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.
 A: For real $y,$ we need $$64-25x^2\ge0$$
What if $64-25x^2=0?$
else
Method$\#1:$
$$y^2(64-25x^2)=(x+5)^2$$
$$\iff x^2(1+25y^2)+10x+25-64y^2=0$$
As the discriminant of the quadratic equation in $x,$  must be $\ge0,$(why?)
$$10^2\ge4(25y^2+1)(25-64y^2)$$
$$0\ge-1600y^4+561y^2=y^2(561-1600y^2)$$
Either $y=0$ or $y^2\le561/1600$
Method$\#2:$
WLOG $5x=8\sin2t$ where $-\dfrac\pi2<2t<\dfrac\pi2$
$y=\dfrac{8\sin2t+25}{40\cos2t}$
Use the Weierstrass substitution to form a Quadratic Equation in $\tan t$ which is real
So, the discriminant must be $\ge0$
A: Hint
Let $$f(x)=\frac{x+5}{\sqrt{64-25x^2}}.$$
You have that $$\lim_{x\to \pm \frac{8}{5}^{\mp}}f(x)=+\infty .$$
According to Intermediate value theorem, $$\text{Im}(f)=\left[\min_{(-\frac{8}{5},\frac{8}{5})}f,\infty \right).$$
I nevertheless let you compute the minimum.
A: For the denominator we need
$$|x|<\frac85$$
and
$$\lim_{x\to \frac 85^-} f(x)=\infty$$
$$\lim_{x\to -\frac 85^+} f(x)=\infty$$
then by EVT the function attains a positive minimum in the interval which can be easily found by first derivative
$$f’(x)=\frac{125x+64}{\sqrt{ (64-25x^2)^2}}=0$$
A: 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:
$$\left[\frac{\sqrt{561}}{40},+\infty\right).$$
