Why is function domain of fractions inside radicals not defined for lower values than those found by searching for domain of denominator in fraction? Consider function $y = \sqrt{\frac{1-2x}{2x+3}}$. To find the domain of this function we first find the domain of denominator in fraction:
$2x+3 \neq 0$
$2x \neq -3$
$x \neq - \frac{3}{2}$
So, the domain of $x$ (for fraction to be valid) is $x \in \left(- \infty, - \frac{3}{2}\right) \cup \left(- \frac{3}{2}, + \infty\right)$.
Then we find the domain for whole fraction:
$\frac{1-2x}{2x+3} \ge 0$
$1-2x \ge 0$
$-2x \ge -1$
$x \le \frac{1}{2}$
My textbook says that the (real) domain of the whole $y$ function is $x \in \left(- \frac{3}{2}, \frac{1}{2}\right]$.
I understand why the function is not defined in values larger than $\frac{1}{2}$ (because condition is $x \le \frac{1}{2}$), but I don't understand why it can't be have values less than $- \frac{3}{2}$ (because condition says only $x \neq - \frac{3}{2}$).
I checked the domain of this function and the domain given in the textbook is correct. Function has imaginary values for $x$ values less than $- \frac{3}{2}$ or bigger than $\frac{1}{2}$. It is undefined in $- \frac{3}{2}$.
Real values only in $\left(- \frac{3}{2}, \frac{1}{2}\right]$ domain.
 A: $\frac ab > 0$ means either 1) !!BOTH!! $a > 0; b>0$ OR 2) !!BOTH!! $a < 0; b< 0$.
And as $\frac ab$ existing means $b \ne 0$ then $\frac ab > 0$ means either 1) both $a \ge 0; b < 0$ or both $a \le 0; b < 0$.
So $\frac{1-2x}{2x+3} \ge 0$ means 
1) $1 - 2x \ge 0;2x + 3 > 0$
2) $1-2x \le 0; 2x + 3 < 0$.
Number 1) yeilds:
$x \le \frac 12$ !!AND!! $x > -\frac 32$ or $-\frac 32 < x \le \frac 12$.
Number 2) yeilds the inconsistent
$x \ge \frac 12$ and $ x < -\frac 32$.
So we have 1) and domain of $f$ is $(-\frac 32, \frac 12]$.
Note if we had a consistent 
$g(x) = \sqrt{\frac{2x-1}{2x +3}}$ you would have gotten the domain as $[\frac 12, \infty)$ when it should be
$(-\infty, -\frac 32) \cup [\frac 12, \infty)$.
A: If $x<-\frac{3}{2}$, then $2x+3<0$ and $1-2x>0$, so $\frac{1-2x}{2x+3}<0$, which means you can't take the square root and get a real answer.
A: note that $$\frac{1-2x}{2x+3}\geq 0$$ is hold if $$1-2x\geq 0$$ and $$2x+3>0$$ or
$$1-2x\le 0$$ and $$2x+3<0$$ you have to solve These two cases
A: 
Find the domain of $y=\sqrt{\frac{1-2x}{2x+3}}$

Draw a simple table:
$$
\begin{array}{c|ccccc}
x & \text{under $-3/2$} & \text{$-3/2$} & \text{$]-3/2;1/2[$} & \text{$1/2$} & \text{over $1/2$} \\
\hline
1-2x & + & + & + & 0 & - \\
2x+3 & - & 0 & + & + & + \\
\frac{1-2x}{2x+3} & - & /// & + & 0 & - \\
\sqrt{\frac{1-2x}{2x+3}} & /// & /// & + & 0 & /// \\
\end{array}
$$
In the above table : /// means "not defined" (easy to sketch).

$y$ is defined for any $x$ in $]\frac{-3}{2};\frac{1}{2}]$.

A: Note that $\frac{1-2x}{2x+3} \geq 0$ implies $1-2x \geq 0$ only if $2x+3>0$.
You will get $1-2x\leq 0$ for $2x+3<0$, because what you're doing basically is multiplying $\frac{1-2x}{2x+3} \geq 0$ by a negative number.
A: ($1-2x\geq 0\text{ and }2x+3>0$) or ($1-2x\leq0\text{ and }2x+3<0$). From the first condition we get $x\in (\frac{-3}{2},\frac{1}{2}]$. Second condition isn't possible.
