Why the Domain Matters
A derivative calculation can produce a list of inputs where the derivative is zero or fails to exist. That list is not automatically the list of critical numbers. Each candidate must also be an input at which the function itself is defined.
This condition prevents a common error: treating a hole, a denominator zero, or another excluded input as a critical number simply because a formula involving the derivative behaves unusually there. Critical numbers belong to the function’s domain.
Definition of a Critical Number
The domain of a function \(f\), written \(\operatorname{Dom}(f)\), is the set of inputs at which \(f\) is defined. For this tutorial, derivatives are understood in the usual two-sided sense. In particular, a derivative may fail to exist at a domain endpoint even though the function is defined there.
There are two ways for an input in the domain to meet the derivative condition: the derivative exists and equals zero, or the derivative does not exist. But domain membership is required in both cases. If \(f(c)\) is undefined, then \(c\) is not a critical number, regardless of what a simplified expression or an algebraic calculation might suggest.
When looking for critical numbers, it is therefore useful to keep three questions separate:
- Is the input in the domain of the original function?
- At domain inputs, where does the derivative equal zero?
- At which other domain inputs does the derivative fail to exist?
The first question is a gatekeeper: an input excluded by the original function cannot pass the definition.
Why an Excluded Input Cannot Be Critical
The derivative at \(c\) is defined, when the relevant limit exists, by the difference quotient
This expression uses \(f(c)\). If \(c\notin\operatorname{Dom}(f)\), then \(f(c)\) is not defined, so this difference quotient does not define a derivative at \(c\). The phrase “the derivative does not exist” in the definition of a critical number applies to an input where the function is defined but its derivative is not; it does not turn an excluded input into a critical number.
There is an important contrast here. At a domain endpoint, \(f(c)\) may be defined while the ordinary two-sided derivative is not. Such an input can qualify as critical under the definition above. At an excluded point, \(f(c)\) itself is unavailable, so it cannot qualify.
Worked Example: Zero Derivatives of a Polynomial
Worked Example: Zero Derivatives of a Polynomial
Find the critical numbers of \(f(x)=x^3-3x\).
Step 1: Check the domain. A polynomial is defined for every real input, so \(\operatorname{Dom}(f)=\mathbb{R}\).
Step 2: Differentiate.
Step 3: Solve \(f'(x)=0\).
Both inputs belong to the domain. The derivative is defined everywhere, so there are no additional candidates from derivative nonexistence.
Check: \(f'(-1)=3(-1)^2-3=3-3=0\), and \(f'(1)=3(1)^2-3=3-3=0\). Therefore the critical numbers are \(-1\) and \(1\). The function values are \(f(-1)=-1+3=2\) and \(f(1)=1-3=-2\), which confirms that the function is defined at both inputs.
When the Function Is Defined but the Derivative Is Not
A critical number does not have to make the derivative equal to zero. It can also be an input in the domain where the derivative fails to exist. To apply this part of the definition, verify that the function value exists first.
Worked Example: A Domain Input with No Derivative
Worked Example: A Domain Input with No Derivative
Find the critical numbers of \(g(x)=x^{2/3}\), interpreted for real \(x\).
Step 1: Check the domain. The cube root of every real number is real, and squaring that value is defined. Thus \(\operatorname{Dom}(g)=\mathbb{R}\), including \(x=0\).
Step 2: Differentiate away from zero. For \(x\ne 0\), the power rule gives
This derivative is never zero where it is defined. At \(x=0\), the displayed derivative formula has a zero denominator, so it does not give a derivative there. We must check the derivative definition because \(g(0)\) is defined.
As \(h\to0^+\), \(1/h^{1/3}\) grows without bound, so the difference quotient has no finite limit. Therefore \(g'(0)\) does not exist.
Check: The input \(0\) is in the domain and its derivative does not exist, so \(0\) is a critical number. For \(x\ne0\), the derivative is defined and nonzero, so there are no other critical numbers.
Cancellation Does Not Change the Original Domain
Algebraic simplification can make a formula look defined at an input where the original function was not. The domain must be determined from the function as given, before cancellation. A simplified expression can help calculate values or derivatives on the domain, but it does not automatically add missing inputs.
