Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$ 
Find the range of $\frac{\sqrt{(x-1)(x+3)}}{x+2}$

My Attempt:$$y^2=\frac{x^2+2x-3}{x^2+4x+4}=z\\\implies  x^2(z-1)+x(4z-2)+4z+3=0$$
Discriminant greater than equal to zero, so, $$(4z-2)^2-4(z-1)(4z+3)\ge0\\\implies z\le\frac43\\\implies -\frac2{\sqrt3}\le y\le\frac2{\sqrt3}$$
But the answer given is $[-\frac2{\sqrt3},1]$
What am I missing?
 A: First notice that
$${\rm Dom}(f)=\left]-\infty,-3\right]\cup \left[1,+\infty\right[$$
Now, setting
$$y=f(x)\implies y=\frac{\sqrt{(x-1)(x+3)}}{x+2}\implies y^{2}=\frac{(x-1)(x+3)}{(x+2)^{2}}$$
Then,
$$y^{2}(x+2)^{2}=(x-1)(x+3)$$
Expanding and collecting all in terms of $x$, we have
$$(1-y^{2})x^{2}+(2-4y^{2})x+(-3-4y^{2})=0,\quad (*)$$
Using the quadratic formula we get
$$x=\frac{1-2y^{2}\pm \sqrt{4-3y^{2}}}{y^{2}-1},\quad (**)$$
Then,

*

*If $y=-1$, then $x=-\frac{7}{2}\in {\rm Dom}(f)$ using $(*)$ so not problem.


*If $y\not=\pm 1$ so only we need $4-3y^{2}\geqslant 0$ so $-\frac{2}{\sqrt{3}}\leqslant y\leqslant \frac{2}{\sqrt{3}}$. $[◇]$
Joint all, we have
$${\rm range}(f)=\left[-\frac{2}{\sqrt{3}},1\right[$$
as desired.
Fixed and more details:
Notice with $y=1$, we have $1=\frac{\sqrt{(x-1)(x+3)}}{x+2}$  which don't have solutions over $\mathbb{R}$. Then exclude the $1$ of the interval.
Well, by $(**)$ we get
$$y\in \left[-\frac{2}{\sqrt{3}},-1\right[\cup \left]-1,1\right[\cup \left]-1,\frac{2}{\sqrt{3}}\right[$$
then we have the conclusion give in $[◇]$. However the interval $\left]-1,\frac{2}{\sqrt{3}}\right[$ is a strange solution (just study $f$ using the domain and the conclusion as follows), so we remove it. In my previous post I included the value $1$ in the range, however it was a mistake because as said above. Therefore the range of $f$ over the natural domain is given by
$${\rm range}(f)=\left[-\frac{2}{\sqrt{3}},-1\right[\cup \left]-1,1\right[=\left[-\frac{2}{\sqrt{3}},1\right[.$$
A: What you have shown is that $|y| \leq \dfrac{2}{\sqrt{3}}$.  While that does imply that $-\dfrac{2}{\sqrt{3}} \leq y \leq \dfrac{2}{\sqrt{3}}$, you cannot necessarily conclude that the range is $\left[-\dfrac{2}{\sqrt{3}}, \dfrac{2}{\sqrt{3}}\right]$ since you obtained that result by squaring both sides of the equation, which is not a reversible step.
We are given the function
\begin{align*}
f(x) & = \frac{\sqrt{(x + 3)(x - 1)}}{x + 2}\\
     & = \frac{\sqrt{x^2 + 2x - 3}}{x + 2}
\end{align*}
Observe that its domain of the function $f$ is $(-\infty, -3] \cup [1, \infty)$.
