(x)= 1/((1/x)-3) , find the domain of f(f(x))
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document.write( "In order for the function f(x) be defined, every occurred denominator should/must be different from zero.\r\n" );
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document.write( "So, we write f(x) =  , and we see that \r\n" );
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document.write( " x must be different from 0: (x =/= 0);\r\n" );
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document.write( " and\r\n" );
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document.write( "  must be different from 0, which implies \r\n" );
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document.write( " =/= 0, 1 =/= 3x, x =/= 1/3.\r\n" );
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document.write( "In this way, we obtain the first necessary condition: \r\n" );
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document.write( " function f(x) is defined over the set of all real numbers, except of x= 0 and/or x= 1/3.\r\n" );
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document.write( " | At this point, half of the problem is solved, |\r\n" );
document.write( " | and the domain of f(x) is determined. |\r\n" );
document.write( " | Now I will work to solve the second half, |\r\n" );
document.write( " | which is finding the domain for f(f(x)). |\r\n" );
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document.write( "Now, in order for f(f(x)) be defined, these two additional conditions must be satisfied: \r\n" );
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document.write( " f(x) =/= 0 and f(x) =/= 1/3.\r\n" );
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document.write( "If f(x) is defined, then f(x) = =  .\r\n" );
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document.write( "Since x= 0 is not in the domain, we see that if f(x) is defined, then it is never equal to zero;\r\n" );
document.write( "so this case is over without giving new restrictions.\r\n" );
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document.write( "If f(x) is defined and f(x) = 1/3, then\r\n" );
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document.write( " =  ,\r\n" );
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document.write( " 3x = 1 - 3x ---> 3x + 3x = 1 ---> 6x = 1 ---> x = 1/6.\r\n" );
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document.write( "Thus, the domain of f(f(x)) is the set of all real numbers except of 0, 1/3 and 1/6.\r\n" );
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document.write( " Now the problem is solved completely.\r\n" );
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document.write( "ANSWER. The domain of f(f(x)) is the set of all real numbers except of 0, 1/3 and 1/6.\r\n" );
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document.write( "Solved completely.\r
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