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for example ...

$\dfrac{x^2-1}{x+2} \ge 0$

critical values are $x \in \{-2,-1,1\}$

the three critical values partition the set of x-values into four regions ...

$-\infty < x < -2$,

$-2 < x < -1$,

$-1 < x < 1$,

and $x > 1$

The "equals to" part of the original inequality occurs at $x = \pm 1$

The "greater than" occurs over the intervals $-2 < x < -1$ and $x > 1$

So, the solution set is all $x$ such that $-2 < x \le -1$ or $x \ge 1$

- Jan 30, 2018

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