How do you find the domain and range of #f(x) = 6/(x-4)#?

2 Answers
Apr 4, 2017

Answer:

see explanation.

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be.

#"solve " x-4=0rArrx=4#

#rArr"domain is " x inRR,x!=4#

#"Rearrange " f(x)" making x the subject"#

#y=6/(x-4)#

#rArry(x-4)=6#

#rArrxy-4y=6rArrxy=6+4y#

#rArrx=(6+4y)/y#

To obtain the range apply same reasoning as used for domain.

#rArr"range is " y inRR,y!=0#

Apr 4, 2017

Answer:

Domain: #D = {x in RR | x ne 4}#
Range: #R = {y in RR | y ne 0}#

Explanation:

Let us first deal with the domain. The domain of a function are all the values for which the function is defined. In your example, it is all the #x#-values that you decide are valid.

In calculus courses, "unless stated otherwise" you want the domain to be as a large part of the real line (denoted #RR#) as possible. In this case, we see that all real #x#-values are valid, except #x=4#, since this leads to division of #0#. #D = {x in RR | x ne 4}# is just a fancy way of writing this. (In English a literal translation could be "The domain #D# are all the #x# in the real numbers, for which #x# is not #4#").

Let us continue with the range of the function. The range is the set of all the values that #f(x)# can take for any #x# in the domain. In your case, we can find the range by somewhat sloppy reasoning as follows:

For very small #x# (as #x# tends to #-infty#), #f(x)# tends to #0#, because we divide #6# by something very large. As #x# becomes larger and approaching #4# from below, we divide #6# by something which is negative and almost #0#. As a result #f(x)# approaches #-infty#.

Since #f(x)# is continuous everywhere except at #x=4#, we know that #f(x)# assumes all values between #- infty# and #0# for #x in (-infty, 4)#. (This is a consequence of the Intermediate value theorem ).

By similar reasoning, we can consider all #x# between #4# and #+infty# to find that #f(x)# assumes all values between #0# and #+infty#.

In total, we see that #f(x)# can assume all real values except #0#. Letting #f(x) = y#, we can then write the range as #R = {y in RR | y ne 0}#.

The reason why #0# is not in the range, is because there is no #x# such that #f(x) = 0#. Remember that we only get closer to #0# as #x# approaches #4#, but never quite get to #0#.

Trying to plot the function can give you an idea of the range and domain. But be warned, plotting tools sometimes do not show correct results, and rarely give you the whole picture.

graph{6/(x -4) [-35.75, 44.25, -20.12, 19.88]}