Question

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

Answer 1

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.

Answer 2

`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`