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:

Maths; Functions; KS5 Year 13; Modulus inequalities

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.

\begin{aligned}-2(x+2)+3&=1\\-2x-4+3&=1\\-2x&=2\\x&=-1\end{aligned}

\begin{aligned}2(x+2)+3&=1\\2x+4+3&=1\\2x&=-6\\x&=-3\end{aligned}

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.

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