Why Domain Restrictions Matter
A formula can look meaningful for many inputs but fail at particular values. For example, the expression \(\frac{1}{x-2}\) has a real value for most real numbers \(x\), but not for \(x=2\): substituting \(2\) makes the denominator zero. That input must be excluded from the domain.
For a polynomial, every real input is allowed. A rational expression is different because it involves division by a polynomial, and division by zero is undefined. The first task when working with a rational expression is therefore to identify the real inputs that make its denominator zero.
Rational Expressions and Their Domains
A polynomial is an expression formed from real constants and nonnegative integer powers of a variable using addition, subtraction, and multiplication. A rational expression in \(x\) is a quotient of two polynomials, written \(\frac{P(x)}{Q(x)}\), where \(P\) and \(Q\) are polynomials and \(Q\) is not the zero polynomial.
When we treat this expression as a real-valued function, its natural domain is the set of real inputs for which the displayed quotient is defined. The denominator must be nonzero. The numerator may be zero; that does not make the quotient undefined.
This definition concerns the expression as it is written. Finding the domain is not the same task as finding where the numerator is zero, nor is it a question about the graph’s appearance. Begin with the denominator and determine exactly which real inputs make it zero.
Why the Denominator Determines the Domain
The definition follows directly from what division means. At an input \(x=a\), the value of \(\frac{P(x)}{Q(x)}\) would have to be the quotient of \(P(a)\) by \(Q(a)\). Division by a real number is defined when the divisor is nonzero. Thus, if \(Q(a)\ne 0\), the quotient is defined; if \(Q(a)=0\), it is not.
This gives a complete domain test: solve \(Q(x)=0\) over the real numbers, then remove those solutions from \(\mathbb{R}\). If the denominator has no real zeros, there are no real inputs to exclude.
A Reliable Procedure
For a rational expression in one real variable, use this sequence:
Do not use the numerator to decide which inputs are excluded.
Solve this equation over the real numbers. Factoring can help when the denominator is a polynomial that factors.
Substitute each candidate into the original denominator and verify that it gives zero.
Write the real numbers with all denominator zeros excluded, or list the allowed values explicitly.
Writing the restriction down immediately is useful even if you intend to perform more algebra afterward. It preserves the information about where the original expression was defined.
Worked Examples
Worked Example: A Difference of Squares in the Denominator
Find the real domain of \(\frac{x+2}{x^2-9}\).
Step 1: Identify the denominator. The denominator is \(x^2-9\). It must not equal zero.
Step 2: Find its zeros. Factor using the difference of squares:
Set each factor equal to zero. From \(x-3=0\), we obtain \(x=3\). From \(x+3=0\), we obtain \(x=-3\).
Step 3: Check both candidates in the original denominator.
Both values make the denominator zero, so both must be excluded. The numerator does not change these restrictions.
Domain:
Check an allowed input. At \(x=0\), the denominator is \(0^2-9=-9\), which is nonzero, so the expression is defined there. The exclusions are exactly the two denominator zeros found above.
Worked Example: A Denominator with No Real Zeros
Find the real domain of \(\frac{5}{x^2+4}\).
Step 1: Identify the denominator. It is \(x^2+4\).
Step 2: Determine whether it can be zero for real \(x\). For every real \(x\), \(x^2\ge 0\). Therefore,
So \(x^2+4\) is never zero for real \(x\), and there are no real denominator zeros to exclude.
Domain:
Check. For example, at \(x=0\), the denominator is \(0^2+4=4\), so the quotient is \(\frac{5}{4}\), a defined real number. The inequality above verifies that the denominator remains nonzero for every real input, not only this test value.
Worked Example: A Numerator Zero Does Not Remove a Restriction
Find the real domain of \(\frac{x-3}{x^2-3x+2}\).
Step 1: Use the denominator only. The denominator must be nonzero, so solve \(x^2-3x+2=0\).
Step 2: Factor and solve.
The equation \((x-1)(x-2)=0\) gives \(x=1\) or \(x=2\).
Step 3: Check the denominator at both values.
Thus both values are excluded. Notice that the numerator is zero at \(x=3\), but this does not exclude \(x=3\): the numerator can be zero while the denominator is nonzero.
Domain:
Check a numerator zero that is allowed. At \(x=3\), the numerator is \(3-3=0\), while the denominator is \(3^2-3(3)+2=9-9+2=2\). The quotient is \(0/2=0\), which is defined. This confirms that numerator zeros are not domain restrictions.
State Restrictions Before Further Algebra
Domain restrictions should be recorded from the original denominator before you simplify an expression or use it in an equation. The reason is that any later algebra is performed on an expression whose original domain has already been determined. If the restriction is forgotten, a later answer may include an input at which the original expression had no value.
For instance, in the third example, \(x=1\) is excluded because it makes the displayed denominator zero. The fact that the numerator also vanishes there does not make division by zero defined. A numerator value of zero is permitted; a denominator value of zero is not.
When an expression is part of an equation, first record the restrictions on each rational expression in that equation. Any proposed solution must be among the allowed inputs for the original expressions. This is a domain check, separate from the algebra used to solve the equation.
Common Mistake: Checking the Numerator Instead
A frequent error is to exclude the zeros of the numerator instead of the zeros of the denominator. Consider \(\frac{x-3}{x^2-3x+2}\). A mistaken calculation might set the numerator equal to zero, find \(x-3=0\), and report that \(x=3\) is excluded. That is incorrect: a zero numerator produces a quotient of zero when the denominator is nonzero.
The correction is to set the denominator equal to zero:
Substitution confirms that the denominator is zero at \(1\) and \(2\), while at \(3\) it equals \(2\). Therefore, \(1\) and \(2\) are excluded, and \(3\) is allowed. Always ask: Which values make the denominator zero?
Summary
A rational expression is defined at a real input precisely when its denominator is nonzero there. To find its natural real domain, identify the original denominator, solve the equation that sets it equal to zero, check the candidate values, and exclude those values from \(\mathbb{R}\). Zeros of the numerator do not by themselves restrict the domain.
Check Your Understanding
For each question, focus on the original denominator and state any real inputs that must be excluded.
- What condition must hold for \(\frac{P(x)}{Q(x)}\) to be defined at a real input \(x=a\)?
- Find the real domain of \(\frac{2x+1}{x^2-16}\).
- Find the real domain of \(\frac{7}{x^2+1}\). Explain why there are no real exclusions.
- For \(\frac{x-5}{x^2-5x+6}\), which values are excluded, and is \(x=5\) allowed?
- Why should restrictions be recorded before further algebra is performed with a rational expression?
