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

Mar 16, 2016

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

Explanation:

We see from the fraction that $x \ne - 3$ or we divide by zero.
Otherwise:
$9 - {x}^{2}$ can be factored into $\left(3 - x\right) \left(3 + x\right) = \left(3 - x\right) \left(x + 3\right)$ and we get:
$\left\mid \frac{\left(3 - x\right) \cancel{x + 3}}{\cancel{x + 3}} \right\mid = \left\mid 3 - x \right\mid$

This gives no restriction on the domain, except the earlier one:
So:
Domain: $x \ne - 3$
Range: $f \left(x\right) \ge 0$