Think of a recipe that tells you what to do with the number you choose. Each instruction has to be possible. A fraction asks you to divide, so its bottom number cannot be zero. An even root asks which number has a certain even power, so the number inside cannot be negative when you work with real numbers. Your domain is the collection of inputs that pass every instruction. Start by considering the whole number line, then cross out the choices that break something. A choice that passes one instruction but breaks another still cannot go into this recipe.
The endpoint 7 stays because it produces a real square root of zero.
Reminder
Solving a linear equation. 2 − x = 0 becomes −x = −2, then x = 2. This finds the input to exclude.
Inequality reversal. −x ≥ −7 becomes x ≤ 7 because division by −1 reverses order.
Fraction arithmetic.0−7 = 0 is defined, but 30 is undefined.
Why it works. Division undoes multiplication. To divide 1 by zero you would need a number whose product with zero is 1, and every such product is zero. Even powers fail in a different way: a real number squared, or raised to another positive even integer power, cannot be negative. Zero causes no trouble for a square root itself because 02 = 0. If that zero root becomes a denominator, division fails. These reasons tell you exactly where to use a nonstrict inequality and where to exclude an endpoint.
RuleEvery denominator must be nonzero; every even-root radicand must be ≥ 0. Keep inputs satisfying all restrictions, plus any supplied domain or contextual conditions.
The same idea, five ways
Say it
Say an input belongs only when every arithmetic instruction works.
Write it
Remove inputs causing division by zero or an even root of a negative number.
In math
Denominator ≠ 0
Even-root radicand ≥ 0
For 7 − x: x ≤ 7
{x | x ≤ 7}
(−∞, 7]
Graph words: include 7 and shade to the left
Like
A recipe works only when every instruction can be carried out.
See itThe endpoint 7 stays because it produces a real square root of zero.
The same idea, other ways
As a recipe inspection
Check every instruction before using an input. A fraction must divide by a nonzero number, and an even root must start from a nonnegative number. A successful input passes both checks when both appear.
With boundary numbers
7 − x works at x = 7 because its inside is 0. It fails at x = 8 because its inside is −1. Substitution gives a concrete reason for the bracket at 7.
Why negative division flips
After 7 − x ≥ 0 becomes −x ≥ −7, division by −1 turns the number line around. The side that was greater becomes smaller, so x ≤ 7. Addition and positive division do not turn it around.
Reflection changes 2 < 7 into −2 > −7, which explains the inequality reversal.
As a checklist
For a polynomial, ordinary sums and products work for every real input. An odd root allows negative radicands, and absolute value measures a distance for any real input. Still inspect anything inside those expressions: ∛(1x) excludes zero because its interior divides by x.
.1Expressions with no arithmetic restriction
A polynomial is made from finitely many additions, subtractions and products of constants and nonnegative whole-number powers of the input. These instructions accept any real number. Absolute value and an odd root also accept any real number supplied to them. An undefined expression inside one of them remains undefined.
Polynomial domain: (−∞, ∞), unless another domain or a context is specified.
Absolute value domain: (−∞, ∞) for |x| itself.
Odd root domain: (−∞, ∞) for 3x itself; (−2)3 = −8 explains a negative cube root.
A nonzero constant denominator imposes no restriction. Always inspect nested operations.
The window shows part of x2 − 1, but the formula accepts real inputs beyond either edge.
Reminder
Solving a linear equation. 2 − x = 0 becomes −x = −2, then x = 2. This finds the input to exclude.
Inequality reversal. −x ≥ −7 becomes x ≤ 7 because division by −1 reverses order.
Fraction arithmetic.0−7 = 0 is defined, but 30 is undefined.
The same idea, five ways
Say it
Every real input works for the basic polynomial.
Write it
The polynomial x2 − 1 accepts every real input.
In math
x is real
{x | x is real}
(−∞, ∞)
Graph words: the horizontal shadow has no finite end
Like
A copying or multiplying machine can process any real number handed to it.
See itThe window shows part of x2 − 1, but the formula accepts real inputs beyond either edge.
Worked exampleA polynomial accepts every real input
Find the domain of f(x) = x2 − 1.
The window shows part of x2 − 1, but the formula accepts real inputs beyond either edge.
We need every input for which the stated rule is defined.State what is being found before choosing the calculation.
Identify the operations: square the input, then subtract 1.These are multiplication and subtraction, with no variable denominator or even root.
Keep all real inputs.Both operations are defined for every real number.
