Question 1031982
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Re-write so that the inequality can be decomposed into a piece-wise function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |3x\ +\ 1|\ -\ |2x\ +\ 3|\ -\ 2\ <\ 0]


This inequality has two critical points; *[tex \Large x] values that make one of the expressions inside of the absolute value bars equal to zero, namely *[tex \Large -\frac{1}{3}] and *[tex \Large -\frac{3}{2}]


Hence there are three intervals to consider when creating a piece-wise function definition: *[tex \Large \left(-\infty,-\frac{3}{2}\right),\ ]*[tex \Large \left(-\frac{3}{2},\,-\frac{1}{3}\right),\ ]and *[tex \Large \left(-\frac{1}{3}\,\infty\right)]


Since *[tex \Large 3x\ +\ 1\ <\ 0] and *[tex \Large 2x\ +\ 3\ <\ 0] when *[tex \Large x\ < -\frac{3}{2}], we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ -3x\ -\ 1\ -\ (-2x\ -\ 3)\ -\ 2] if *[tex \LARGE x\ < -\frac{3}{2}]


So we need to find the interval that satisfies both *[tex \Large -3x\ -\ 1\ -\ (-2x\ -\ 3)\ -\ 2\ <\ 0] and *[tex \Large x\ < -\frac{3}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -3x\ -\ 1\ -\ (-2x\ -\ 3)\ -\ 2\ <\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ >\ 0].


But the intervals *[tex \Large (0,\,\infty)] and *[tex \Large \left(-\infty,-\frac{3}{2}\right)] are disjoint.  Hence, there is no part of the solution set in the interval *[tex \Large \left(-\infty,-\frac{3}{2}\right)].


Next, for values in *[tex \Large \left(-\frac{3}{2},\,-\frac{1}{3}\right)],


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\  -3x\ -\ 1\ -\ (2x\ +\ 3)\ -\ 2]


So, set the function less than zero and solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -3x\ -\ 1\ -\ (2x\ +\ 3)\ -\ 2\ <\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ >\ -\frac{6}{5}]


Hence, the function is true on the interval *[tex \Large \left(-\frac{6}{5},\,-\frac{1}{3}\right)]


Finally, for values in *[tex \Large \left(-\frac{1}{3},\,-\infty\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ +\ 1\ -\ (2x\ +\ 3)\ -\ 2\ <\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ >\ 4]


So the inequality holds for values in the interval *[tex \Large \left(-\frac{1}{3},\,4\right)]


That leaves the value *[tex \Large -\frac{1}{3}] to check which we can do in the original inequality:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |3\left(-\frac{1}{3}\right)\ +\ 1|\ -\ |2\left(-\frac{1}{3}\right)\ +\ 3|\ -\ 2\ <\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{13}{3} <\ 0]


So now we can specify the union of the two valid intervals which include the *[tex \Large -\frac{1}{3}] endpoint, and 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-\frac{6}{5},\,4\right)]


Is the complete solution set interval.


Compare this result to the portion of the graph of *[tex \Large f(x)] that is below the *[tex \Large x]-axis.


*[illustration Absolute_Value_Inequality_Solution.jpg]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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