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

2 Answers
May 29, 2017

The domain is #RR-{1/4}#
The range is #RR-{-1/4}#

Explanation:

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

As you cannot divide by #0#, #=>#, #1-4x!=0#

So,

#x!=1/4#

The domain is #RR-{1/4}#

To find the range, we calculate the inverse function #y^-1#

We interchange #x# and #y#

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

We express #y# in terms of #x#

#x(1-4y)=4+y#

#x-4xy=4+y#

#y+4xy=x-4#

#y(1+4x)=x-4#

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

The inverse is #y^-1=(x-4)/(1+4x)#

The range of #y# is #=# to the domain of #y^-1#

#1+4x!=0#

The range is #RR-{-1/4}#

May 29, 2017

#x inRR,x!=1/4#
#y inRR,y!=-1/4#

Explanation:

#"the domain is defined for all real values of x, except"#
#"those values which make the denominator zero"#

#"to find excluded values equate the denominator to zero"#
#"and solve for x"#

#"solve " 1-4x=0rArrx=1/4larrcolor(red)"excluded value"#

#rArr"domain is " x inRR,x!=1/4#

#"to find any excluded values in the range, change the subject"#
#"of the function to x"#

#y(1-4x)=4+x#

#rArry-4xy=4+x#

#rArr-4xy-x=4-y#

#rArrx(-4y-1)=4-y#

#rArrx=(4-y)/(-4y-1)#

#"the denominator cannot equal zero"#

#rArr-4y-1=0rArry=-1/4larrcolor(red)" excluded value"#

#rArr"range is " y inRR,y!=-1/4#