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" 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.

Please help thank you!

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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 $2<x<3$ 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.

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  • $\begingroup$ Wow thanks a lot!! $\endgroup$ Commented Apr 13, 2020 at 8:36

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