Difficulty: Foundational to Intermediate | Prerequisites: Arithmetic and Algebraic Operations notes.
Inequalities show up everywhere in calculus, from domains and ranges to epsilon-delta proofs. You need to solve absolute value inequalities, quadratic inequalities, and rational inequalities by hand, and express solutions using interval notation.
Absolute value (|x|)
The distance of x from zero on the number line. Formally: |x| = x when x ≥ 0, and |x| = -x when x < 0. Think of it as "always positive" (or zero).
Open interval (a, b)
The set of all real numbers x with a < x < b. The endpoints are excluded.
Closed interval [a, b]
The set of all real numbers x with a ≤ x ≤ b. The endpoints are included.
Compound inequality (AND statement)
Two inequalities that must both be true at once, e.g. -2 ≤ x - 1 ≤ 2. In simple terms, x is squeezed between two bounds.
Disjunction (OR statement)
A condition where x satisfies at least one of two inequalities, e.g. x < -4 or x > 0. The solution set is two separate intervals.
Triangle inequality
For any real numbers a and b: |a + b| ≤ |a| + |b|. Also: |a - b| ≥ ||a| - |b||. These crop up in rigorous calculus proofs.
|x| < a means -a < x < a. This is a single interval.
Example: |x| < 2 gives the open interval (-2, 2).
|x - 1| ≤ 2 means -2 ≤ x - 1 ≤ 2, so -1 ≤ x ≤ 3.
To find a from -1 < x < 5 written as |x - a| < 3: the centre of the interval is (5 + (-1))/2 = 2, so a = 2.
|x| > a means x < -a or x > a. This gives two separate intervals.
Example: |y + 2| > 2 means y + 2 < -2 or y + 2 > 2, so y < -4 or y > 0. The two intervals are (-∞, -4) and (0, ∞).
x² < 1 is equivalent to |x| < 1, giving (-1, 1).
(z - 1)² < 1 means |z - 1| < 1, so 0 < z < 2.
z² - z < 6: rearrange to z² - z - 6 < 0, factor as (z - 3)(z + 2) < 0, giving (-2, 3).
For 6/x > 2 where x > 0: multiply both sides by x (positive, so inequality direction stays) to get 6 > 2x, so x < 3. Combined with x > 0, the solution is (0, 3).
Be careful: if you do not know the sign of x, you cannot simply multiply through.
For f(x) = x³ - 2x² - 5x + 6 > 0 with x < 3: since f(1) = 0, factor out (x - 1). The polynomial factors as (x - 1)(x - 3)(x + 2). Test the signs on each interval. With x < 3, f(x) > 0 on (-2, 1).
For positive a < b, taking reciprocals flips the inequality: 1/b < 1/a. This is true.
For non-zero a < b (no sign restriction), 1/b < 1/a is false in general. Counterexample: a = -1, b = 1 gives 1/1 < 1/(-1), which is 1 < -1, false.
If x ≠ 0, then x² > 0. True.
|a| · |b| = |ab| for all real a, b. True.
a < 0 and b > 0 implies ab < 0. True.
If a < b and c > 0, then ac < bc. True (multiplying by a positive preserves direction).
If xz < yz and z < 0, then x > y. True (dividing by a negative flips the inequality).
|a + b| ≤ |a| + |b|. True (triangle inequality).
|a - b| ≥ ||a| - |b||. True (reverse triangle inequality).
If x < -1, then x² > 1. True.
If x > -1, then x² > 1. False (try x = 0).
If a < b, then a² < b². False (try a = -3, b = 1).
Students often think -x > a means x > -a. It does not. Multiplying both sides by -1 flips the inequality: -x > a means x < -a.
Students often think |x + 1| > 0 implies x < 0. It does not. |x + 1| > 0 is true for all x except x = -1.
Students often think a < b implies a² < b² for all real numbers. It does not. Squaring can flip relationships when negatives are involved. For example, -3 < 1, but (-3)² = 9 > 1.
Students often think a < b and c ≤ 0 implies ac ≤ bc. It does not. Try a = 1, b = 2, c = -1: (1)(-1) = -1 is not ≤ (2)(-1) = -2.
⚠️ Absolute value inequalities define epsilon-delta neighbourhoods, which underpin the formal definition of a limit.
⚠️ Knowing when multiplying or dividing by a variable flips the inequality is essential for solving related rates and optimisation constraints.
⚠️ The triangle inequality (|a + b| ≤ |a| + |b|) appears in formal proofs and is tested directly in some AP Calculus courses.
True or False: if x ≠ 0, then x² > 0. (Answer: True.)
True or False: |a| · |b| = |ab| for all real a, b. (Answer: True.)
Fill in the blank: |x - 3| ≤ 4 gives the interval ___ ≤ x ≤ ___. (Answer: -1 ≤ x ≤ 7.)
True or False: if ab > 0, then either a > 0 or b > 0. (Answer: False. Both could be negative.)
True or False: a + b < 2c whenever a < c and b < c. (Answer: True.)
Q: The set of real numbers x with x² < 1 is an interval a < x < b. Find a and b.
A: a = -1, b = 1. Since x² < 1 means |x| < 1.
Q: Express the solution to |x - 3| ≤ 4 as a closed interval.
A: [-1, 7]. Rewrite as -4 ≤ x - 3 ≤ 4, then add 3.
Q: The solution to z² - z < 6 is an open interval (a, b). Find a and b.
A: a = -2, b = 3. Factor z² - z - 6 = (z - 3)(z + 2) < 0.
Q: True or False: the set of x satisfying both |x + 2| < 4 and |x - 5| < 4 is the empty set.
A: False. The first gives -6 < x < 2, the second gives 1 < x < 9. The overlap is (1, 2), which is not empty.
Q: True or False: if a < b for positive a, b, then 1/b < 1/a.
A: True. Reciprocating positive numbers flips the inequality.
Q: The solution to |x + 1| - |x - 1| > 0 is (a, ∞). Find a.
A: a = 0. Sketch the two absolute value graphs; |x + 1| exceeds |x - 1| for all x > 0.
Inequalities connect directly to finding domains of functions (where is the expression under a square root non-negative?), to the definition of continuity and limits (epsilon-delta proofs rely on absolute value inequalities), and to optimisation (constraints are inequalities). The triangle inequality reappears in real analysis and multivariable calculus.
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