<s>[INST] How to find the range of $y=\frac{x+5}{\sqrt{64-25x^2}}$? How to find the range of $y=\frac{x+5}{\sqrt{64-25x^2}}$?
I am sure that the range is $y \ge 0$ but I don't know the basic process of finding the range without using calculus.
Any comments or suggestion sill be much appreciated.
 [/INST] Since for $64-25x^2>0$ we have $x+5>0$, by AM-GM we obtain:$$\frac{x+5}{\sqrt{64-25x^2}}=\frac{2\sqrt{17\cdot33}(x+5)}{2\sqrt{17(8-5x)\cdot33(8+5x)}}\geq$$
$$\geq\frac{2\sqrt{561}(x+5)}{17(8-5x)+33(8+5x)}=\frac{\sqrt{561}}{40}.$$
The equality occurs for $$17(8-5x)=33(8+5x)$$ or $$x=-\frac{64}{125},$$ which says that we got a minimal value.
Also, since $$\lim_{x\rightarrow\frac{8}{5}^-}\frac{x+5}{\sqrt{64-25x^2}}=+\infty$$ and we work with a continuous function on $\left(-\frac{8}{5},\frac{8}{5}\right),$ we got the range:
$$\left[\frac{\sqrt{561}}{40},+\infty\right).$$
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