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Given $-3<x<0$ find the values of $a,b$ defining the minimal interval for $$a<\dfrac{2x-1}{1-x}<b.$$

I'm having some difficulties on solving for intervals like that. I will present below 2 approaches. Wanna know why the second one doesn't work.

Approach 1 (correct): First, notice that $$\dfrac{2x-1}{1-x}=-2+\dfrac{1}{1-x}.$$ Therefore $$-3<x<0\Leftrightarrow 0<-x<3\Leftrightarrow 1<1-x<4\Leftrightarrow \frac{1}{4}<\frac{1}{1-x}<1$$ Now adding $-2$ to all terms of last inequality, leads to $$-\frac{7}{4}<-2+\frac{1}{1-x}<-1 \Leftrightarrow -\frac{7}{4}<\dfrac{2x-1}{1-x}<-1.$$ Therefore $(a,b)=(-7/4,-1)$. This result appears correct.

Approach 2 (incorrect): Starting from $-3<x<0$ we can easily get to the intervals $$-7<2x-1<-1~~\text{and}~~\frac{1}{4}<\frac{1}{1-x}<1$$ Now is where my problem begins. If I'm trying to use a rule I found in a book on inequalities stating that for $a<x<b$ and $c<y<d$ it is possible to get the interval for $xy$ as $m<xy<M$, where $m =\min[ac,ad,bc,bd]$ and $M =\max[ac,ad,bc,bd]$, without any constraint on the signs for $a,b,c,d$.

If I use this rule in this case, I will get to $$-7<\dfrac{2x-1}{1-x}<-\frac{1}{4}.$$ a result that appears wrong (the result from Approach 1 is correct).

Question: What am I missing in these last steps in Approach 2? Is the rule in the book correct or there are constraints on its application that should be considered?

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Your second answer isn't wrong. The interval $(-7, -\frac14)$ in Approach $\color{blue}2$ is just 'wider' than the one obtained in Approach $\color{red}1$:$$\color{blue}{-7}<-\color{red}{\frac74}<\frac{2x-1}{1-x}<\color{red}{-1}<\color{blue}{-\frac14}$$

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  • $\begingroup$ It is reassuring that at least Approach 2 is not wrong but just leads to a larger interval. The rule used is indeed correct? Then in general what approach should I use in problems like that, which will ensure me the shortest interval. If this question appears in an exam or math context, for instance? $\endgroup$ – bluemaster Dec 14 '17 at 19:43
  • $\begingroup$ One issue that I've noticed in this last solution, is that if I try to solve it for the interval for $x$ (using WolframAlpha), with the solution in Approach 1 I get the original interval $-3<x<0$ (wolframalpha.com/input/?i=solve+-7%2F4%3C(2x-1)%2F(1-x)%3C-1), the second solution leads to $x<3/7$ or $x>6/5$ (wolframalpha.com/input/?i=solve+-7%3C(2x-1)%2F(1-x)%3C-1%2F4), so there might be a problem indeed. because the original intervals for $x$ are not consistent. $\endgroup$ – bluemaster Dec 14 '17 at 19:51
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    $\begingroup$ The second interval contains the first one. This is consistent with the results from WolframAlpha since $(-3, 0)$ is contained within $x<\frac37$. $\endgroup$ – TheSimpliFire Dec 14 '17 at 19:55
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    $\begingroup$ Suppose that given $a_1<x<a_2$ we find $m,n$ such that $$m<\frac{\alpha x+\beta}{\gamma x+\delta}<n$$Then Approach $1$ gives the interval$$\left(\frac{\alpha a_2+\beta}{\gamma a_2+\delta},\frac{\alpha a_1+\beta}{\gamma a_1+\delta}\right)$$Note $a_1$ and $a_2$ may swap places depending on whether they are positive or negative. $\\$ Let $A=\{(\alpha a_1+\beta)(\gamma a_1+\delta),(\alpha a_1+\beta)(\gamma a_2+\delta),(\alpha a_2+\beta)(\gamma a_1+\delta),(\alpha a_2+\beta)(\gamma a_2+\delta)\}$. Approach $2$ gives the interval $$(\min A, \max A)$$ So I think it depends for different values. $\endgroup$ – TheSimpliFire Dec 14 '17 at 20:06
  • $\begingroup$ Great. I understand better now what the issues are. Then a possible improvement for approach 2 is not using the min, max rule and but testing for the interval pair which has minimal lenght. $\endgroup$ – bluemaster Dec 14 '17 at 20:18

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