Compound Inequalities in Interval Notation

Most students assume that solving compound inequalities interval notation problems means memorizing a different rule for every symbol. The truth is simpler: every answer, whether it comes from an "and" inequality, an "or" inequality, or an absolute value inequality interval notation exercise, is just a set of real numbers written with endpoints, parentheses, brackets, and possibly the union symbol. Once you stop treating each problem as a new puzzle and start seeing the underlying set, the notation writes itself. Solve compound inequality problems and express the result, then check the common failure points where most answers go wrong.

When a compound inequality uses the word "and," as in x > 2 and x < 5, you are looking for numbers that satisfy both conditions at the same time. The solution is the intersection of the two individual solution sets. On the number line, shade the region where the two shaded segments overlap. In interval notation, that overlap becomes a single interval, here (2,5).

Endpoint Checks That Save Points

Write the smaller endpoint on the left and the larger on the right, always. If the inequality is strict, use a parenthesis. If it is non-strict, use a bracket. For x ≥ -1 and x < 3, the answer is [-1,3). A common mistake is to write (-1,3] because you misread which side carries the equals sign. Check each endpoint against its own inequality before you write anything.

When the two conditions cannot both be true, the intersection is empty. For x < 2 and x > 5, no real number is both less than 2 and greater than 5. The solution is the empty set, written as ∅ or { }. Do not write [ ] or ( ). Those look like intervals and they are not standard notation. Many calculators and textbooks use ∅, and you should too.

Double Inequalities as Compact And Statements

For a double inequality like -3 ≤ 2x+1 < 7, you are looking at an "and" statement in disguise. It says -3 ≤ 2x+1 and 2x+1 < 7. Solve both sides at once by subtracting 1 from all three parts, giving -4 ≤ 2x < 6, then divide by 2 to get -2 ≤ x < 3. The interval is [-2,3). This single operation, applying the same step to all three parts, is the fastest way to solve compound inequality problems of this shape.

When a compound inequality uses "or," as in x < 1 or x > 4, you are looking for numbers that satisfy at least one of the conditions. The solution is the union of the two individual solution sets. In interval notation, join the two intervals with the union symbol ∪, so the answer is (-∞,1) ∪ (4,∞). The union symbol ∪ means "or" in this context, not "and." A number belongs to the union if it belongs to the first interval OR the second.

The Union Symbol Is Not Optional

The most common error here is writing (-∞,1)(4,∞) with no symbol between the intervals. That is not valid notation. Separate intervals require the union symbol. Without it, the expression is unreadable. Always check that you have placed a ∪ between any two intervals that do not touch or overlap.

When the intervals do overlap, simplify the union. For x > 2 or x > 5, every number greater than 2 already satisfies the second condition, so the union is just (2,∞). Writing (2,∞) ∪ (5,∞) is not wrong, but it is redundant. The shortest correct interval is the one you should give.

For x < -1 or x ≥ 3, the answer is (-∞,-1) ∪ [3,∞). Notice the bracket on the 3 because the second inequality is non-strict. The parenthesis on -1 stays because the first inequality is strict. This mixture of parentheses and brackets within a single union is normal and must be preserved.

Absolute value inequalities follow two patterns, and confusing them is the single biggest source of lost points on this subject. For |x| < a, where a is positive, the solution is a bounded interval: -a < x < a. In interval notation, that is (-a,a). For |x| > a, the solution is two unbounded intervals: x < -a or x > a, written as (-∞,-a) ∪ (a,∞).

Distance From Zero Drives Everything

The reasoning is direct. The absolute value of x measures distance from zero. The inequality |x| < a says the distance from zero is less than a, which means x sits strictly between -a and a. The inequality |x| > a says the distance from zero is greater than a, which pushes x to one side or the other, hence the union.

When the inequality is non-strict, such as |x| ≤ a, use brackets: [-a,a]. When it is |x| ≥ a, use brackets on the outside endpoints of the union: (-∞,-a] ∪ [a,∞). The endpoint itself is included because the equals sign is part of the condition.

When the Expression Is Not Just X

A frequent failure case appears when the expression inside the absolute value is not just x. For |2x-3| < 5, you must first write -5 < 2x-3 < 5, then solve the double inequality by adding 3 to all parts to get -2 < 2x < 8, then divide by 2 to get -1 < x < 4. The interval is (-1,4). Do not skip the middle step. Students who try to jump straight to the interval often flip a sign or drop a constant.

For |x| < 0, there is no solution, because absolute value is never negative. The answer is ∅. For |x| > -2, every real number satisfies the inequality, because the absolute value is always at least 0, which is greater than -2. The answer is (-∞,∞), which is the same as ℝ.

Interval notation always reads left to right, smallest endpoint to largest. The left endpoint is always the smaller number. Writing (5,2) is wrong before you even consider the symbols. The pair (a,b) means the set of all real numbers between a and b, not a point on a plane. That confusion, reading (2,5) as the coordinate of a point, is a mistake that disappears once you remember the context is a solution set on a single number line.

