What is the domain and range of #f(x)= (x+7)/(2x-8)#?

2 Answers
Jun 1, 2017

Answer:

Domain: #={x|x!=4}#
Range #={y|y!=0.5}#

Explanation:

Disclaimer : My explanation may be missing some certain aspects due to the fact that I am not a professional mathematician.

You can find both the Domain and Range by graphing the function and seeing when the function is not possible. This may be a trial and error and take some time to do.
You can also try the methods below

Domain
The domain would be all the values of #x# for which the function exists. Hence, we can write for all the values of #x# and when #x!=# a certain number or numbers. The function will not exist when the denominator of the function is 0. Hence we need to find when it does equal 0 and say that the domain is when #x# does not equal the value we find:
#2x-8=0#

#∴2x=8#

#∴x=8/2#

#∴x=4#

When #x=4#, the function is not possible, as it becomes #f(x)=(2+7)/0# which is undefined, hence not possible.

Range
To find the range, you can find the domain of the inverse function, to do this, rearrange the function to get x by itself. That would get quite tricky.

or

We can find the range by finding the value of y for which #x# approaches #oo# (or a very big number). In this case we will get
#y=(1(oo)+7)/(2(oo)-8)#

As #oo# is a very big number the #+7# and the #-8# wont change it much, Hence we can get rid of them. We are left with:
#y=(1(oo))/(2(oo))#
The #oo#'s can cancel out, and we are left with
#y=1/2#
Hence the function is not possible for when #y=1/2#

A short way to do this is to get rid of everything except for the constants for the variables (the numbers in front of the #x#'s)
#∴y=x/(2x)->1/2#

Hope that's helped.

Jun 1, 2017

Answer:

#x inRR,x!=4#
#y inRR,y!=1/2#

Explanation:

#"y = f(x) is defined for all real values of x, except for any"#
#"that make the denominator equal zero"#

#"equating the denominator to zero and solving gives"#
#"the value that x cannot be"#

#"solve " 2x-8=0rArrx=4larrcolor(red)" excluded value"#

#"domain is " x inRR,x!=4#

#"to find any excluded values in the range, rearrange"#
#"f(x) making x the subject"#

#rArry(2x-8)=x+7larrcolor(blue)" cross-multiplying"#

#rArr2xy-8y=x+7#

#rArr2xy-x=7+8y#

#rArrx(2y-1)=7+8y#

#rArrx=(7+8y)/(2y-1)#

#"the denominator cannot equal zero"#

#"solve " 2y-1=0rArry=1/2larrcolor(red)" excluded value"#

#"range is " y inRR,y!=1/2#