The Modulus Function

Graph and solve equations involving the modulus function at A-Level. Handle |f(x)| and f(|x|).

The modulus (absolute value) function x|x| returns the non-negative value of xx. It creates V-shaped graphs and requires careful handling when solving equations and inequalities.

Core Concepts

Definition

x=x|x| = x if x0x \geq 0, x=x|x| = -x if x<0x < 0.

Graphs

  • y=f(x)y = |f(x)|: reflect the negative parts of f(x)f(x) in the x-axis.
  • y=f(x)y = f(|x|): keep the graph for x0x \geq 0, reflect it in the y-axis.

Solving f(x)=g(x)|f(x)| = g(x)

Method 1: Sketch both graphs, find intersections. Method 2: Solve f(x)=g(x)f(x) = g(x) and f(x)=g(x)f(x) = -g(x), then check validity.

Solving f(x)=g(x)|f(x)| = |g(x)|

Square both sides: [f(x)]2=[g(x)]2[f(x)]^2 = [g(x)]^2.

Worked Example: Example 1

Problem

Solve 2x3=5|2x - 3| = 5.

2x3=52x - 3 = 5x=4x = 4. 2x3=52x - 3 = -5x=1x = -1.

Solution

Worked Example: Example 2

Problem

Solve x+1=2x|x + 1| = 2x.

x+1=2xx + 1 = 2xx=1x = 1 (check: 2=2|2| = 2 ✓). x+1=2xx + 1 = -2xx=13x = -\frac{1}{3} (check: 23=23|\frac{2}{3}| = -\frac{2}{3} ✗ — rejected).

Solution: x=1x = 1 only.

Solution

Worked Example: Example 3

Problem

Solve 3x1<5|3x - 1| < 5.

5<3x1<5-5 < 3x - 1 < 54<3x<6-4 < 3x < 643<x<2-\frac{4}{3} < x < 2.

Solution

Practice Problems

    1. Sketch y=x24y = |x^2 - 4|.
    1. Solve 4x+1=2x3|4x + 1| = |2x - 3|.
    1. Solve x2>3|x - 2| > 3.

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Key Takeaways

  • f(x)|f(x)|: reflect negative parts above x-axis.

  • f(x)f(|x|): reflect right side to create symmetry about y-axis.

  • Solve by considering both cases and checking validity.

  • For f(x)<a|f(x)| < a: a<f(x)<a-a < f(x) < a.

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