An excluded value is any input that breaks the original expression, the number you are not allowed to plug in because it makes the expression undefined. The restriction comes from the original expression, not the simplified one, and that is the single point almost everyone misses. Simplifying can quietly erase a division that was truly there, so a value that broke the original keeps breaking it even after the algebra looks clean. If you want to watch where a restriction appears, paste any expression into the simplify calculator and check each step against your own.
Most domain restrictions come from three places: a denominator that could be zero, an even root that could sit over a negative, or a logarithm that could take a non-positive argument. This guide finds all of them, keeps the ones that cancelled factors leave behind, and writes the finished domain in both inequality and interval notation.
It helps to keep two words straight. The excluded values are the specific inputs you throw out. The domain is everything that survives, the full set of inputs the expression accepts. You find the domain by finding the excluded values first and then taking every real number that is not on that list. So the work always runs the same direction: hunt down what breaks, then describe what is left.
Find every excluded value by factoring the denominator
A single denominator is the easy case. For 5/(x − 2) 5/(x-2) , the one thing that breaks it is a bottom of zero, so set x − 2 = 0 and get x = 2 . That value, and only that value, is excluded.
The trap is a denominator that factors. Values that make the denominator zero hide inside each factor, so you have to break the denominator all the way down before you can list them. Take 7/(x² − x − 6) 7/(x^2-x-6) . Left as one lump, it looks like it has a single problem spot. Factor it, x² − x − 6 = (x − 3)(x + 2) (x-3)(x+2) , and two problem spots appear: the product is zero when x = 3 or when x = −2 .
That is the rule for every rational expression. A quadratic denominator gives two excluded values, one from each factor, and a cubic can give three. Miss the factoring and you miss half the restrictions. The same care matters inside complex fractions, where a small denominator can sit buried three levels down and still break the whole thing.
The reason each factor counts on its own is the zero-product rule: a product is zero exactly when at least one of its pieces is zero. So (x − 3)(x + 2) (x-3)(x+2) hits zero whenever the first factor is zero, at x = 3 , or whenever the second is, at x = −2 . You solve the two little equations separately, and every solution you find is an input the denominator cannot survive. Nothing about the numerator changes this step; the excluded values live entirely in the bottom of the fraction, so that is the only part you factor to find them.
Why a cancelled factor still leaves its restriction
Here is where the “read the original” rule earns its keep. Look at (x² − 9)/(x² − x − 6) (x^2-9)/(x^2-x-6) . Factor both parts: the top is (x − 3)(x + 3) (x-3)(x+3) and the bottom is (x − 3)(x + 2) (x-3)(x+2) . The (x − 3) x-3 cancels, and the expression collapses to a tidy (x + 3)/(x + 2) (x+3)/(x+2) .
Now the tempting mistake: that clean answer has only one obvious problem, x = −2 . But the original had a (x − 3) x-3 on the bottom, and at x = 3 that denominator was zero. Plugging x = 3 into the original still gives 0/0, which is undefined. Cancelling the factor did not un-break it. So the answer carries both restrictions, x ≠ 3 and x ≠ −2 .
The graph makes it concrete. At x = −2 , where the factor did not cancel, the curve shoots off to infinity: a vertical asymptote. At x = 3 , where the factor cancelled, the curve looks continuous but has a single missing point, a hole, one pixel punched out where the function has no value. Same undefined status, different picture. The hole is exactly the trace of a restriction that a cancellation tried to erase.
Why the two look so different comes down to what cancellation does. A factor that survives on the bottom drives the fraction toward a division by something tiny, and dividing by tiny numbers blows the output up: that is the asymptote. A factor that cancels no longer drives the output anywhere, so near x = 3 the curve behaves just like the simplified (x + 3)/(x + 2) (x+3)/(x+2) , which is perfectly well behaved there. The graph would be continuous if it were allowed to touch that point, but the original expression forbids it, so the point is lifted out. If you sketch it, you draw the smooth curve and then erase a single dot at x = 3 . That erased dot is the whole story: the value is excluded, the graph shows it, and no amount of tidy algebra puts it back.
Restrictions that do not come from a denominator
Denominators are not the only source. Two others show up constantly, and both live in the expression itself rather than under a fraction bar.
An even root cannot sit over a negative number and stay real. So a square root needs its radicand to be at least zero. For √(x − 1) sqrt(x-1) , you need x − 1 ≥ 0 , which gives x ≥ 1 . The endpoint is included here, because the root of zero is a perfectly good zero. Odd roots, like a cube root, accept every real number and add no restriction at all, which is why “even” matters in the rule.
A logarithm is stricter. It needs its argument strictly positive, never zero, because the log of zero is undefined and the log of a negative is not real. For log(x − 5) log(x-5) , you need x − 5 > 0 , so x > 5 , with the endpoint excluded. That single difference, at least zero for a root versus greater than zero for a log, decides whether an endpoint gets a bracket or a parenthesis later.
Two of these sources can stack, and when they do the stricter one wins. Put a square root in a denominator, like 1/√(x − 3) 1/sqrt(x-3) , and the radicand rule would allow x = 3 , but the denominator rule forbids it, because a root of zero is zero and you cannot divide by that. The condition tightens from x ≥ 3 to x > 3 . Whenever the same value is allowed by one rule and banned by another, the ban stands.
