Finding range of rational function with absolute value without use of calculus

Question

How can one find range of a function $$f(x) = \left| \frac{x-2}{x+3}\right|$$ without use of the calculus methods, using purely precalculus algebra?

My best bet was finding inverses of this function in each interval and then finding it's domain. But this yields me a wrong result. How can I solve it?

Answer 1

you have a vertical asymptote at $x=-3$

That tells you that you have no upper bound.
Lower bound, $f(x)$ cannot be less than $0.$ Because of the absolute value.

But $f(2) = 0.$ So we have our lower bound.

Is $f(x)$ continuous in $(-3,2]$?

$\frac {f(x)}{g(x)}$ is continuous over $(a,b)$ if $f(x)$ is continuous and $g(x)$ is continuous, and $g(x) \ne 0$ anywhere in $(a,b)$

And liner functions are continuous.

And the IVT (which requires a theory of real numbers, which is not formally proven until after calculus) says that $f(x)$ must pass though every number in between the upper bound and the lower bound.

Real numbers greater than or equal to 0.

Answer 2

Let $r\geq 0$ then see if the equation $f(x)=r$ has at least a solution $x\not=-3$.

Now we have that $$\left| \frac{x-2}{x+3}\right|=r\Leftrightarrow x-2=\pm r(x+3) \Leftrightarrow x=\frac{\pm 3r+2}{1\mp r}.$$ What may we conclude?