How do you find the domain and range for #f(x)=(2x+1)/(x-3)#?

Question

17564 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-29T06:39:44

The domain of #=RR-{3}#
The range of #=RR-{2}#

As we cannot divide by #0#, #x!=3#

The domain of #f(x)# is #D_f(x)=RR-{3}#

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

Then,

#yx-3y=2x+1#

#yx-2x=3y+1#

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

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

Therefore,

#f^-1(x)=(3x+1)/(x-2)#

The domain of #x# is the range of #y#

The range of #f(x)# is #R_f(x)=RR-{2}#
graph{(2x+1)/(x-3) [-28.86, 28.9, -14.43, 14.43]}