Infinity Is Not a Number

Infinity is not a number. It cannot be included with a bracket, so you always write (-∞,a) or (a,∞) or (-∞,∞). The parenthesis next to ∞ is non-negotiable. Treating ∞ like a very large number and writing [a,∞] is the most common error in this entire subject. The bracket would claim the interval includes ∞, which is impossible.

Bounded intervals have two finite endpoints. Unbounded intervals have at least one infinite endpoint. A half-open interval includes one endpoint but not the other, like [1,5) or (1,5]. The length of an interval is the difference between the endpoints, so the length of [1,5) is 4, and the length of (-∞,3) is infinite.

For discrete sets like {1,2,3}, interval notation cannot represent them because interval notation only describes continuous sets of real numbers. You would have to write {1,2,3} explicitly or use set-builder notation. If a problem asks for interval notation and the answer is a finite list of individual numbers, you have likely misread the inequality.

Most students meet interval notation for the first time in algebra, but it becomes unavoidable in precalculus and calculus because the domain and range of a function are sets of real numbers. The domain of a square root function requires the radicand to be ≥ 0.

Domain for Logs and Rational Functions

The domain of a logarithmic function requires the argument to be > 0.A rational function like f(x)=1/(x²-4) needs the denominator to be nonzero, so x²-4 ≠ 0, which excludes x = -2 and x = 2. The domain is (-∞,-2) ∪ (-2,2) ∪ (2,∞). Missing the middle interval is the classic error here, and it happens because students write the two excluded points but forget that the entire region between them is also part of the domain.

Range in Interval Notation

For range, you are asking what y-values the function can output.Writing the range in interval notation is the same skill as writing the domain, just applied to the output set.

When you have a solution set, ask two questions. First, does the set include the endpoint? If yes, use a bracket. If no, use a parenthesis. Second, does the set extend without bound in either direction? If yes, use ∞ or -∞, always with a parenthesis. Then write the smaller endpoint on the left and the larger on the right, joining any separate pieces with the union symbol ∪.

For x < 2 or x ≥ 5, the left piece is (-∞,2), the right piece is [5,∞), and the union is (-∞,2) ∪ [5,∞). For -1 < x ≤ 4, it is (-1,4]. For all real numbers, it is (-∞,∞). The last one is the interval notation for the set of all real numbers, which you may also see written as ℝ.

The empty set is the one exception to the two-endpoint rule. It has no endpoints at all and is written as ∅ or { }. Some calculators display an empty set symbol when no real number satisfies the inequality. If your calculator outputs ∅, it is telling you the solution is empty, not that the problem is broken.

Inequality	Meaning	Interval Notation
x > a	Strict, unbounded right	(a,∞)
x ≥ a	Non-strict, unbounded right	[a,∞)
x < a	Strict, unbounded left	(-∞,a)
x ≤ a	Non-strict, unbounded left	(-∞,a]
a < x < b	Strict, bounded	(a,b)
a ≤ x ≤ b	Non-strict, bounded	[a,b]
a < x ≤ b	Half-open, strict left	(a,b]
a ≤ x < b	Half-open, strict right	[a,b)
|x| < a	Distance from zero less than a	(-a,a)
|x| > a	Distance from zero greater than a	(-∞,-a) ∪ (a,∞)

The intersection of two intervals is the set of numbers in both, which is what "and" means. The union is the set of numbers in either, which is what "or" means. The symbols ∩ and ∪ look similar, and swapping them is a fatal error. A good check is to test a specific number. For x > 2 and x < 5, the number 3 works, but 6 does not. For x < 2 or x > 5, the number 6 works, but 3 does not.

When intervals touch, the union may collapse into a single interval. For x < 3 or x > 3, the solution is (-∞,3) ∪ (3,∞), which is all real numbers except 3. The union does not include 3 because both inequalities are strict. If one were non-strict, such as x ≤ 3 or x > 3, the union would be (-∞,∞).

In interval notation calculus, which deals with intervals of increase, decrease, and concavity, the same union symbol does the work. A function that increases on (-∞,0) and (2,∞) has that set as its interval of increase. The notation does not change because the subject is calculus.

When your interval does not match any option, do not erase your work. Check the endpoints first. Look at each inequality symbol and confirm whether the endpoint carries a bracket or a parenthesis. A single swapped symbol, writing (2,5) instead of [2,5], will usually appear in the options because it is the most common wrong answer.

Check the union next. If your answer has two intervals, confirm the ∪ symbol is present. A missing union symbol is invisible in your own handwriting but will not match any option. Then check the infinity signs. A bracket next to ∞, such as [2,∞], is a guaranteed distractor.

If the answer still does not match, test a number. Pick a value inside your proposed interval and plug it into the original inequality. Then pick a value outside and confirm it fails. This takes thirty seconds and catches nearly every error, including sign flips and reversed endpoints.

When the normal route of solving step by step is closed, for instance on a timed test, solve by testing the answer choices instead. Plug the endpoints of each option into the original inequality. The correct interval will have both endpoints satisfying the non-strict condition and neither endpoint violating the strict one. This works for both compound and absolute value inequalities interval notation problems, and it does not require you to trust your own algebra under pressure.