Solving modulus equations
In a nutshell
You can solve modulus equations graphically and algebraically.
Solving a modulus equation
By understanding the domain, range, transformation and modulus of a function you can solve modulus equations. Most modulus functions have some sort of symmetry.
Example
Given the equation f(x)=2∣x−1∣−1, sketch a graph of y=f(x). Find its domain and range. Use this to find the points where the graph of y=f(x) intersects the graph of g(x)=x+1.
This function is like ∣x∣, but it has been vertically stretched with scale factor 2, translated one unit to the right and then translated one unit downwards. So it looks like this:
Its domain is x∈R and its range is [−1,+∞).
Solving the equation can be done graphically or algebraically.
The two points of intersection are (0,1) and (4,5). To show this, create two equations: one where the argument of the modulus is positive and one where it is negative.
−2(x−1)−1−2x+2−1−2x+1xf(0)=1=x+1=x+1=x+1=0⟹(0,1) | 2(x−1)−12x−2−12x−3xf(4)=5=x+1=x+1=x+1=4⟹(4,5) |
These are the points of intersection.