Question 26640
to be honest, looking at the "function"... the existence of the square root means that the fundamental definition of what a function is may well be broken by this. In short, i do not believe this is a function, strictly.


However, putting that aside! What values of x are we allowed to put into the equation? Well avoid anything that makes the square root negative is nothing less than x=-2. Also the denominator should not be zero: when it is, the equation will become infinite..an asymptote.


This happens when 3x-(1/2) = 0
--> 3x = 1/2
--> x = 1/6


So, at a first glance the domain is any real value of x greater or equal to -2 but not 1/6.


Ask your teacher though to clarify if this is really a function! I think it isn't :-)


jon.