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]}