Differentiating gives
\begin{align*}
f'(x) & = \frac{\frac{1}{2\sqrt{x^2 + 2x - 3}}(2x + 2)(x + 2) - \sqrt{x^2 + 2x - 3}}{(x + 2)^2}\\
      & = \frac{\frac{(x + 1)(x + 2)}{\sqrt{x^2 + 2x - 3}} - \sqrt{x^2 + 2x - 3}}{(x + 2)^2}\\
      & = \frac{\frac{x^2 + 3x + 2 - (x^2 + 2x - 3)}{\sqrt{x^2 + 2x - 3}}}{(x + 2)^2}\\
      & = \frac{x + 5}{(x + 2)^2\sqrt{x^2 + 2x - 3}}
\end{align*}
Setting the derivative equal to zero yields
\begin{align*}
f'(x) & = 0\\
\frac{x + 5}{(x + 2)^2\sqrt{x^2 + 2x - 3}} & = 0\\
x + 5 & = 0\\
x & = -5
\end{align*}
Thus, the function has a critical point at $x = -5$.
Observe that the denominator of the derivative function is always positive within the domain of the function. Therefore, the sign of the derivative depends only on the numerator.  Hence, $f'(x) < 0$ if $x < -5$, $f'(-5) = 0$, and $f'(x) > 0$ if $-5 < x < -3$.  Since the sign of the derivative changes from negative to positive at the critical point $x = -5$, the function has a relative minimum at $x = -5$ by the First Derivative Test.
Its relative minimum value is
\begin{align*}
f(-5) & = \frac{\sqrt{(-5)^2 + 2(-5) - 3}}{-5 + 2}\\
      & = \frac{\sqrt{25 - 10 - 3}}{-3}\\
      & = \frac{\sqrt{12}}{-3}\\
      & = \frac{2\sqrt{3}}{-3}\\
      & = -\frac{2}{\sqrt{3}}
\end{align*}
Also, note that $f'(x) > 0$ when $x > 1$.  Therefore, $f(x)$ is strictly increasing on the interval $[1, \infty)$.  Observe that $f(1) = 0$.  Moreover,
\begin{align*}
\lim_{x \to \infty} f(x) & = \lim_{x \to \infty} \frac{\sqrt{x^2 + 2x - 3}}{x + 2}\\
& = \lim_{x \to \infty} \frac{|x|\sqrt{1 + \frac{2}{x} - \frac{3}{x^2}}}{x\left(1 + \frac{2}{x}\right)}\\
& = \lim_{x \to \infty} \frac{x\sqrt{1 + \frac{2}{x} - \frac{3}{x^2}}}{x\left(1 + \frac{2}{x}\right)}\\
& = \lim_{x \to \infty} \frac{\sqrt{1  + \frac{2}{x} - \frac{3}{x^2}}}{1 + \frac{2}{x}}\\
& = 1 
\end{align*}
Since the function $f$ is continuous on the interval $[1, \infty)$, it assumes every value in the interval $[0, 1)$ by the Intermediate Value Theorem.
Observe that $f(-3) = 0$.  Since $f$ assumes the value $-\dfrac{2}{\sqrt{3}}$ at $x = -5$ and is continuous on the interval $[-5, 0]$, it assumes every value in the interval $\left[-\frac{2}{\sqrt{3}}, 0\right]$ in the interval $[-5, 0]$.  Moreover,
\begin{align*}
\lim_{x \to -\infty} f(x) & = \lim_{x \to -\infty} \frac{\sqrt{x^2 + 2x - 3}}{x + 2}\\
& = \lim_{x \to -\infty} \frac{|x|\sqrt{1 + \frac{2}{x} - \frac{3}{x^2}}}{x\left(1 + \frac{2}{x}\right)}\\
& = \lim_{x \to -\infty} \frac{-x\sqrt{1 + \frac{2}{x} - \frac{3}{x^2}}}{x\left(1 + \frac{2}{x}\right)}\\
& = \lim_{x \to -\infty} \frac{-\sqrt{1  + \frac{2}{x} - \frac{3}{x^2}}}{1 + \frac{2}{x}}\\
& = -1 
\end{align*}
Since $f$ is continuous on the interval $(-\infty, -5]$, it assumes every value in the interval $\left[-\dfrac{2}{\sqrt{3}}, -1\right)$ in the interval $(-\infty, -5]$.
