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?


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

  • $\begingroup$ 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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