How do you find the domain and range of # y = (x + 3) / (x -5)#?

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Answer 1 · 2017-06-09T15:16:53

The domain is #x in (-oo,5)uu(5,+oo)#
So, the range is #y in (-oo,1)uu(1,+oo)#

As you cannot divide by #0#, #x!=5#

The domain of #y# is #x in (-oo,5)uu(5,+oo)#

To find the range, we need #y^-1#

#y=(x+3)/(x-5)#

Interchange #y# and #x#

#x=(y+3)/(y-5)#

Express #y# in terms of #x#

#(y-5)x=y+3#

#xy-5x=y+3#

#y(x-1)=5x+3#

#y=(5x+3)/(x-1)#

This is #y^-1#

The domain of #y^-1# is the range of #y#

So, the range is #y in (-oo,1)uu(1,+oo)#