Answer
Domain: (−∞, ∞).
Check For x = −2, 0 and 12, the outputs are 3, −1 and −34. Different input signs and a fraction all work.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: The domain of ∛(1x) includes zero because odd roots accept negative numbers.
The inside division fails at zero before the cube root is taken.
Inspect the inside operations even when the outside rule is an odd root or absolute value.
.2A denominator excludes its zeros
The denominator is the number below the fraction bar. It tells you what to divide by. A negative denominator is allowed, and a zero numerator is allowed when the denominator is nonzero. Find the inputs that make the bottom zero and remove those inputs. Cancellation means dividing the numerator and denominator by the same nonzero factor. For x − 6x − 6, that gives 1 only when x − 6 is nonzero. At x = 6 the original formula is 00. Every proposed quotient multiplied by 0 gives 0, so division cannot choose a unique answer. Removing a factor from the written formula never adds that forbidden input back to the original domain.
Denominator check: set the denominator equal to zero, solve, then exclude the solutions.
Do not set the numerator equal to zero to find restrictions.
Keep an original denominator restriction even if later algebra cancels a factor.
The two allowed intervals stop before the forbidden denominator input 2.
Reminder
Solving a linear equation. 2 − x = 0 becomes −x = −2, then x = 2. This finds the input to exclude.
Inequality reversal. −x ≥ −7 becomes x ≤ 7 because division by −1 reverses order.
Fraction arithmetic.0−7 = 0 is defined, but 30 is undefined.
The same idea, five ways
Say it
The bottom of a fraction must stay nonzero.
Write it
For the fraction with denominator 2 − x, the input 2 is excluded.
In math
2 − x ≠ 0
x ≠ 2
x < 2 or x > 2
{x | x ≠ 2}
(−∞, 2) ∪ (2, ∞)
Like
Dividing an amount asks how much goes in each equal group.
See itThe two allowed intervals stop before the forbidden denominator input 2.
Worked examplea denominator
Find the domain of f(x) = x + 12 − x.
Domain: (−∞, 2) ∪ (2, ∞). The endpoint symbols record which limits belong.
We need every input for which the stated rule is defined.State what is being found before choosing the calculation.
Set 2 − x = 0.A denominator equal to zero makes division undefined.
Subtract 2 from both sides: −x = −2.The same subtraction preserves the equality.
Divide both sides by −1: x = 2.This equation finds the input to exclude, not the input to keep.
Keep every real number except 2.The numerator x + 1 is defined for all real inputs, and no other input makes the denominator zero.
Write (−∞, 2) ∪ (2, ∞).Both intervals stop before the forbidden value, and together they include all other inputs.
Answer
Domain: (−∞, 2) ∪ (2, ∞).
Check At x = 1 the output is 21 = 2; at x = 3 it is 4−1 = −4. At x = 2 the numerator is 3 but the denominator is zero, so that input fails.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Remove an input whenever the numerator is zero.
A zero numerator over a nonzero denominator has the valid quotient zero.
✓ Instead: For x + 2x − 5, keep −2 because 0−7 = 0.
Tips and tricks
The bottom decides division's restriction; the top may legitimately equal zero.
.3Even roots allow zero unless division forbids it
A square root asks for the nonnegative number that squares to its inside. Since a real square cannot be negative, a negative inside has no real answer. Zero does have an answer. Fourth roots and other even roots have the same restriction because an even number of negative factors also produces a nonnegative value.
Radicand means the expression inside a radical, or root symbol.
An even root requires radicand ≥ 0.
If the entire denominator is A, require A > 0.
If the denominator is something larger such as 1 + A, inspect that whole expression rather than automatically rejecting A = 0.
With roots in both numerator and denominator, inspect every root separately. The numerator may equal zero; the whole denominator may not.
Every real square lies on or above zero, so a negative number has no real square root.
Reminder
Solving a linear equation. 2 − x = 0 becomes −x = −2, then x = 2. This finds the input to exclude.
Inequality reversal. −x ≥ −7 becomes x ≤ 7 because division by −1 reverses order.
Fraction arithmetic.0−7 = 0 is defined, but 30 is undefined.
The same idea, five ways
Say it
The inside of an even root is zero or more.
Write it
A root used as the entire denominator also has to be nonzero.
In math
Radicand ≥ 0
For denominator x − 3: x − 3 > 0
x > 3
{x | x > 3}
(3, ∞)
Like
A square tile's area cannot be negative, and its side length can be zero.
