IP AMaths Notes (Upper Sec, Year 3-4): 11) Modulus Functions

Modulus expressions measure distance from zero. Convert them into piecewise linear forms to analyse turning points and solve inequalities.

Fundamental identities

• \( |x| = \begin{cases} x & x \geq 0 \ -x & x < 0 \end{cases} \).
• \( |ax + b| = |a| \cdot \bigl|x + \tfrac{b}{a}\bigr| \) for \( a \neq 0 \).
• Solve \( |f(x)| = c \) by considering \( f(x) = c \) and \( f(x) = -c \).

Worked example 1 — Piecewise sketch

Sketch \( y = |2x - 3| - 1 \).

1. Set \( 2x - 3 = 0 \) → \( x = \tfrac{3}{2} \).
2. For \( x \geq \tfrac{3}{2} \), \( y = 2x - 3 - 1 = 2x - 4 \).
3. For \( x < \tfrac{3}{2} \), \( y = -(2x - 3) - 1 = -2x + 2 \).
4. Plot both lines; the vertex occurs at \( \bigl(\tfrac{3}{2}, -1\bigr) \).

Worked example 2 — Modulus inequality

Solve \( |x - 4| \leq 3 \).

1. Write double inequality: \( -3 \leq x - 4 \leq 3 \).
2. Add \( 4 \) throughout: \( 1 \leq x \leq 7 \).

Now solve \( |2x + 5| > 9 \).

1. Either \( 2x + 5 > 9 \) or \( 2x + 5 < -9 \).
2. First case: \( 2x > 4 \) → \( x > 2 \).
3. Second case: \( 2x < -14 \) → \( x < -7 \).

Solution: \( x < -7 \) or \( x > 2 \).

Try this

Rewrite \( y = |x + 1| + |x - 3| \) as piecewise linear segments and determine the minimum value of \( y \).