# the range of y

#### Albert

##### Well-known member
$a,y \in R$

$y=\sqrt {a^2+a+1} - \sqrt {a^2-a+1}$

please find the range of y

#### MarkFL

Staff member
My solution:

Computing the derivative of $y$ with respect to $a$, we find:

$$\displaystyle \frac{dy}{da}=\frac{(2a+1)\sqrt{a^2-a+1}-(2x-1)\sqrt{a^2+a+1}}{2\sqrt{a^2-a+1}\sqrt{a^2+a+1}}$$

Analysis of the discriminants of the radicands in the denominator reveal no critical values there. Equating the numerator to zero, we find:

$$\displaystyle (2a+1)\sqrt{a^2-a+1}=(2a-1)\sqrt{a^2+a+1}$$

Squaring, we find:

$$\displaystyle \left(4a^2+4a+1 \right)\left(a^2-a+1 \right)=\left(4a^2-4a+1 \right)\left(a^2+a+1 \right)$$

$$\displaystyle 4a^4+a^2+3a+1=4a^4+a^2-3a+1$$

$$\displaystyle a=-a$$

$$\displaystyle a=0$$

A check reveals that this is an extraneous root, thus the original function is monotonic. And since $y'(0)>0$ we know the function is strictly increasing.

We can also see that the function is odd, so its range will be symmetric about $y=0$.

So, looking at:

$$\displaystyle L=\lim_{a\to\infty}y(a)=\lim_{a\to\infty}\frac{2a}{\sqrt{a^2+a+1}+\sqrt{a^2-a+1}}=\lim_{a\to\infty}\frac{2}{\sqrt{1+\frac{1}{a}+\frac{1}{a^2}}+\sqrt{1-\frac{1}{a}+\frac{1}{a^2}}}=1$$

We then conclude that:

$$\displaystyle -1<y<1$$.

#### Albert

##### Well-known member
$a,y \in R$

$y=\sqrt {a^2+a+1} - \sqrt {a^2-a+1}$

please find the range of y
$y=\sqrt{[a+(1/2)]^2+(\sqrt 3/2)^2} -\sqrt{[a-(1/2)]^2+(\sqrt 3/2)^2}$

$*A(\dfrac{-1}{2},\dfrac {\sqrt 3}{2} ) \,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\ \,\,\,\,\,\, \,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\ \,\,\,\,\,\,\,\,*B(\dfrac{1}{2},\dfrac {\sqrt 3}{2} )$

$\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\ \,\,\,\,\,\,*P(a,0)$

we can consider y =$\begin{vmatrix} AP \end{vmatrix} - \begin{vmatrix} BP \end{vmatrix}<\begin{vmatrix} AB \end{vmatrix}=1$
$\therefore -1<y<1$