Question 1120785
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Multiply both sides of the inequality by the denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 2\ <\ 3x\ -\ 12]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2x\ <\ -14]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ >\ 7]


Thus *[tex \Large x\ =\ 7] is one critical point.  Other critical points are where the function is not defined, i.e. where the denominator is zero. In this case at *[tex \Large x\ =\ 4]


With two critical points we divide the *[tex \Large x]-axis into three open intervals:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (-\infty,4), (4,7), (7,\infty)]


Choose a value from each interval that is NOT an end point:  I choose 3, 5, and 8


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(3)\ +\ 2}{3\ -\ 4}\ =\ \frac{5}{-1}\ <\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(5)\ +\ 2}{5\ -\ 4}\ =\ \frac{7}{1}\ >\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(8)\ +\ 2}{8\ -\ 4}\ =\ \frac{10}{4}\ <\ 3]


Thus the solution set is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \{x\ \in\ \mathbb{R}\ |\ -\infty\ <\ x\ <\ 4\ \text{or}\ 7\ <\ x\ <\ \infty}]


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