How do you find the domain and range of #Y = g(x) = (x-3)/(x+1)#?

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Answer 1 · 2018-04-02T11:54:16

The domain is #x in RR-{-1}#. The range is #y in RR-{1}#

The denominator must #!=0#

Therefore,

#x+1!=0#

#x!=-1#

The domain of #g(x)# is #x in RR-{-1}#

To find the range, proceed as follows

#y=(x-3)/(x+1)#

#yx+y=x-3#

#yx-x=-3-y#

#x(y-1)=-(3+y)#

#x=-(3+y)/(y-1)#

The denominator is #!=0#

#y-1!=0#

#y!=1#

The range of #g(x)# is #y in RR-{1}#

graph{(x-3)/(x+1) [-22.8, 22.83, -11.4, 11.4]}