Why the Domain Matters
For a polynomial, the derivative is defined at every real input. A rational function has a different feature: its denominator excludes some inputs from its domain. Those exclusions matter when finding critical numbers. A value that makes the derivative’s numerator zero is only a candidate; it must also belong to the domain of the original function.
This distinction is especially important when algebraic factors cancel. Simplifying an expression can make it look as though a missing input is allowed, even though it remains excluded in the function as originally given.
Definition and the Rational-Function Test
Let \(f(x)=P(x)/Q(x)\), where \(P\) and \(Q\) are polynomials and \(Q(x)\ne 0\) on the domain under consideration. At each point of this domain, the quotient rule gives
Both polynomials are differentiable everywhere, and the quotient rule applies wherever \(Q(x)\ne 0\). Thus \(f'\) exists at every point in the domain of this rational expression. Its critical numbers are precisely the domain values where its derivative equals zero.
The condition \(Q(x)\ne 0\) is not optional. It is part of the original function’s domain and must be checked after solving the numerator equation.
Why the Test Works
The quotient rule expresses the derivative as a numerator divided by \(Q(x)^2\). For every input in the domain, \(Q(x)\ne 0\), so \(Q(x)^2>0\). Dividing by this nonzero quantity cannot turn a nonzero numerator into zero, or a zero numerator into a nonzero value. Therefore, at each domain point, the derivative is zero exactly when \(N(x)=0\).
Since the quotient rule also guarantees that the derivative exists at every point where \(Q(x)\ne 0\), there are no additional critical numbers arising from a derivative that fails to exist at a domain point. A zero of \(Q\), however, is not a critical number: it is not in the function’s domain in the first place.
A Reliable Procedure
Find where the original denominator is zero. These inputs are excluded.
Use the quotient rule and identify the numerator of \(f'\).
Find all real solutions of the equation \(N(x)=0\).
Keep only solutions that belong to the original domain.
Substitute each retained value into the derivative numerator and confirm the original denominator is nonzero.
Worked Examples
Worked Example: Two Valid Critical Numbers
Find the critical numbers of \(f(x)=\frac{x^2+1}{x-1}\).
Step 1: Record the domain. The denominator is zero when \(x-1=0\), so \(x=1\) is excluded. The domain is all real numbers except \(1\).
Step 2: Differentiate. Set \(P(x)=x^2+1\) and \(Q(x)=x-1\). Then \(P'(x)=2x\) and \(Q'(x)=1\). The quotient rule gives
Step 3: Solve the numerator equation.
Step 4: Check the domain. Neither value is \(1\), since \(1+\sqrt{2}\ne 1\) and \(1-\sqrt{2}\ne 1\). Both are in the domain.
Verification. For \(x=1\pm\sqrt{2}\), substituting into the numerator gives
Also, \(x-1=\pm\sqrt{2}\ne 0\), so the derivative is defined at both inputs. The critical numbers are \(1-\sqrt{2}\) and \(1+\sqrt{2}\).
Worked Example: A Numerator Root That Must Be Rejected
Find the critical numbers of \(f(x)=\frac{x^2-1}{x-1}\), using the function as given.
Step 1: Record the domain. The original denominator vanishes at \(x=1\), so the domain excludes \(1\).
Step 2: Differentiate using the quotient rule. Here \(P(x)=x^2-1\), \(Q(x)=x-1\), \(P'(x)=2x\), and \(Q'(x)=1\). Therefore
For every input in the domain, this derivative equals \(1\). The numerator equation \((x-1)^2=0\) produces the candidate \(x=1\), but that candidate is excluded from the domain.
Verification and conclusion. For every allowed \(x\),
Although the expression simplifies to \(f(x)=x+1\) when \(x\ne 1\), the original function is still undefined at \(1\). The candidate is not a critical number, and there are no critical numbers.
Worked Example: A Critical Number Between Two Exclusions
Find the critical numbers of \(f(x)=\frac{x^2}{x^2-1}\).
Step 1: Record the domain. The denominator factors as
Thus \(x=-1\) and \(x=1\) are excluded.
Step 2: Differentiate. Take \(P(x)=x^2\) and \(Q(x)=x^2-1\), so \(P'(x)=2x\) and \(Q'(x)=2x\). Then
Step 3: Solve the numerator equation. The equation \(-2x=0\) gives \(x=0\).
Step 4: Check and verify. Since \(0\ne -1\) and \(0\ne 1\), the candidate is in the domain. Substitution confirms
Therefore, \(0\) is the only critical number. The derivative is defined there, and the original function is defined there.
Common Mistakes
A frequent error is to count every solution of the derivative-numerator equation as a critical number. The second example shows why that fails: the equation \((x-1)^2=0\) gives \(x=1\), but the original denominator is also zero there. That input is not in the domain, so it cannot be a critical number.
Another error is to cancel a factor and then silently enlarge the domain. In the expression \(\frac{x^2-1}{x-1}\), cancellation yields \(x+1\) only for \(x\ne 1\). The canceled expression describes the same values as the original quotient on its domain; it does not assign a value to the original function at the missing input.
Why This Distinction Matters
Critical numbers are used to identify inputs where a function’s local behavior may need closer examination. For a rational function, the domain may split into separate intervals at poles or holes. Those excluded inputs can be important when studying the function, but they are not themselves critical numbers because the function has no value there.
The calculation therefore has two logically separate parts: use the derivative to identify possible zero-derivative inputs, and use the original domain to determine which candidates are valid. Keeping those checks separate prevents algebraic simplification from changing the function being analyzed.
Check Your Understanding
For each question, pay attention to both the derivative equation and the original domain.
- For \(f(x)=\frac{x+2}{x-3}\), what equation identifies the candidates for critical numbers, and which input must be excluded before solving?
- Why does a real solution of \(P'(x)Q(x)-P(x)Q'(x)=0\) fail to be a critical number when \(Q(x)=0\) there?
- For \(f(x)=\frac{x^2-4}{x-2}\), does cancellation make \(x=2\) part of the original domain? Explain.
- What property of \(Q(x)^2\) on the domain lets us conclude that \(f'(x)=0\) exactly when the derivative numerator is zero?
- Find the critical numbers of \(f(x)=\frac{x}{x^2+1}\), and verify that each answer is in the domain.
