Question 1208183
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If f(x)= 1/((1/x)-3) , find the domain of f(f(x))
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<pre>
In order for the function f(x) be defined, every occurred denominator should/must be different from zero.


So, we write  f(x) = {{{1/((1/x)-3)}}},  and we see that 


         x  must be different from 0:  (x =/= 0);

    and

         {{{1/x-3}}}  must be different from 0,  which implies  

                      {{{1/x-3}}} =/= 0,  1 =/= 3x,  x =/= 1/3.


In this way, we obtain the  first necessary condition: 

    function f(x) is defined over the set of all real numbers, except of x= 0  and/or  x= 1/3.


    +--------------------------------------------------+
    |   At this point, half of the problem is solved,  |
    |      and the domain of f(x) is determined.       |
    |    Now I will work to solve the second half,     |
    |     which is finding the domain for f(f(x)).     |
    +--------------------------------------------------+


Now, in order for f(f(x)) be defined, these two additional conditions must be satisfied: 

    f(x) =/= 0  and f(x) =/= 1/3.


If f(x) is defined,  then  f(x) = {{{1/((1/x)-3)}}} = {{{x/(1-3x)}}}.


Since x= 0 is not in the domain, we see that if f(x) is defined, then it is never equal to zero;
so this case is over without giving new restrictions.


If f(x) is defined and  f(x) = 1/3,  then

    {{{x/(1-3x)}}} = {{{1/3}}},

    3x = 1 - 3x  --->  3x + 3x = 1  --->  6x = 1  --->  x = 1/6.


Thus, the domain of f(f(x)) is the set of all real numbers except of 0, 1/3 and 1/6.


                    Now the problem is solved completely.


<U>ANSWER</U>.  The domain of f(f(x)) is the set of all real numbers except of 0, 1/3 and 1/6.
</pre>

Solved completely.