<s>[INST] Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$ 
Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$

My work so far:
1) $$y =\frac{2x+9}{\sqrt{-x^2+4x+12}+\sqrt{-x^2+2x+3}} $$
2) I used a derivative and found the answer ($y=\sqrt3$ at $x=0$). Is there any other way?
 [/INST] $$y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$$
$$\implies y=\sqrt{16-4-x^2+4x}-\sqrt{4-1-x^2+2x}$$
$$\implies y=\sqrt{16-(4+x^2-4x)}-\sqrt{4-(1+x^2-2x)}$$
$$\implies y=\sqrt{16-(2-x)^2}-\sqrt{4-(1-x)^2}$$
From the first radical, we have that $$-4 \le 2-x\le 4 \implies -2 \le x\le 6$$
And from the second radical, we have that $$-2 \le 1-x\le 2 \implies -1 \le x\le 3$$
Net domain will be $[-1,3]$.
Further $y(-1)=\sqrt7$ , $y(0)=\sqrt{12}-\sqrt3$ , $y(1)=\sqrt{15}-2$, $y(2)=4-\sqrt{3}$ and $y(3)=0$ 
See if this helps.
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