How do you find the domain and range of #p(x) = 1/(x+5) -9#?

1 Answer
Dec 22, 2016

Answer:

#"Domain: " {x|x!="-5"}#;
#"Range: " {y|y!="-9"}#.

Explanation:

Domain:

The domain of a function is simply the set of all possible #x#-values for which the function produces an output. For any number #x#, as long as #p(x)# is also a number, that #x# is in the domain of #p(x)#.

Conversely, any number #x# which #p(x)# does not know what to do with is not in the domain of #p(x)#. This alternate approach is often easier.

In this example, #p(x)=1/(x+5)-9#, and we can immediately see where a potential problem is. Since #x# appears in a denominator, #x# cannot take on any value that makes the denominator 0; that would create division by 0, which is undefined. So we simply need to keep #x# from making #x+5=0#. (All other #x# values are okay.)

The only #x# that would make that true is #x="-5"# (which we get by solving #x+5=0# for #x#). That means, our domain is all numbers other than -5, or:

#"Domain: " {x|x!="-5"}#
("all #x# such that #x# is not #"-5"#")

Range:

Similar to domain, the range of a function is all the possible output values that function can take on. If #x# is in the domain, and #p(x)=y#, then that #y# is in the range of #p#.

And, conversely, if there is no #x# such that #p(x)=y#, then that #y# is not in the range of #p#. So, can we find some values that are impossible for the function to be?

Yes we can, since it is not possible for #1/(x+5)# to be 0. No matter how large (positive or negative) #x# is, we cannot divide 1 by a real number and get 0.

(Side note: we can, however, make #1/(x+5)# as close to #+-oo# as we like, by taking #x# close to #"-5"#. So #1/(x+5)# can be anything else, just not 0.)

Since #1/(x+5)# cannot be 0, we know that #1/(x+5)-9# cannot be -9. Thus, our range is all #y#-values except -9:

#"Range: " {y|y!="-9"}#
("all #y# such that #y# is not #"-9"#")

graph{1/(x+5)-9 [-22.47, 6, -12.71, 1.53]}