2. The Modulus Function

Video coming soon3 worked examples

Theory

The modulus (or absolute value) of $x$ is its size, ignoring sign. Formally it is defined piecewise:

|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}

The second line often looks wrong at first glance, but $-x$ is positive precisely when $x$ is negative — so both branches return a non-negative value.

Graphically, $y = |f(x)|$ takes the graph of $y = f(x)$ and reflects every part below the $x$ -axis up into the first and second quadrants. Anything already above the axis is unchanged. The result never dips below $y = 0$ , and typically has sharp corners where the original graph crossed the axis.

Be careful to distinguish this from $y = f(|x|)$ , which is a different transformation: it keeps the graph for $x \ge 0$ and reflects that in the $y$ -axis, discarding the original left-hand side.

To solve an equation or inequality containing a modulus, split it into cases at the points where the expression inside changes sign, solve each case separately, and then check each solution back in the original.

The Golden Rule: $|A| = B$ means $A = B$ or $A = -B$ — always two cases. Solve both, then verify each answer in the original equation, because the case analysis can introduce solutions that do not actually work.

⚠️ Common Examiner Traps

Only taking the positive case: forgetting $A = -B$ loses half the solutions.
Not checking back: squaring or splitting cases can produce values that fail the original equation. Substitute every candidate.
Confusing the two graph transformations: $|f(x)|$ reflects the parts below the $x$ -axis upwards; $f(|x|)$ reflects the right-hand side across the $y$ -axis.
Assuming the graph is smooth: $y = |f(x)|$ has a sharp corner wherever $f(x) = 0$ , so it is not differentiable at those points.

Worked examples

Example 1

Solve $|2x - 1| = 7$ .

Step 1: The expression inside the modulus could be either $7$ or $-7$ , so split into two cases:

2x - 1 = 7 \qquad \text{or} \qquad 2x - 1 = -7

Step 2: Solve each separately:

2x = 8 \implies x = 4 \qquad \text{or} \qquad 2x = -6 \implies x = -3

Step 3: Check both in the original equation:

|2(4) - 1| = |7| = 7 \ \checkmark \qquad |2(-3) - 1| = |-7| = 7 \ \checkmark

Both solutions are valid: $x = 4$ and $x = -3$ .

Example 2

Solve the inequality $|x - 3| < 5$ .

Step 1: Read the inequality as a statement about distance: $|x-3|$ is the distance from $x$ to $3$ , so we need every point less than $5$ away.

Step 2: That gives a double inequality — for a “less than” modulus the solution is always a single interval:

-5 < x - 3 < 5

Step 3: Add $3$ throughout:

-2 < x < 8

Check: at $x = 0$ we get $|-3| = 3 < 5$ ✓, while at $x = 9$ we get $|6| = 6 \not< 5$ ✓ — the endpoint behaves as expected.

Note that $|x-3| > 5$ would instead give two separate intervals, $x < -2$ or $x > 8$ .

Example 3

Describe how to sketch $y = |x^2 - 4|$ from the graph of $y = x^2 - 4$ , and state the coordinates of any sharp corners.

Step 1: Start with $y = x^2 - 4$ , a parabola with minimum at $(0, -4)$ , crossing the $x$ -axis where:

x^2 - 4 = 0 \implies x = -2 \ \text{ or } \ x = 2

Step 2: The parabola lies below the $x$ -axis between $x = -2$ and $x = 2$ . That is exactly the section the modulus reflects upwards.

Step 3: Outside that interval the graph is already at or above zero, so it is unchanged.

Step 4: The reflected section turns the minimum at $(0,-4)$ into a maximum at $(0, 4)$ , giving a shape like a hump between the two roots.

Step 5: The sharp corners occur where the original graph met the axis — at $(-2, 0)$ and $(2, 0)$ . The curve is not differentiable at either point, since the gradient jumps sign there.