Question 553110
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{x^2\ -\ 1}\ \geq\ -3]


Multiply both sides by *[tex \Large x^2\ -\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ \geq\ -3\left(x^2\ -\ 1\right)]


Multiply both sides by *[tex \Large -\frac{1}{3}].  Don't forget to reverse the sense of the inequality because of multiplying by a number less than zero.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 1\ \geq\ -\frac{2}{3}]


Add 1 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ \geq\ \frac{1}{3}]


Now when you take the root of bothsides, recognize that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |x|\ \geq\ \frac{\sqrt{3}}{3}]


Which is to say that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ \geq\ \frac{sqrt{3}}{3}]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ \leq\ -\frac{sqrt{3}}{3}]


Draw a number line.  Put a filled in dot at *[tex \Large \frac{sqrt{3}}{3}].  Make a fat arrow going to the right with an arrowhead indicating it goes on forever.  Put a filled in dot at *[tex \Large -\frac{sqrt{3}}{3}].  Make a fat arrow going to the left with an arrowhead indicating it goes on forever.  


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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