Worked Example: A Hole Is Not a Critical Number
Worked Example: A Hole Is Not a Critical Number
Find the critical numbers of \(r(x)=\frac{x^2-1}{x-1}\).
Step 1: Identify the domain of the original function. Its denominator is zero at \(x=1\), so
Step 2: Simplify only for inputs in the domain. Since \(x^2-1=(x-1)(x+1)\), for \(x\ne1\) we have
Thus \(r'(x)=1\) for every \(x\ne1\). This derivative is never zero, and it exists at every input in the domain. There are no critical numbers.
Check the excluded input: The simplified expression \(x+1\) would give \(2\) at \(x=1\), but the original function has denominator \(1-1=0\) there. Hence \(r(1)\) is undefined, so \(1\) is not a critical number. Cancellation does not fill the hole in the original function.
Worked Example: Zero Derivatives and an Excluded Input
Worked Example: Zero Derivatives and an Excluded Input
Find the critical numbers of \(q(x)=\frac{(x-1)^2}{x-2}\).
Step 1: Check the domain. The denominator is zero when \(x=2\), so \(\operatorname{Dom}(q)=\mathbb{R}\setminus\{2\}\).
Step 2: Differentiate on the domain. Using the quotient rule,
Factor the numerator:
Therefore, for \(x\ne2\),
Step 3: Find and check the candidates. The derivative is zero when \(x=1\) or \(x=3\). Both are in the domain. At \(x=2\), the original function is undefined; it is not an additional critical number.
Check: \(q(1)=0\) and \(q(3)=\frac{(3-1)^2}{3-2}=4\), so both candidate inputs are in the domain. Substitution into the derivative gives \(q'(1)=\frac{(1-1)(1-3)}{(1-2)^2}=0\) and \(q'(3)=\frac{(3-1)(3-3)}{(3-2)^2}=0\). Thus the critical numbers are \(1\) and \(3\), not \(2\).
A Reliable Check for Candidate Inputs
For a function whose formula has restrictions, keep the domain and derivative checks visible throughout the calculation. A derivative formula may have a denominator, root, or other restriction of its own, and its algebraic candidates still have to be checked against the original function.
Record inputs excluded by the given function before simplifying its formula.
Use the derivative formula only on inputs where it applies.
Solve \(f'(x)=0\), then discard any input outside the original domain.
Include an input only if the function is defined there but its derivative is not.
Common Mistake: Keeping a Denominator Zero as a Critical Number
Consider \(r(x)=\frac{x^2-1}{x-1}\). A mistaken approach cancels \(x-1\), obtains \(r(x)=x+1\), and then says, “The derivative is \(1\) except where the original denominator was zero; therefore the derivative does not exist at \(x=1\), so \(1\) is critical.” The conclusion is wrong: \(r(1)\) is not defined, so the definition of a critical number rules it out.
The correction is to preserve the original domain, \(\mathbb{R}\setminus\{1\}\). On that domain \(r(x)=x+1\) and \(r'(x)=1\). There are no critical numbers. An excluded input is not a point where the derivative fails for a domain value; it is a point where the function itself has no value.
Why the Domain Check Is Useful
Critical numbers are inputs at which the derivative condition in the definition is met while the function remains defined. This makes the domain check essential whenever a formula has restrictions or has been simplified. It also keeps two different situations distinct: a function may have a defined value but no derivative there, or it may have no value at all. Only the first situation can contribute a critical number.
When using critical numbers in later analysis, this distinction ensures that candidate lists refer to actual inputs of the function. A derivative equation is a tool for finding candidates, not a substitute for the definition.
Check Your Understanding
For each question, apply both parts of the definition: the derivative condition and membership in the function’s domain.
- State the definition of a critical number, including its domain requirement.
- If \(f(c)\) is undefined, can \(c\) be a critical number merely because a derivative formula is undefined there? Explain.
- For \(p(x)=x^2\), find all critical numbers and verify them using \(p'(x)\).
- For \(s(x)=\frac{x^2-4}{x-2}\), explain why \(x=2\) is or is not a critical number.
- Can a function be defined at an input where its derivative does not exist? Explain how that input is treated by the definition.
