Question

What is the domain and range of `f(x) = abs((9-x^2)/(x+3))`?

Answer 1

In this case the range is pretty clear. Because of the absolute bars `f(x)` can never be negative

Explanation:

We see from the fraction that `x!=-3` or we divide by zero.
Otherwise:
`9-x^2` can be factored into `(3-x)(3+x)=(3-x)(x+3)` and we get:
`abs(((3-x)cancel(x+3))/cancel(x+3))=abs(3-x)`

This gives no restriction on the domain, except the earlier one:
So:
Domain: `x!=-3`
Range: `f(x)>=0`