Modulus inequalities

In a nutshell

Like any other function, a modulus can be part of an inequality which can be solved.

Solving modulus inequalities

The solution of an inequality is an interval: a range of numbers that satisfy the conditions.

Therefore, given that modulus functions tend to have symmetries, the solution can consist of more than one interval.

Example 1

Given the function $$f(x)=-2\vert x+2\vert +3$$:

a) Find the domain and range.

b) Solve the inequality $$f(x) \gt 1$$.

The easiest way to solve this problem is by sketching the function and the inequality:

The domain is $$\underline{x \in \R}$$​ and the range is $$\underline{(-\infty,3]}$$.

To algebraically find the interval which satisfies the inequality $$f(x)\gt 1$$, you have first to locate the coordinates where the two functions meet.

To solve the equation $$-2\vert x+2\vert +3=1$$, create two new equations: one where the argument inside the modulus is positive and another where it is negative.

The two functions therefore intersect at $$x=-3$$​ and $$x=-1$$​. The sketch of the two functions shows that $$f(x)\gt 1$$ between these two points.

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Therefore, the solution to the inequality $$-2\vert x+2\vert +3\gt 1$$ is $$\underline{-3\lt x \lt -1}$$.