Solving inequalities (§5.4)

There is a very general technique for solving inequalities in one variable that applies to expressions built using pretty much all of the functions that we consider in this course. Specifically, it applies to all piecewise-continuous functions. Exactly what that means is generally explained in a Calculus course, but I can already tell you what examples we have of these: any function made of the following operations is piecewise-continuous: This is a long list, but there are potential exceptions here: if you want to solve (−2)x < 1, for example, then it can be done, but not directly by this method; the problem is that the base is not positive and the exponent is not constant.

Here is the method:

  1. Turn the inquality into an equation and solve it.
  2. Besides these solutions, also find when the expressions in the original inequality are undefined.
  3. Finally, if you have a piecewise-defined function in the problem, find all of the endpoints in the intervals of the pieces' conditions.
  4. Using the numbers found in the Steps 1–3, pick one number between each pair of consecutive numbers, as well as one number on either side, as long as the function is defined there.
  5. For each of the numbers found in the Steps 1–4, check whether the inequality is true or false there.
  6. Now you can read off the answer, letting each number found in Step 4 speak for all of the numbers in the open interval from which it was chosen.
This works because the only way for the inequality to shift from true to false is by going through a place where the equation is true or undefined or by switching from one piece to another in piecewise-defined examples. For rational functions, this method is in the textbook, but it still applies to other expressions, such as those inolving roots (of constant index) or logarithms.
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This web page was written in 2015 and 2016 by Toby Bartels, last edited on 2016 May 25. Toby reserves no legal rights to it.

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