What is the domain and range of `f(x) = (x+9)/(x-3)`?

Domain

The domain of a function is the set of points in which the function is defined. With numeric function, as you probably know, some operations are not allowed - namely division by `0`, logarithms of non-positive numbers and even roots of negative numbers.

In your case, you have no logarithms nor roots, so you only have to worry about the denominator. When imposing `x - 3 \ne 0`, you'll find the solution `x \ne 3`. So, the domain is the set of all real numbers, except `3`, which you can write as `\mathbb{R}\setminus{3}` or in the interval form `(-\infty, 3) \cup (3, \infty)`

Range

The range is an interval whose extrema are the lowest and highest possible values reached by the function. In this case, we already notices that our function has a point of non-definition, which leads to a vertical asymptote. When approaching vertical asymptotes, functions diverge towards `-infty` or `infty`. Let's study what happens around `x=3`: if we consider the left limit we have

`lim_{x \to 3^-}\frac {x+9}{x-3} = \frac{12}{0^-} = -\infty`

In fact, if `x` approaches `3`, but is still less than `3`, `x-3` will be slightly less than zero (think, for example, at `x` assuming values like `2.9, 2.99, 2.999,...`

By the same logic,

`lim_{x \to 3^+}\frac {x+9}{x-3} = \frac{12}{0^+} = \infty`

Since the function approaches both `-infty` and `infty`, the range is `(-\infty, infty)`, which of course is equivalent to the whole real numbers set `\mathbb{R}`.

The denominator of f)x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be.

`"solve "x-3=0rArrx=3larrcolor(red)"excluded value"`

`"domain "x in(-oo,3)uu(3,oo)`

`"let "y=(x+9)/(x-3)`

`"rearrange making x the subject"`

`y(x-3)=x+9`

`xy-3y=x+9`

`xy-x=9+3y`

`x(y-1)=9+3y`

`x=(9+3y)/(y-1)`

`"solve "y-1=0rArry=1larrcolor(red)"excluded value"`

`"range "y in(-oo,1)uu(1,oo)`
graph{(x+9)/(x-3) [-10, 10, -5, 5]}