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?
 A: Rewriting the function that way only complicates things.
First let's determine the domain:
\begin{cases}
-x^2+4x+12\ge0\\
-x^2+2x+3\ge0
\end{cases}
reduces to $-1\le x\le3$.
The derivative is
$$
y'=
\frac{-x+2}{\sqrt{-x^2+4x+12}}-
\frac{-x+1}{\sqrt{-x^2+2x+3}}
$$
(which is undefined at $-1$ and $3$) and we want to see where it vanishes, that is,
$$
(2-x)\sqrt{-x^2+2x+3}=(1-x)\sqrt{-x^2+4x+12}
$$
We need either $-1<x<1$ or $2<x<3$ so that both terms are either positive or negative. Now we can square safely:
$$
(2-x)^2(-x^2+2x+3)=(1-x)^2(-x^2+4x+12)
$$
becomes $12x^2-16x=0$, so the only critical point is $0$ (because $4/3$ doesn't satisfy the above limitations).
We also have
$$
f(-1)=\sqrt{7},\qquad
f(0)=\sqrt{3},\qquad
f(3)=\sqrt{15}
$$
Since maxima and minima are at critical points or at the extremes of the domain, we can conclude.

A: Domain gives $-1\leq x\leq3$.
If $x=0$ so $y=\sqrt3$.
We'll prove that it's a minimal value of $y$.
Id est, we need to prove that 
$$\sqrt{-x^2+2x+3}+\sqrt3\leq\sqrt{-x^2+4x+12}$$ or after squaring of the both sides we need to prove that
$$\sqrt{3(-x^2+2x+3)}\leq x+3$$ or since $x+3>0$, we need to prove that
$$3(-x^2+2x+3)\leq(x+3)^2,$$
which is $x^2\geq0$.
Done!
A: $$y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$$
$$\implies y=\sqrt{16-4-x^2+4x}-\sqrt{4-1-x^2+2x}$$
$$\implies y=\sqrt{16-(4+x^2-4x)}-\sqrt{4-(1+x^2-2x)}$$
$$\implies y=\sqrt{16-(2-x)^2}-\sqrt{4-(1-x)^2}$$
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.
A: find the first derivative $dy/dx$ and set it equal to zero. Study the variation in case you are looking for a "local" minimum.
