What Values of x Satisfy the Inequality \(\frac{x-1}{x+1} > 0\)?

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SUMMARY

The inequality \(\frac{x-1}{x+1} > 0\) is satisfied for values of \(x\) in the intervals \((- \infty, -1)\) and \((1, \infty)\). This conclusion is derived from analyzing the conditions under which both the numerator and denominator are either positive or negative. Specifically, \(x\) must be greater than 1 or less than -1, excluding \(x = -1\) where the expression is undefined. The solution can be expressed in set builder notation as \(\{x : \frac{x-1}{x+1} > 0\} = (-\infty, -1) \cup (1, \infty)\).

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Homework Statement



Find all x for which \frac{x-1}{x+1}>0 \qquad(1)

Homework Equations



(2) AB > 0 if A,B >0 OR A,B < 0

(3) 1/Z > 0 => Z > 0

The Attempt at a Solution



Since (1) holds if:

(x-1) &gt; 0 \text{ and } (x+1) &gt; 0 \qquad x\ne -1

then we must have x>1 AND x>-1

and since (1) also will hold if:

(x-1) &lt; 0 \text{ and } (x+1) &lt; 0 \qquad x\ne -1

then we must have x<1 AND x<-1

So that the solution is x on the interval (-\infty,-1) \cup (1,\infty).

What is the proper way to write the solution using set builder notation?
 
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You could write, for example:

\left\{x : \frac{x-1}{x+1} &gt; 0\right\} = (-\infty, -1) \cup (1, \infty)

Or:

\frac{x-1}{x+1} &gt; 0 \iff x \in (-\infty, -1) \cup (1, \infty)
 

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