Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$

Question

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?

Answer 1

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.

Answer 2

$$y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$$ $$\implies y=\sqrt{16-4-x^2+4x}-\sqrt{4-1-x^2+2x}$$ $$\implies y=\sqrt{16-(4+x^2-4x)}-\sqrt{4-(1+x^2-2x)}$$ $$\implies y=\sqrt{16-(2-x)^2}-\sqrt{4-(1-x)^2}$$

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.

Answer 3

find the first derivative $dy/dx$ and set it equal to zero. Study the variation in case you are looking for a "local" minimum.

Answer 4

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!