What is the domain and range of #f(x)=(2x-1)/(3-x)#?

2 Answers
Aug 20, 2017

Answer:

#x inRR,x!=3#
#y inRR,y!=-2#

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be.

#"solve "3-x=0rArrx=3larrcolor(red)" excluded value"#

#"domain is "x inRR,x!=3#

To find any excluded values in the range rearrange f(x) making x the subject.

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

#rArry(3-x)=2x-1larrcolor(blue)" cross-multiplying"#

#rArr3y-xy=2x-1#

#rArr-xy-2x=-3y-1larrcolor(blue)" collecting terms in x together"#

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

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

#"the denominator cannot equal zero"#

#"solve "-y-2=0rArry=-2larrcolor(red)" excluded value"#

#rArr"range is "y inRR,y!=-2#

Aug 20, 2017

Answer:

The domain is #x in (-oo,3) uu (3,+oo)#. The range is # y in (-oo,-1) uu(-1,+oo)#

Explanation:

The function is #f(x)=(2x-1)/(3-x)#

The denominator must be #!=0#

So,

#3-x!=0#, #=>#, #x!=3#

The domain is #x in (-oo,3) uu (3,+oo)#

Let,

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

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

#3y-yx=2x-1#

#2x+yx=1+3y#

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

#2+y!=0#

#y!=-1#

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

graph{(y-(2x-1)/(3-x))=0 [-58.53, 58.54, -29.26, 29.24]}