Hello!
The range of a function is the set of its possible values.
The given expression for the function `f` is not so simple, let's try to simplify it. But first find the domain. There is a denominator, and it has not to be zero:
`x^2 - x - 6 != 0,` which gives `x != -2` and `x != 3` (easy to guess).
Therefore `x^2 - x - 6 = (x+2)(x-3).` The numerator has something in common:
`(x^2-4)(x-3) = (x-2)(x+2)(x-3).`
Thus the function becomes `f(x) = ((x-2)(x+2)(x-3))/((x+2)(x-3)) = x-2,` a simple expression. But remember that `x != -2` and `x != 3.`
Now we can easily find the range. For `x-2` when all x's are possible, it is the entire `RR.` Because `x=-2` and `x=3` are forbidden, the values `y=-2-2=-4` and `y=3-2=1` are impossible. So the range of `f` is `RR\{-4,1},` which is
`(-oo,-4) uu (-4,1) uu (1,+oo)`
in interval notation.
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