What is the domain and range of # f(x) = 4/(9-x) #?

1 Answer
Feb 3, 2016

Answer:

domain: #{x in RR | x != 9}#

range: #{x in RR | x != 0}#

Explanation:

The domain of a function is the set of possible values you can input into it. In this case, the only value that cannot be entered into #f(x)# is #9#, as that would result in #f(9) - 4/(9-9) = 4/0#. Thus the domain of #f(x)# is #{x in RR|x != 9}#

The range of #f(x)# is the set of all possible outputs of the function. That is, it is the set of all values which can be obtained by inputting something from the domain into #f(x)#. In this case, the range consists of all real numbers besides #0#, as for any nonzero real number #y in RR#, we can input #(9y-4)/y# into #f# and obtain

#f((9y-4)/y) = 4/(9-(9y-4)/y) = (4y)/(9y - 9y + 4) = (4y)/4 = y#

The fact that this works shows that #f^(-1)(y) = (9y-4)/y# is actually the inverse function of #f(x)#. It turns out that the domain of the inverse function is the same as the range of the original function, meaning that the range of #f(x)# is the set of possible values you can input into #f^(-1)(y) = (9y-4)/y#. As the only value that cannot be entered into this is zero, we have the desired range as
#{x in RR|x!=0}#