Keep a short mental checklist of where restrictions come from:
| Source | Condition | Example | Excluded set |
|---|---|---|---|
| Denominator | must not be zero | 1/(x − 4) 1/(x-4) | x ≠ 4 |
| Even-root radicand | must be ≥ 0 | √(x − 1) sqrt(x-1) | x < 1 |
| Logarithm argument | must be > 0 | log(x − 5) log(x-5) | x ≤ 5 |
Read the last column as the inputs you throw away: everything below 1 for the root, everything at or below 5 for the log. What remains is the domain.
Three worked examples, easy to exam-level
Watch all three sources come together. Each step names the rule it uses, and each answer states its restriction against the original.
Excluded values, three levels
5/(x-2) - Set the denominator equal to zero: x − 2 = 0 .
- Solve: x = 2 .
- That value makes the original undefined, so exclude it: x ≠ 2 . In interval form, (−∞, 2) ∪ (2, ∞) .
(x^2-9)/(x^2-x-6) - Factor the original denominator: x² − x − 6 = (x − 3)(x + 2)
(x-3)(x+2), giving excluded values x = 3 and x = −2 . - Factor the numerator: x² − 9 = (x − 3)(x + 3)
(x-3)(x+3). - Cancel the common (x − 3)
x-3, leaving (x + 3)/(x + 2)(x+3)/(x+2). - But x = 3 stays excluded, since it came from the original denominator; it is now a hole in the graph. The simplified form still carries x ≠ 3 and x ≠ −2 .
(x+3)/(x+2) Restriction x ≠ 3, x ≠ −2
sqrt(x-1)/(x-4) - The radicand needs x − 1 ≥ 0 , so x ≥ 1 .
- The denominator needs x − 4 ≠ 0 , so x ≠ 4 .
- Combine both: x ≥ 1 with x ≠ 4 .
- In interval notation, punch the 4 out of the ray: [1, 4) ∪ (4, ∞) .
Restriction [1, 4) ∪ (4, ∞)
Writing the domain in both notations
Once you know the excluded values, the domain is everything else, and there are two standard ways to write it.
Inequality notation describes the allowed inputs with signs. The warm-up domain is “all real numbers except 2,” which you can write as x ≠ 2 . The exam-level domain is x ≥ 1 and x ≠ 4 . This form reads naturally and is easy to check one value at a time.
Interval notation describes the same set as a run of numbers on the line. A square bracket means the endpoint is in, a parenthesis means it is out, and infinity always takes a parenthesis, since you never actually reach it. A single excluded value in the middle of a range splits it into two pieces joined by the union symbol. So x ≠ 2 becomes (−∞, 2) ∪ (2, ∞) , and x ≥ 1 and x ≠ 4 becomes [1, 4) ∪ (4, ∞) . Notice the bracket on the 1, since the root allows it, and the parentheses around the 4, since the denominator forbids it. The bracket-versus-parenthesis choice is where the ”≥ for roots, > for logs” distinction cashes out.
A handy way to read it: each excluded value in the middle of the line acts like a scissor cut. One cut makes two intervals, two cuts make three, and each cut point gets a parenthesis on both sides because the value itself is gone. So the two excluded values of x/(x² − 16) x/(x^2-16) , namely x = 4 and x = −4 , cut the line into three pieces: (−∞, −4) ∪ (−4, 4) ∪ (4, ∞) . Count the cuts and you know how many intervals to expect before you write a single bracket. For the formal definition of a function’s domain, see Wikipedia’s Domain of a function.
How do you know you are finished?
Excluded-value work has a short, specific checklist. You are done when all of these are true:
- Every source is checked. You looked at the denominator, at any even root, and at any logarithm, not just the first one you noticed.
- The denominator is fully factored. A quadratic bottom gave you its two candidates, not one.
- The restriction is stated for the original. Any factor that cancelled still contributed its excluded value, kept as a hole.
- The domain is written in interval form, with brackets and parentheses matching whether each endpoint is included.
If any of those is missing, a restriction is probably hiding. The habit that saves the most marks is the second one: even after the expression simplifies, the excluded values come from the version you started with. The same instinct pays off in negative exponents, where a factor sitting in a denominator carries information the tidy answer can lose.
Practice: excluded values
Practice: excluded values
Try each one on paper first, then open the answers to check your working.
- State the excluded value of 3/(x + 7)
3/(x+7). - State the excluded values of x/(x² − 16)
x/(x^2-16). - State the excluded values of (x + 2)/(x² + 5x + 6)
(x+2)/(x^2+5x+6). - State the domain of √(x + 5)
sqrt(x+5). - State the domain of 1/√(x − 3)
1/sqrt(x-3). - State the excluded values of (x − 1)/((x − 1)(x + 4))
(x-1)/((x-1)(x+4)).
Show answers
- Set x + 7 = 0 : x ≠ −7 .
- Factor x² − 16 = (x − 4)(x + 4)
(x-4)(x+4): x ≠ 4, x ≠ −4 . - Factor x² + 5x + 6 = (x + 2)(x + 3)
(x+2)(x+3); the (x + 2)x+2cancels but its restriction survives: x ≠ −2, x ≠ −3 . - Radicand x + 5 ≥ 0 : x ≥ −5 , interval [−5, ∞) .
- Under a root and in a denominator, so strict: x > 3 , interval (3, ∞) .
- Factor the bottom fully; the (x − 1)
x-1cancels but stays excluded: x ≠ 1, x ≠ −4 .
The restriction travels with the answer
Simplifying changes how an expression looks, never which inputs it was allowed to take. Factor the original denominator all the way, check the roots and the logs, keep every value a cancellation tried to hide, and write the survivors as a domain in both notations. Do that and your answer arrives with its full set of caveats, the way it should.
A hole in the graph is nothing more than an excluded value that a cancellation almost got away with.