See itEvery real square lies on or above zero, so a negative number has no real square root.
Worked examplea square root used as the denominator
Find the domain of f(x) = 1x − 3.
Domain: (3, ∞). The endpoint symbols record which limits belong.
We need every input for which the stated rule is defined.State what is being found before choosing the calculation.
Require x − 3 ≥ 0.The square root must be a real number before you can divide by it.
Also require x − 3 ≠ 0.This entire square root is the denominator, and division by zero is undefined.
Combine the checks to require x − 3 > 0.Zero would satisfy the root check but fail the denominator check.
Add 3: x > 3, so write (3, ∞).Adding the same number preserves the strict inequality and the excluded endpoint.
Answer
Domain: (3, ∞).
Check x = 4 gives 11 = 1. x = 3 gives a zero denominator, and x = 2 gives a negative radicand.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this:0 is undefined.
Zero is the nonnegative number whose square is zero.
✓ Instead:0 = 0. A separate division by this output would fail.
Tips and tricks
Check the root and any denominator as two separate operations.
.4Build a formula for a requested domain
Imagine designing an entrance gate instead of reading one that already exists. You want the gate to admit exactly the inputs in the requested set. Different arithmetic operations can supply different checks. A denominator can reject one input. An even root can reject all inputs on one side of a boundary. A root used as the entire denominator also rejects the boundary itself. Build the check into a formula, then verify its actual domain rather than guessing from its appearance.
For a nonstrict ray x ≥ a, x − a has exactly that domain.
For a strict ray x > a, 1x − a has exactly that domain.
For all real inputs except a, 1x − a has exactly that domain.
This is a construction strategy: many formulas can have the same domain.
Construct the formula to admit exactly this strict ray.
Reminder
Root versus divisor.0 = 0 works as a value, but 10 divides by zero.
The same idea, five ways
Say it
Say create a formula that accepts exactly the requested inputs.
Write it
For a strict rightward ray, a square root in the entire denominator rejects the boundary and everything below it.
In math
x > 14
{x | x > 14}
(14, ∞)
F(x) = 1x − 14
Graph words: hollow dot at 14, shade right
Like
Build an entrance gate with the checks required by the visitor list.
See itThe designed rule must match this entire allowed set.
Worked exampleBuild a function whose domain is a strict ray
Construct one real-valued function with exactly the domain x > 14. You are creating a formula that permits every input above 14 and rejects 14 and all smaller inputs.
The desired domain is open at 14 and includes every greater real input.The constructed rule works for a sample input above its excluded boundary.
Choose an inside expression x − 14.It is positive precisely when x > 14, locating the required boundary.
A square root x − 14 alone would allow x = 14.A square root accepts inside zero, so this would give a nonstrict domain and miss the requested exclusion.
Put that root in the entire denominator and choose a nonzero constant numerator: F(x) = 1x − 14.The root must be real and nonzero, combining the two arithmetic checks into x − 14 > 0.
Add 14 to x − 14 > 0 to get x > 14.Solving verifies that the constructed formula has exactly the requested domain.
At x = 14, the denominator is 0 = 0; at x = 13, its radicand is −1. Every input above 14 gives a positive denominator.Plugging the boundary and both sides into the original formula confirms that all and only the requested inputs work.
Answer
One possible function: F(x) = 1x − 14.
Domain: {x | x > 14} = (14, ∞).
Check At x = 15 the result is 1. At x = 18 it is 12. Many different formulas could have this same domain; the question requires one valid construction.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Use x − 14 for the requested strict domain x > 14.
At x = 14 the root equals zero, which is allowed for a root by itself.
✓ Instead: Use 1x − 14 so division also rejects the boundary.
Tips and tricks
Verify the constructed domain in both directions: every requested input must work, and every excluded input must fail.
Strategy: step by step
1. Start with all real inputs unless a domain or context is supplied.
2. Find every denominator. Set the whole denominator equal to zero to locate exactly the inputs where division would fail. Solve that equation, substitute each answer into the original denominator to confirm zero, and exclude those inputs.
3. Find every even-root radicand, require it to be ≥ 0, and solve the inequality.
4. If a root is part of a denominator, check the whole denominator as well as the radicand.
5. Keep only inputs that meet every condition. Write intervals from smaller to larger endpoints.
6. Substitute each boundary or excluded value to see which operation permits or rejects it.
Strategy
Find a formula's domain
1
Is there a denominator?
