Question

Question #a3ceb

Answer 1

Given `y=(4x)/(x-3)`

Add 0 to the numerator in the form of `-12+12`:

`y=(4x-12+12)/(x-3)`

Separate into two fractions:

`y=(4x-12)/(x-3)+12/(x-3)`

Factor the first numerator:

`y=(4(x-3))/(x-3)+12/(x-3)`

The first fraction becomes 4:

`y=4+12/(x-3)`

Subtract 4 from each side:

`y-4=12/(x-3)`

Multiply both sides by `(x-3)/(y-4)`:

`x-3=12/(y-4)`

Add 3 to both sides:

`x=12/(y-4)+ 3`

`x=f(y) = 12/(y-4)+ 3`

The term `12/(y-4)` can never become 0, therefore, range of `f(y)` is:

`f(y) !=3`