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I'm currently studying for an admission test, and have encountered problem with one of the "warm-up" questions in one of the preparation compendiums, unfortunately enough. The problem can be presented as:

Solve the inequality:

$ \frac{1}{x-1} \ge \frac{a}{x+1}: a > 1 $

Then, designate the largest solution to the prevailing inequality;

C.A. $ \frac{a+1}{a-1} $;

****** (R.H.S = Right Hand Side (of equal sign), L.H.S ....) ******

My attempt to solve this problem, is mainly that I multiply the denominator of R.H.S and thereof obtain the following expression of the variable a:

$ \frac{x+1}{x-1} \ge a $

Thereafter, I'm basically stuck understanding-wise. The biggest value of what? I comprehend that the value of a of the original equation can be considered, or is, the value of the R.H.S equation crossing the Y-axis. Also that the new expression on the L.H.S is bigger than 1, since a > 1. Which it is for every x > 1? Taking the limit of the expression, regarding plus minus infinity with L'hôp. (as it is indefinite form), the intervall of L.H.S being bigger than 1 is (1, R) : R goes to infity... Well, I can only say that I'm lost...

Does anyone have a clue for me to build on, am I missing something obvious and/or crucial? Am I in need to go back to the basics of the principals of ineq.? It seems that I'm way off.

I'm very thankful for response //CARL

* MY PROCESS / EDIT *

$ \frac{x+1}{x-1} \ge a $

$ x+1 \ge a(x-1) $

$ 0\ge x(a-1)-a-1 $

$ a+1 \ge x(a-1) $

$ \frac{a+1}{a-1} = xMAX $ (apoligizes for informal designation of xmax, couldn't find the code of indexing in MathJax)

Thus the biggest value corresponds as C.A in the formulated problem.

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  • $\begingroup$ When you multiply both sides by $x-1$, you have to worry about whether you've multiplied by a negative, which would reverse the inequality. Your solution has an error because it says, for instance, that $x=0$ should be a solution. But if you test $x=0$ in the original inequality, it fails. $\endgroup$ – B. Goddard Apr 1 '17 at 13:27
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There's a technique called the "split point method" for solving inequalities. You find all the places where the graph of $LHS - RHS$ could cross the $x$-axis, plot them on a number line and then test each of the intervals you just created. Each interval is part of the solution set or not (because the graph can't cross the $x$-axis inside the interval.) This is a good technique because you can deal with an equation rather than an inequality and not have to worry about cases and whether to flip the direction of the inequality because you might have divided by a negative. And also because it's used often in calculus to find intervals of increase/decrease and concavity of graphs.

The splits points are all solutions to the equation plus any discontinuities. Your inequality has two discontinuity points $x=1$ and $x=-1$. Then solving the equation:

$$x+1 = a(x-1)$$

$$ax-x = a+1$$

$$x=\frac{a+1}{a-1}$$

That gives 3 split points, which cuts the number line into 4 intervals. Note that since $a>1$, we have $\frac{a+1}{a-1}>1$. The four intervals are:

$$(-\infty,-1), \; (-1,1), \; (1,\frac{a+1}{a-1}), \; (\frac{a+1}{a-1},\infty).$$

Now you just have to decide which intervals are in the solution set. You can do this by testing one point from each interval in the original inequality. Try $x=-2$ for the first interval and get $\frac{1}{-3} \geq \frac{a}{-1}$ which will be true since $a>1$. So all values of $x$ in $(\infty, -1)$ satisfy the inequality.

Then try $x=0$ to get $\frac{1}{-1} \geq \frac{a}{1}$. This is false, so the entire interval $(-1,1)$ is not in the solution set.

I'm getting lazy now, so I'm going to point out that this has to work no matter what $a$ is, so I'm going to let $a=2$. Then if I test $x=2$, I find the 3rd interval to be part of the solution, and if I test $x=4$, I find the 4th interval to be not in the solution. Then the right hand endpoint of the 3rd interval is the largest possible solution to the inequality.

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  • $\begingroup$ Yes, thanks! I've noticed in my post, that I had accidentally put the inequality sign instead of the equal sign for the maximum value (which also was poorly entitled). Although, my calculations weren't motivated, and I appreciate that you took time introducing S.P.M and a motivated solution! $\endgroup$ – OSCAR Apr 1 '17 at 14:06
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we solving this by doing case work: 1) assuming $$x>1$$ then we have $$x+1\ge ax-a$$ this is equivalent to $$x(1-a)\geq -(1+a)$$ since we have $$a>1$$ we get $$x\le -\frac{1+a}{1-a}$$ 2) assuming $$-1<x<1$$ then we have $$x\geq -\frac{1+a}{1-a}$$ 3) the case $$x>1$$ and $$x<-1$$ is impossible 4) we assume $$x<-1$$ then we get $$x+1\geq ax-a$$ or $$x\le -\frac{1+a}{1-a}$$

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Rewrite the inequality as $$ \frac{1}{x-1}-\frac{a}{x+1}\ge0 $$ or $$ \frac{(1-a)x+(1+a)}{(x-1)(x+1)}\ge0 $$ or, taking into account $a>1$, $$ \frac{(a-1)x-(1+a)}{(x-1)(x+1)}\le0 $$ The numerator vanishes at $$ x=b=\frac{a+1}{a-1}=\frac{a-1+2}{a-1}=1+\frac{2}{a-1}>1 $$ so you can make the following diagram, where the positivity or negativity of the factor is listed (the open circle means the point cannot be considered, because it makes the expression invalid). $$ \begin{array}{lcccccccccccc} & & -1 && 1 && b \\ x-b\qquad & <0 &\circ& <0 &\circ& <0 &=0& >0 \\ x-1 & <0 &\circ& <0 &\circ& >0 &>0& >0 \\ x+1 & <0 &\circ& >0 &\circ& >0 &>0& >0 \end{array} $$ so you see that the fraction is $\le0$ for $$ x<-1\qquad\text{or}\qquad1<x\le \frac{a+1}{a-1} $$ Alternatively, observe that the expression has the same sign as $$ f(x)=(x+1)(x-1)(x-b) $$ which can flip sign only at $-1$, $1$ and $b$; since $$ f(0)=b>0 $$ we can get the same conclusion as before: start from what you know, that is, $$ f(x)\begin{cases} & \text{for $x<-1$}\\ =0 & \text{for $x=-1$}\\ >0 & \text{for $-1<x<1$}\\ =0 & \text{for $x=1$}\\ & \text{for $1<x<b$}\\ =0 & \text{for $x=b$}\\ & \text{for $x=b$} \end{cases} $$ and flip sign each time you jump over one of the points where $f$ vanishes: $$ f(x)\begin{cases} <0 & \text{for $x<-1$}\\ =0 & \text{for $x=-1$}\\ >0 & \text{for $-1<x<1$}\\ =0 & \text{for $x=1$}\\ <0 & \text{for $1<x<b$}\\ =0 & \text{for $x=b$}\\ >0 & \text{for $x=b$} \end{cases} $$

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