Therefore, the range of $f$ is
$$\left[-\frac{2}{\sqrt{3}}, -1\right) \cup \left[-\frac{2}{\sqrt{3}}, 0\right] \cup [0, 1) = \left[-\frac{2}{\sqrt{3}}, 1\right)$$
A: The problem is that squaring an equation may introduce extraneous solutions. You either have to check each solution (by substituting it into the original equation), or you use equivalent equations throughout the solution process which ensures extraneous solutions won't occur. Taking the second approach, your equation $y=\sqrt{(x-1)(x+3)}/(x+2)$ is equivalent to the system
$$\left\{
\begin{array}{l}
(y^2-1)x^2+2(2y^2-1)x+4y^2+3=0\\
y(x+2)\ge 0\\
x+2\ne 0
\end{array}
\right. $$
The range of $f$ is those $y$ for which the system has a solution. For $y^2-1=0\Leftrightarrow y=\pm1$, the equation is $2x+7=0\Leftrightarrow x=-7/2$. With $y=1$ the first inequality isn't satisfied while $y=-1$ is ok. So $y=-1$ is in the range of $f$ and $y=1$ isn't. For $y^2-1\ne 0$, the quadratic formula gives
$$x=\frac{1-2y^2\pm\sqrt{4-3y^2}}{y^2-1},\quad\quad
y\in\left[-\frac{2}{\sqrt 3},\frac{2}{\sqrt 3}\right]\setminus\{-1,1\} $$
and now you have to solve the inequalities $y(x+2)\ge 0,x+2\ne 0$ with each of the two roots for $x$. That is, you have to solve
$$\frac{y\left(-1-\sqrt{4-3y^2}\right)}{y^2-1}\ge0\\-1-\sqrt{4-3y^2}\ne 0 $$
and
$$\frac{y\left(-1+\sqrt{4-3y^2}\right)}{y^2-1}\ge0\\-1+\sqrt{4-3y^2}\ne 0 $$
The solutions to these are, respectively, $y\in\left[-\frac{2}{\sqrt{3}},-1\right)\cup\left[0,1\right)$ and $y\in\left[-\frac{2}{\sqrt{3}},-1\right)\cup\left(-1,0\right]$. Combining the two sets and also adding $y=-1$ (but not $y=1$), we conclude that the range of $f$ is
$$\left[-\frac{2}{\sqrt{3}},1\right)$$
Edit: For the second pair of inequalities, you can use
$$-1+\sqrt{4-3y^2}
=\frac{\left(-1+\sqrt{4-3y^2}\right)\left(1+\sqrt{4-3y^2}\right)}{1+\sqrt{4-3y^2}}
=\frac{-3(y^2-1)}{1+\sqrt{4-3y^2}} $$
A: HINT
A little bit different approach from the other users.
To begin with, notice that
\begin{align*}
\frac{\sqrt{(x - 1)(x + 3)}}{x + 2} & = \frac{\sqrt{x^{2} + 2x - 3}}{x + 2}\\\\
& = \frac{\sqrt{(x^{2} + 4x + 4) - (2x + 4) - 3}}{x + 2}\\\\
& = \frac{\sqrt{(x + 2)^{2} - 2(x + 2) - 3}}{x + 2}\\\\
& = \frac{\sqrt{[(x + 2) - 1]^{2} - 4}}{x + 2}\\\\
& = \frac{\sqrt{(x + 1)^{2} - 4}}{(x + 1) + 1}
\end{align*}
Now we can make the change of variable $u = x + 1$ and study the range of the transformed expression:
\begin{align*}
y = \frac{\sqrt{u^{2} - 4}}{u + 1} \Longleftrightarrow
\begin{cases}
y(u + 1) = \sqrt{u^{2} - 4}\\\\
|u| \geq 2
\end{cases} \Longleftrightarrow
\begin{cases}
(1 - y^{2})u^{2} - 2y^{2}u - y^{2} - 4 = 0\\\\
|u| \geq 2
\end{cases}
\end{align*}
Can you take it from here?
