Why the Domain Must Come First
A derivative equation can produce values that are not inputs to the function being studied. Those values cannot be critical numbers, even if a derivative formula can be written or evaluated there using an algebraic expression.
For example, the expression \(x^2-4x\) is defined for every real number. But if the function is restricted to inputs between 3 and 5, then values such as \(x=2\) are not part of that function’s domain. Finding a critical number therefore requires two checks: identify where the derivative is zero or does not exist, and then verify that each candidate belongs to the domain.
A critical number is an input value, not the corresponding output \(f(c)\). The distinction matters: \(c\) identifies a location in the domain, while \(f(c)\) is the function value there.
A Reliable Search Procedure
There are two possible sources of critical numbers:
- domain inputs where the derivative equals zero;
- domain inputs where the derivative does not exist.
The domain is a filter on both sources. A zero obtained by solving \(f'(x)=0\) counts only if \(f\) is defined at that input. Likewise, an input where a derivative formula is undefined counts only if the original function is defined there.
Use the actual function and its restrictions, not just the domain of a formula that resembles it.
Use the appropriate differentiation rules on the function’s domain.
Keep only solutions that are in the domain.
Inspect domain inputs where a formula for the derivative is undefined or where the function has a corner or another failure of differentiability.
Combine the qualifying values without including excluded inputs.
Why Solving the Derivative Equation Is Only Part of the Work
The definition itself explains the procedure. At any input \(c\) in the domain, the definition says that \(c\) is critical precisely when at least one of two conditions holds: the derivative is zero, or the derivative does not exist. Thus solving \(f'(x)=0\) finds one type of candidate, but cannot find points where the derivative fails. Checking the domain is necessary because the definition requires \(c\) to be an input of \(f\).
Here is a direct derivation that illustrates why restrictions matter. Let \(f(x)=x^2-4x\) for \(3<x<5\). For any \(x\) in this open interval, sufficiently small nonzero \(h\) keeps \(x+h\) in the interval. The difference quotient can therefore be simplified using values of this restricted function:
This derivative calculation applies to inputs in the stated domain. The expression \(2x-4\) is also algebraically defined outside that interval, but those values are not derivatives of this restricted function at inputs outside its domain.
Worked Examples
Worked Example: A Derivative Zero Outside the Restricted Domain
Find the critical numbers of \(f(x)=x^2-4x\), with domain \(D=(3,5)\).
Differentiate on the domain. The derivative formula is \(f'(x)=2x-4\). Set it equal to zero:
But \(2\notin(3,5)\), so it is not an input of this function and cannot be a critical number. The derivative formula \(2x-4\) is defined for every real number; in particular, the derivative of the restricted function exists at every \(x\in(3,5)\), so there are no domain points where it fails.
Check: The only solution to \(f'(x)=0\) is \(2\), and \(2\) is outside \(D\). Every point of \(D\) has a derivative, so there are no other candidates. The function has no critical numbers on this restricted domain.
Worked Example: A Domain Point Where the Derivative Does Not Exist
Find the critical numbers of \(f(x)=\sqrt{x-1}\) on its natural real domain.
The square root requires \(x-1\geq 0\), so the domain is \(D=[1,\infty)\). For \(x>1\), the chain rule gives
This derivative is never zero for \(x>1\). At \(x=1\), the function is defined, but the derivative has no finite value. This can also be checked from the difference quotient available from within the domain:
As \(h\) approaches zero through positive values, \(1/\sqrt{h}\) grows without bound, so there is no finite derivative at \(1\). Because \(1\in D\), it is a critical number.
Check: The derivative is positive and finite for every \(x>1\), so it provides no zero or failure points in the interior. At \(x=1\), the function is defined and its derivative does not exist. Therefore the critical-number set is \(\{1\}\).
Worked Example: A Derivative Formula Fails Only at an Excluded Input
Find the critical numbers of \(f(x)=\dfrac{x+2}{x-1}\).
The denominator cannot be zero, so the domain is \(D=\mathbb{R}\setminus\{1\}\). For \(x\in D\), the quotient rule gives
The numerator is \(-3\), so \(f'(x)\) is never zero wherever it is defined. The displayed derivative formula is undefined at \(x=1\), but \(1\) is excluded from the domain of \(f\). It is not a critical number.
Check: For each \(x\in D\), \((x-1)^2>0\), making \(f'(x)=-3/(x-1)^2\) defined and nonzero. The sole excluded input is not eligible. Hence this function has no critical numbers.
Worked Example: A Corner Inside the Domain
Find the critical numbers of \(f(x)=|x-2|\), with domain \(D=(0,4)\).
For \(x<2\), \(x-2<0\), so \(f(x)=2-x\) and \(f'(x)=-1\). For \(x>2\), \(x-2>0\), so \(f(x)=x-2\) and \(f'(x)=1\). Neither derivative value is zero.
At \(x=2\), the function is defined. The left-hand difference quotient is \(-1\), while the right-hand difference quotient is \(1\):
Since the two one-sided limits differ, the derivative at \(2\) does not exist. The point \(2\) is in the domain, so it is a critical number.
Check: On \(0<x<2\), the derivative is \(-1\), and on \(2<x<4\), it is \(1\); neither interval contributes a zero or a derivative failure. At \(x=2\), the function is defined but not differentiable. Thus the critical-number set is \(\{2\}\).
Common Mistake: Keeping Every Root or Undefined Value
A frequent error is to differentiate an expression, solve the equation, and report every algebraic answer without comparing it to the function’s domain. Another is to call a point critical merely because the derivative formula is undefined there, even when the original function is not defined there.
For the restricted function \(f(x)=x^2-4x\), \(3<x<5\), the incorrect calculation is:
The conclusion is wrong because \(2\notin(3,5)\). The corrected conclusion is that \(x=2\) is an algebraic solution of the derivative equation but not a critical number of this restricted function. Since the derivative exists throughout its domain and has no zero there, the function has no critical numbers on \((3,5)\).
Why Critical Numbers Matter
Critical numbers identify domain inputs where the derivative-based description of a function’s behavior may need special attention. A derivative may be zero at a point, or the function may be defined there but not differentiable. Finding these inputs is useful in later work on extrema and derivative behavior.
However, a critical number is not automatically a local maximum or minimum. The definition identifies candidates; further reasoning is needed to classify what happens around them. Conversely, a domain endpoint may be relevant to an optimization problem even if it is not treated as an interior critical number. Always follow the definition and the specific question’s instructions.
Before reporting an answer, check these questions:
- Is the candidate actually in the domain?
- Does the derivative equal zero there?
- If not, does the derivative fail to exist there?
- Was a denominator-zero input incorrectly included even though the function is undefined there?
- Is the question asking for critical numbers, or does it also require checking endpoints separately?
Check Your Understanding
For each question, use the function’s stated domain when deciding which values qualify.
- Find the critical numbers of \(f(x)=x^2-6x\) on \(D=(4,8)\).
- For \(g(x)=\sqrt{5-x}\) on its natural real domain, identify the domain and find its critical numbers.
- For \(h(x)=\dfrac{1}{x+3}\), explain why \(x=-3\) is not a critical number.
- Find the critical numbers of \(p(x)=|x+1|\) on \(D=(-2,2)\), and justify whether the derivative exists at each candidate.
- In one sentence, explain why a solution of \(f'(x)=0\) outside the domain is not a critical number.
