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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.

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  • $\begingroup$ It is formally wrong to deduce that $1-2x\geq0$ from $\frac{1-2x}{2x+3}\geq0$. In order for the fraction to be nonnegative, you need numerator and denominator to have the same sign: either $1-2x\geq0$ and $2x+3>0$, OR $1-2x<0$ and $2x+3<0$. The first case gives $x\in(-\frac{3}{2},\frac{1}{2}]$. The second case gives no solutions. $\endgroup$ – symplectomorphic Dec 18 '17 at 18:20
  • $\begingroup$ I have no idea, they teach as this way in school. $\endgroup$ – Hanlon Dec 18 '17 at 18:23
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    $\begingroup$ No they don't. You've misinterpreted something you think you've learned. From $\frac{a}{b}>0$ it does not follow that $a>0$. You cannot multiply both sides by $b$, because you don't know the sign of $b$. $\endgroup$ – symplectomorphic Dec 18 '17 at 18:24
  • $\begingroup$ Maybe I did. Actually, I probably did since I usually can't remember how we do math in class and I later have to figure out things on my own. $\endgroup$ – Hanlon Dec 18 '17 at 18:28
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$\frac ab > 0$ 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

$(-\infty, -\frac 32) \cup [\frac 12, \infty)$.

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    $\begingroup$ Shouldn't the second line have $\frac{a}{b} \ge 0$ instead of $\frac{a}{b} \gt 0$? And $a \ge 0;b \gt 0$ instead of $a \ge 0; b \lt 0$? $\endgroup$ – Hanlon Dec 18 '17 at 19:10
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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.

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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

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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 $]\frac{-3}{2};\frac{1}{2}]$.

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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.

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($1-2x\geq 0\text{ and }2x+3>0$) 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.

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