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?
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 $-1&lt;x&lt;1$ or $2&lt;x&lt;3$ 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.
find the first derivative $dy/dx$ and set it equal to zero. Study the variation in case you are looking for a "local" minimum.