How do you find the domain and range of #y = 3(x-2)/x#?

Question

Julia Adams · Accepted Answer

Domain: #(-oo, 0) uu (0, + oo)#
Range: #(-oo, 3) uu (3, + oo)# Right from the start, you can say that the domain of the function will not include #x=0#, since that would make the denominator of the fraction equal to zero. This means that the domain of the function will be #RR - {0}#, or #(-oo, 0) uu (0, + oo)#. To find if the range of the function has any restrictions, calculate the inverse of #y# by solving for #x#, then switching #x# with #y# #y = (3x - 6)/x# #y * x = 3x - 6# #x (y-3) = 6 implies x = 6/(y-3)# Therefore, the inverse function will be #y = 6/(x-3)# As you can see, this is not defined for #x=3#, which means that your original function cannot take the value #y=3#. The range of the function will thus be #RR- {3}#, or #(-oo, 3) uu (3, + oo)#. graph{(3(x-2))/x [-10, 10, 5, 5, 10]}

Nathan Axel · Answer

undefined To find the domain of the function $ y = \frac{3(x-2)}{x} $, we need to identify any values of $ x $ that would make the denominator $ x $ equal to zero. Since division by zero is undefined, the domain is all real numbers except for $ x = 0 $. So, the domain is $ x \neq 0 $.

To find the range, we need to analyze the behavior of the function as $ x $ approaches positive infinity and negative infinity. As $ x $ approaches positive infinity, the term $ (x - 2) $ becomes negligible compared to $ x $, so $ y $ approaches $ 3 $. Similarly, as $ x $ approaches negative infinity, $ y $ also approaches $ 3 $. Therefore, the range of the function is all real numbers except $ y = 3 $. So, the range is $ y \neq 3 $.