How do you find the Domain and Range of #f(x)=(5x-3)/(2x+1)#?

Question

9628 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-24T14:39:07

The domain is #x in (-oo,-1/2)uu(-1/2,+oo)#. The range is #y in (-oo, 5/2)uu(5/2,+oo)#

The denominator must be #!=0#

#2x+1!=0#

#x!=-1/2#

The domain is #x in (-oo,-1/2)uu(-1/2,+oo)#

To find the range, proceed as follows :

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

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

#2yx+y=5x-3#

#2yx-5x=-y-3#

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

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

Here,

#2y-5!=0#

#y!=5/2#

The range is #y in (-oo, 5/2)uu(5/2,+oo)#

graph{(5x-3)/(2x1) [-16.02, 16.02, -8.01, 8.01]}