YesSet the whole denominator to zero to find failing division inputs. Solve, substitute each result to confirm zero, and remove those inputs from the domain.
NoContinue to the root check.
↓
2
Is there an even root?
YesRequire its radicand to be ≥ 0 and solve.
NoContinue to any supplied context.
↓
3
Is the entire denominator an even root?
YesCombine the root and nonzero checks, making its radicand > 0.
NoUse the actual whole-denominator condition; do not automatically exclude a zero radicand.
↓
4
Are there several conditions?
YesKeep only inputs passing them all.
NoUse the single condition, or all real inputs if none exist.
↓
5
Does the question ask you to construct a function for a domain?
YesChoose operations that impose the desired checks, then recompute the whole domain to verify an exact match.
NoRead the restrictions from the supplied formula.
1. Start with all real inputs unless a domain or context is supplied.
2. Find every denominator. Set the whole denominator equal to zero to locate exactly the inputs where division would fail. Solve that equation, substitute each answer into the original denominator to confirm zero, and exclude those inputs.
3. Find every even-root radicand, require it to be ≥ 0, and solve the inequality.
4. If a root is part of a denominator, check the whole denominator as well as the radicand.
5. Keep only inputs that meet every condition. Write intervals from smaller to larger endpoints.
6. Substitute each boundary or excluded value to see which operation permits or rejects it.
Worked exampleA polynomial accepts every real input
Find the domain of f(x) = x2 − 1.
The window shows part of x2 − 1, but the formula accepts real inputs beyond either edge.
We need every input for which the stated rule is defined.State what is being found before choosing the calculation.
Identify the operations: square the input, then subtract 1.These are multiplication and subtraction, with no variable denominator or even root.
Keep all real inputs.Both operations are defined for every real number.
Answer
Domain: (−∞, ∞).
Check For x = −2, 0 and 12, the outputs are 3, −1 and −34. Different input signs and a fraction all work.
Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.
Rung 1a denominator
Find the domain of f(x) = x + 12 − x.
Domain: (−∞, 2) ∪ (2, ∞). The endpoint symbols record which limits belong.
We need every input for which the stated rule is defined.State what is being found before choosing the calculation.
Set 2 − x = 0.A denominator equal to zero makes division undefined.
Subtract 2 from both sides: −x = −2.The same subtraction preserves the equality.
Divide both sides by −1: x = 2.This equation finds the input to exclude, not the input to keep.
Keep every real number except 2.The numerator x + 1 is defined for all real inputs, and no other input makes the denominator zero.
Write (−∞, 2) ∪ (2, ∞).Both intervals stop before the forbidden value, and together they include all other inputs.
Answer
Domain: (−∞, 2) ∪ (2, ∞).
Check At x = 1 the output is 21 = 2; at x = 3 it is 4−1 = −4. At x = 2 the numerator is 3 but the denominator is zero, so that input fails.
Rung 2Domain of a function with a square root inside the denominator
Find the domain of g(x) = x + 16 − x − 2. Write your answer in interval notation.
Start with all real numbers as possible inputs.No domain or context is supplied, so only the arithmetic can rule inputs out.
Identify the even root 6 − x and require its radicand to be nonnegative: 6 − x ≥ 0.A square root of a negative number is not a real number.
Solve 6 − x ≥ 0 to get x ≤ 6.Adding x to both sides gives 6 ≥ x. No division by a negative number is needed, so the inequality does not flip.
Identify the whole denominator 6 − x − 2 and set it equal to zero: 6 − x − 2 = 0.Division by zero is undefined. The root sits inside the denominator, so the whole denominator must be checked, not only the radicand.
Solve: 6 − x = 2, square both sides to get 6 − x = 4, so x = 2.Isolating the root and then squaring turns the equation into a linear one.
Substitute x = 2 into the original denominator: 6 − 2 − 2 = 4 − 2 = 2 − 2 = 0. Exclude x = 2.Squaring can create false solutions, so we confirm that x = 2 really makes the denominator zero. It does, so x = 2 is excluded.
Combine the conditions: x ≤ 6 and x ≠ 2. In interval notation this is (−∞, 2) ∪ (2, 6].Only inputs that meet every restriction are kept. Intervals are written from smaller to larger endpoints. Use a parenthesis at 2 because 2 is excluded, and a bracket at 6 because 6 is allowed.
