What is the domain and range of # y =1/(2x-4)#?

2 Answers
Narad T.
May 2, 2017

The domain of #y# is #=RR-{2}#
The range of #y#, #=RR-{0}#

Explanation:

As you cannot divide by #0#,

#2x-4!=0#

#x!=2#

Therefore, the domain of #y# is #D_y=RR-{2}#

To determine the range, we calculate #y^-1#

#y=1/(2x-4)#

#(2x-4)=1/y#

#2x=1/y+4=(1+4y)/y#

#x=(1+4y)/(2y)#

So,

#y^-1=(1+4x)/(2x)#

The domain of #y^-1# is #D_(y^-1)=RR-{0}#

This is the range of #y#, #R_y=RR-{0}#
graph{1/(2x-4) [-11.25, 11.25, -5.625, 5.625]}

Jim G.
May 2, 2017

#"domain " x inRR,x!=2#

#"range " y inRR,y!=0#

Explanation:

The denominator of y cannot be zero as this would make y #color(blue)"undefined".#Equating the denominator to zero and solving gives the value that x cannot be.

#"solve " 2x-4=0rArrx=2larrcolor(red)" excluded value"#

#"domain " x inRR,x!=2#

#"to find excluded value/s in the range"#

#"Rearrange the function making x the subject"#

#rArry(2x-4)=1#

#rArr2xy-4y=1#

#rArr2xy=1+4y#

#rArrx=(1+4y)/(2y)#

#"the denominator cannot be zero"#

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

#"range " y inRR,y!=0#
graph{1/(2x-4) [-10, 10, -5, 5]}

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