How do you find the domain and range of #(5x+3)/(1-2x)#?

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Answer 1 · 2016-06-22T19:54:12

The domain is #RR-(1/2)# and the range is #RR-(-5/2)#

the domain is obtained by finding the values for x so that:

#1-2x!=0`#

or

#x!=1/2#

Then, in order to find the range, let's calculate the inverse function of #y=(5x+3)/(1-2x)#:

You need some steps:

#y(1-2x)=5x+3#

#y-2xy=5x+3#

#5x+2xy=y-3#

#x(5+2y)=y-3#

#x=(y-3)/(5+2y)#

so the range is obtained by the values y:

#5+2y!=0#

or

#y!=-5/2#