Check the boundary and excluded values. At x = 6: the radicand is 6 − 6 = 0, which is allowed, and the denominator is 0 − 2 = −2 ≠ 0, so 6 is included. At x = 2: the denominator is 0, so division rejects 2. At x = 7: the radicand is −1 < 0, so the square root rejects 7.Testing each boundary shows which operation permits or rejects that value.
Answer
(−∞, 2) ∪ (2, 6]
Check Test x = 6: g(6) = 70 − 2 = −72, which is defined, so 6 belongs in the domain. Test x = −3: g(−3) = −29 − 2 = −21 = −2, which is defined. Test x = 2: the denominator is 4 − 2 = 0, so g(2) is undefined. Test x = 7: −1 is not real, so g(7) is undefined. All four results agree with (−∞, 2) ∪ (2, 6].
Work to write
Radicand: 6 − x ≥ 0, so x ≤ 6
Denominator: 6 − x − 2 = 0, so 6 − x = 2, so 6 − x = 4, so x = 2
Check: 4 − 2 = 0, so exclude x = 2
At x = 6: radicand is 0 (allowed) and denominator is −2 ≠ 0, so include 6
Domain: (−∞, 2) ∪ (2, 6]
(−∞, 2) ∪ (2, 6]
Rung 3Domain of a quotient with a square root in the denominator
Find the domain of h(x) = x − 3x + 7 − 3. Write your answer in interval notation.
Start with all real numbers as possible inputs.No domain or context is supplied, so only the arithmetic can restrict x.
Radicand condition: require x + 7 ≥ 0, so x ≥ −7.An even (square) root of a negative number is not a real number.
Denominator condition: set the whole denominator equal to zero: x + 7 − 3 = 0, so x + 7 = 3.Division fails exactly where the entire denominator is zero, not just where the root is zero.
Square both sides: x + 7 = 9, so x = 2. Substitute into the original denominator: 2 + 7 − 3 = 9 − 3 = 3 − 3 = 0. Exclude x = 2.Squaring can create false solutions, so substituting confirms that x = 2 really makes the denominator zero.
Combine the conditions: x ≥ −7 and x ≠ 2. In interval notation this is [−7, 2) ∪ (2, ∞).Inputs must satisfy every restriction. −7 is included because the radicand may equal 0. 2 is removed with open parentheses. Intervals are written from smaller to larger endpoints.
Test the boundary values. At x = −7: 0 − 3 = −3 ≠ 0, so h(−7) = −10−3 = 103, which is allowed. At x = 2 the denominator is 0, so 2 is rejected. At x = −8 the radicand is −1, so −8 is rejected.Substituting each boundary or excluded value shows which operation permits or rejects it.
Answer
Domain of h: [−7, 2) ∪ (2, ∞)
Check Try x = 18: 25 − 3 = 2, so h(18) = 152, which is defined. 18 lies in (2, ∞). Try x = −3: 4 − 3 = −1, so h(−3) = −6−1 = 6, which is defined. −3 lies in [−7, 2). The excluded value x = 2 gives division by zero, and x = −8 gives −1. Both are correctly left out.
Work to write
x + 7 ≥ 0 ⇒ x ≥ −7
x + 7 − 3 = 0 ⇒ x + 7 = 9 ⇒ x = 2
Check: 9 − 3 = 0, so exclude x = 2
x = −7 is allowed: denominator is −3 ≠ 0
Domain: [−7, 2) ∪ (2, ∞)
Domain of h: [−7, 2) ∪ (2, ∞)
Rung 4Domain of a quotient with a square root inside the denominator
Find the domain of k(x) = x − 25 − 2x + 6. Write your answer in interval notation.
Start with all real numbers as candidate inputs.No domain or context is supplied, so only the arithmetic can remove inputs.
Identify the even-root radicand 2x + 6 and require 2x + 6 ≥ 0.A square root of a negative number is not a real number.
Solve 2x + 6 ≥ 0: 2x ≥ −6, so x ≥ −3.Subtracting 6 and dividing by the positive number 2 keeps the inequality direction.
Set the whole denominator equal to zero: 5 − 2x + 6 = 0, so 2x + 6 = 5.Division fails exactly where the entire denominator is zero. The root sits inside the denominator, so the whole expression is checked, not just the radicand.
Square both sides: 2x + 6 = 25, so 2x = 19 and x = 192.Squaring undoes the square root. This value satisfies x ≥ −3, so it is a real candidate.
