# Finding the range of a function without graphing

" Find the range of $$f(x)=\frac{\sqrt{x+2}}{x^2-9}$$ without graphing "

I can see that the range is R when I graph it, but I can't prove it without graph.

I've tried to show that f(x) is decreasing function in each domain.

We must not take roots of negatives and we must not divide by zero, hence the maximal domain of $$f$$in $$\Bbb R$$ is $$[-2,3)\cup(3,\infty]$$.
In its domain, $$f$$ is obviously continuous (as composition of continuous functions $$+,-,\cdot,/,\sqrt{}$$).
We have $$f(x)\to 0$$ as $$x\to\infty$$ and $$f(x)\to +\infty$$ as $$x\to 3^+$$. Then by the intermediate value theorem, the range of $$f$$ contains $$(0,\infty)$$. We have $$f(x)\le 0$$ for all $$x\in [-2,3)$$ and $$f(-2)=0$$ and $$f(x)\to-\infty$$ as $$x\to 3^-$$. Again by the intermediate value theorem, the range of $$f$$ contains $$(-\infty,0]$$. Together, these two findings show that the range of $$f$$ is $$\Bbb R$$.
If you want, you can avoid limits in the arguments of the preceding paragraph (though using them clarifies much better what is behind the result): For $$x>3$$, we have $$f(x)=\frac{\sqrt{x+2}}{x^2-9}<\frac{\sqrt{x+3}}{x^2-9}<\frac{x+3}{x^2-9}=\frac1{x-3}$$ and $$f(x)=\frac{\sqrt{x+2}}{x^2-9}>\frac2{x^2-9}.$$ Assume $$y>0$$. Let $$a=\frac1y+3>3$$ and $$b=\sqrt{\frac2y+9}>3$$. Then $$f(a)>y>f(b)$$ and by continuity of $$f$$ on $$[a,b]\subset(3,\infty)$$, we can apply the inermediate value theorem to deduce the existence of $$c\in (a,b)$$ with $$f(c)=y$$.
You can find similar nice (i.e., explicitly invertible) bounds for $$2 and uses these to show similarly that all negative $$y$$ are in the range. And of course $$f(-2)=0$$ shos that also $$0$$ is in the range.