Absolute Value Calculator — |x| Properties, Equations & Distance
Calculate absolute values and solve equations with |x|. Covers the definition, distance interpretation, properties, solving |ax+b| = c type equations, and graphing basics.
The absolute value of a number is its distance from zero — always non-negative, regardless of sign. It's a deceptively simple concept that becomes the foundation of error analysis, vector magnitudes, and the definition of distance itself. Calculate and solve absolute value expressions with CalcHub.
Definition
$$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$$
Examples:- |7| = 7
- |−5| = 5
- |0| = 0
- |−3.14| = 3.14
- |3 − 8| = |−5| = 5
Geometric Interpretation: Distance on a Number Line
|a − b| = the distance between a and b on the number line.
|7 − 3| = 4 → the numbers 7 and 3 are 4 units apart ✓
|−4 − 2| = |−6| = 6 → the numbers −4 and 2 are 6 units apart ✓
This is why the distance formula is built on squared differences (which are always positive): (x₂−x₁)² = |x₂−x₁|².
Properties of Absolute Value
| Property | Formula | Example | |||||||
|---|---|---|---|---|---|---|---|---|---|
| Non-negativity | \ | x\ | ≥ 0 | \ | −5\ | = 5 ≥ 0 | |||
| Identity | \ | x\ | = 0 ↔ x = 0 | Only zero has | x | = 0 | |||
| Multiplication | \ | ab\ | = \ | a\ | ·\ | b\ | \ | 3·(−4)\ | = 12 |
| Division | \ | a/b\ | = \ | a\ | /\ | b\ | \ | −8/2\ | = 4 |
| Triangle inequality | \ | a+b\ | ≤ \ | a\ | +\ | b\ | \ | 3+(−5)\ | = 2 ≤ 8 |
| Even function | \ | −x\ | = \ | x\ | \ | −7\ | = \ | 7\ | = 7 |
| Power | \ | x²\ | = x² | Always true (x² ≥ 0) |
Solving Absolute Value Equations
Type 1: |x| = k (where k ≥ 0)
Split into two cases: x = k or x = −k Solve |x − 3| = 5: x − 3 = 5 → x = 8 x − 3 = −5 → x = −2 Solutions: x = 8 and x = −2Check: |8−3| = |5| = 5 ✓ and |−2−3| = |−5| = 5 ✓
Type 2: |2x + 1| = 7
2x + 1 = 7 → x = 3 2x + 1 = −7 → x = −4 Solutions: x = 3 and x = −4Type 3: |x| = negative number
No solution. Absolute value is always ≥ 0, so |x| = −3 has no solution.Type 4: |x − 2| = |x + 4|
Two cases: Case 1: x − 2 = x + 4 → −2 = 4 (impossible) Case 2: x − 2 = −(x + 4) → x − 2 = −x − 4 → 2x = −2 → x = −1 Solution: x = −1Absolute Value Inequalities
| Inequality | Equivalent form | Graph | ||
|---|---|---|---|---|
| \ | x\ | < k | −k < x < k | Bounded interval |
| \ | x\ | > k | x < −k or x > k | Two rays outward |
| \ | x − a\ | < r | a−r < x < a+r | Interval centered at a |
Applications
Error and Tolerance: |measured − actual| < 0.01 means the error is within 0.01 of the actual value. This is the standard form for specifying tolerances in engineering. Average Deviation: Mean Absolute Deviation (MAD) = (1/n) Σ|xᵢ − x̄| — a robust measure of spread that's less sensitive to outliers than variance. Norm of a vector: |v| = √(v₁² + v₂² + v₃²) — the length (magnitude) of a vector is its "absolute value" in multiple dimensions.Is absolute value the same as magnitude?
For real numbers, yes — |x| is the magnitude of x on the number line. For complex numbers z = a + bi, the magnitude is |z| = √(a²+b²), which is the distance from the origin in the complex plane. For vectors, the magnitude (length) is the Euclidean norm — the square root of the sum of squared components.
Can you have |x| inside a derivative?
Yes, but carefully. d/dx|x| = 1 for x > 0, and −1 for x < 0. The derivative is undefined at x = 0 because there's a sharp corner. This non-differentiability at zero is why absolute value appears in many robust optimization problems (L1 regularization in machine learning) — the kink at zero promotes sparsity.
How do you remove the absolute value from an expression algebraically?
Replace |f(x)| with f(x) when f(x) ≥ 0 and with −f(x) when f(x) < 0. This case-split approach works for any expression inside absolute value bars. For equations and inequalities, this is how you solve them systematically.