Substitute x = 192 into the original denominator: 2·192 + 6 = 25, 25 = 5, and 5 − 5 = 0. Exclude x = 192.Squaring can create false solutions, so the value is confirmed in the original denominator. It really gives zero.
Combine the conditions: x ≥ −3 and x ≠ 192. Write the result as [−3, 192) ∪ (192, ∞).Only inputs that meet every condition are kept. The intervals are written from smaller to larger endpoints.
Test the boundaries. At x = −3 the radicand is 0 and 0 = 0, so the denominator is 5 and k(−3) = −55 = −1. This value is included. At x = 192 the denominator is 0, so this value is excluded.The square root permits radicand 0, so the bracket is used at −3. Division rejects 192, so parentheses are used on both sides of it.
Answer
[−3, 192) ∪ (192, ∞)
Check Pick x = 5, which lies inside the first interval: 16 = 4 and the denominator is 5 − 4 = 1, so k(5) = 3. This is defined. Pick x = 15, which lies inside the second interval: 36 = 6 and the denominator is 5 − 6 = −1, so k(15) = −13. This is defined. Pick x = −4, which lies outside: the radicand is −2, so the root is not real. This is undefined, as expected.
Solve x − 4 = 0 or x + 4 = 0, giving 4 or −4.A product is zero when at least one factor is zero.
Remove both values and write the three remaining intervals.One interval before −4, one between −4 and 4, and one after 4 are needed.
Answer
Domain: (−∞, −4) ∪ (−4, 4) ∪ (4, ∞).
Check Both 42 and (−4)2 equal 16 and make the denominator zero. At x = 0 the denominator is −16, so a negative denominator itself is allowed.
Rung 7Two roots impose different checks
Find the domain of R(x) = x − 12x − 7. You need every input for which both roots exist and the denominator is nonzero.
The stricter starting input comes from the numerator’s root; 12 itself works.
Require x − 12 ≥ 0; add 12 to both sides to get x ≥ 12.The numerator’s even root must exist. It may equal zero because it is not being used as a divisor.
Require x − 7 > 0; add 7 to both sides to get x > 7.The denominator’s root must exist and stay nonzero, so its inside must be strictly positive.
Keep inputs passing both checks: x ≥ 12.Every number at least 12 is also greater than 7; a number between 7 and 12 fails the numerator check.
At x = 12, the original expression is 05 = 0, so include the starting input.The denominator is positive here and a zero numerator is allowed.
Answer
Domain: [12, ∞).
Check At x = 13 both radicands are positive. At x = 11 the denominator’s root exists but the numerator’s radicand is −1, so that input fails the combined condition.
Rung 8A denominator that contains a square root
Find the domain of f(x) = 11 + x.
Domain: [0, ∞). The endpoint symbols record which limits belong.
We need every input for which the stated rule is defined.State what is being found before choosing the calculation.
Require x ≥ 0.The radicand of x must be nonnegative.
For these inputs, x ≥ 0, so 1 + x ≥ 1.The principal square root never produces a negative output.
The whole denominator can never equal zero on these inputs.Being at least 1 makes it nonzero, including at x = 0.
Keep [0, ∞).There is no additional exclusion after the root condition.
Answer
Domain: [0, ∞).
Check At x = 0 the value is 11 + 0 = 1. This verifies that a root somewhere in a denominator does not automatically force its radicand to be strictly positive.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: 7 − x ≥ 0 gives x ≥ 7.
Dividing −x ≥ −7 by −1 reflects and reverses the order. Substitution of x = 8 would give a negative radicand.
✓ Instead: x ≤ 7, so the domain is (−∞, 7].
✗ Not this: The domain of 11 + x is (0, ∞) because a root is in the denominator.
The whole denominator is 1 + x, which equals 1 at x = 0. The root may be zero without making that denominator zero.
✓ Instead: The domain is [0, ∞).
✗ Not this: A zero numerator must be removed from the domain.
Zero divided by a nonzero number is the valid output zero.
✓ Instead: For x + 2x − 5, keep x = −2 because 0−7 = 0.
Tips and tricks
Write denominator ≠ 0 and radicand ≥ 0 before solving anything.
Memory device: zero below means no go. A zero numerator can still be valid.
Test the endpoint in the original formula to decide whether the interval needs a bracket.
Keep restrictions from the original expression even if later cancellation makes it look unrestricted.
Trap. Changing ≥ 0 to > 0 whenever you see a square root. 0 is allowed. Strict positivity is needed for a square root that must itself be a nonzero denominator, such as 1x − 3.
