Why a Careful Candidate List Matters
A calculus problem may ask for critical numbers on a closed interval, or it may require them as part of a later analysis. The calculation is not complete when you solve \(f'(x)=0\). You must also find points where the derivative does not exist, check that each point belongs to the function’s domain, and remove points outside the interval.
It is also important to distinguish interior critical numbers from the endpoints of the interval. The endpoints are included in a closed-interval analysis for a separate reason: they are part of the interval, even though the interior critical-number condition does not ask you to test them.
This distinction is useful even if an endpoint also meets the general critical-number condition. It remains an endpoint in the closed-interval list and must be considered as an endpoint. The interior procedure below focuses on \(a<c<b\).
A Systematic Procedure
For a function considered on \([a,b]\), first confirm that the function is defined at the points you are examining. Then find all interior points where the derivative is zero or fails to exist. Finally, retain only points in the domain and in \((a,b)\). A candidate that comes from manipulating a derivative formula is not automatically a critical number: it must pass both checks.
Write down the interval \([a,b]\) and identify where the original function \(f\) is defined.
Find \(f'(x)\) on the parts of the domain where the derivative exists.
Solve \(f'(x)=0\), and separately find domain points where \(f'\) does not exist.
Keep only values where \(f\) is defined and that lie strictly between \(a\) and \(b\).
Do not describe \(a\) or \(b\) as interior critical numbers. Keep them visible as endpoints of the closed interval.
The derivative equation and the nondifferentiability check are separate tasks. For example, a cusp may give a critical number even though there is no equation \(f'(x)=0\) to solve at that point. Conversely, a denominator that vanishes in a derivative formula does not give a critical number if the original function is also undefined there.
Why These Interior Points Are the Candidates
The derivative condition is not an arbitrary recipe. At a differentiable interior local maximum or minimum, the derivative must be zero. Here is a short proof for a local maximum. Suppose \(f\) has a local maximum at an interior point \(c\), and \(f'(c)\) exists. For sufficiently small positive \(h\), \(f(c+h)\leq f(c)\), so
For sufficiently small negative \(h\), the numerator is still nonpositive but the denominator is negative, so
Because the derivative exists, both one-sided limits of these difference quotients equal \(f'(c)\). The right-hand quotients are nonpositive and the left-hand quotients are nonnegative, so their common limit must be zero. Thus \(f'(c)=0\). The same reasoning with inequalities reversed proves the result for a local minimum.
This explains why both zero-derivative points and points of nondifferentiability belong in the candidate list. It does not say that every critical number is an extremum. Classifying what happens at those points requires additional analysis.
Worked Example: Finding Zero-Derivative Points
Worked Example: Finding Zero-Derivative Points
Find the interior critical numbers of \(f(x)=x^3-3x\) on \([-2,2]\), and distinguish them from the endpoints.
The polynomial is defined and differentiable for every real \(x\). Differentiate:
Set the derivative equal to zero and solve:
Both values are in the domain and in the interval’s interior, since \(-2<-1<2\) and \(-2<1<2\). Check them in the derivative:
The derivative exists everywhere, so there are no additional candidates from nondifferentiability. The interior critical numbers are \(-1\) and \(1\). The interval endpoints are \(-2\) and \(2\), which are listed separately. As an endpoint check, \(f'(-2)=3(4)-3=9\) and \(f'(2)=3(4)-3=9\); neither endpoint solves the derivative-zero condition.
Worked Example: Including a Point Where the Derivative Does Not Exist
Worked Example: Including a Point Where the Derivative Does Not Exist
Find the interior critical numbers of \(g(x)=|x-1|\) on \([0,3]\).
For \(x<1\), \(x-1<0\), so \(g(x)=1-x\) and \(g'(x)=-1\). For \(x>1\), \(x-1>0\), so \(g(x)=x-1\) and \(g'(x)=1\). The derivative is therefore never zero where these formulas apply.
At \(x=1\), the function is defined, with \(g(1)=0\). To check differentiability there, compute the difference quotient. If \(h<0\), then \(|h|=-h\), and if \(h>0\), then \(|h|=h\):
The left-hand and right-hand limits differ, so \(g'(1)\) does not exist. Since \(1\) is in the domain and \(0<1<3\), it is an interior critical number. There are no other interior candidates. The endpoints \(0\) and \(3\) are kept separate; the one-sided formulas near them have slopes \(-1\) and \(1\), respectively.
Check the Original Domain
A critical number must be a point where the function itself is defined. This matters when formulas involve roots, denominators, or logarithms. Differentiating may produce an expression that is undefined at a value where the original function is also undefined. That value is not a critical number: the derivative condition only applies at points in the function’s domain.
Likewise, a point where the derivative is undefined can still be irrelevant to a particular interval. If it lies outside the interval, it is not an interior critical number for that interval. The next example makes both checks explicit.
Worked Example: Excluding a Derivative Problem Outside the Interval
Find the interior critical numbers of \(h(x)=\dfrac{1}{\sqrt{x+1}}\) on \([0,2]\).
The original function requires \(x+1>0\), because the square root is in the denominator. Thus its domain is \(x>-1\), which includes the entire interval \([0,2]\). Differentiate using the power rule:
The derivative cannot equal zero where it is defined, because its numerator is \(-1\). Its formula is undefined at \(x=-1\), but the original function is also undefined there, and \(-1\) lies outside \([0,2]\). Therefore it is not a critical number on the interval. At every \(x\) in \((0,2)\), the derivative exists and is nonzero. There are no interior critical numbers. The endpoints \(0\) and \(2\) are listed separately.
As a check, \(h'(0)=-\frac12\) and \(h'(2)=-\frac{1}{2(3)^{3/2}}=-\frac{1}{6\sqrt3}\); both are defined and nonzero. These endpoint values do not change the conclusion about the interior.
Common Mistakes and Corrections
A frequent error is to report only solutions of \(f'(x)=0\). For \(g(x)=|x-1|\), that approach finds no solutions because the derivative is \(-1\) to the left of \(1\) and \(1\) to the right. But it misses \(x=1\), where the function is defined and the derivative does not exist. The correction is to check separately for points of nondifferentiability.
Another error is to include every value that makes a derivative expression undefined. For \(h(x)=1/\sqrt{x+1}\), the derivative expression is undefined at \(-1\), but the function is not defined there, and the point is outside \([0,2]\). Including \(-1\) as a critical number would violate both the domain check and the interval check.
A third error is to put endpoints into the interior critical-number list simply because the task concerns a closed interval. The endpoint values matter in a closed-interval analysis, but their role is different: they are included because they are endpoints. Keep the interior critical numbers and the endpoints visibly separate.
Check Your Understanding
For each question, remember to check the original domain and distinguish interior values from endpoints.
- Find the interior critical numbers of \(f(x)=x^2-4x\) on \([0,5]\).
- For \(g(x)=|x+2|\) on \([-3,0]\), which interior point makes the derivative fail to exist?
- For \(h(x)=\sqrt{x}\) on \([1,4]\), is the derivative’s failure to exist at \(x=0\) relevant to the interior critical-number list? Explain.
- Why is a value that makes the derivative formula undefined not automatically a critical number?
- State separately what you would report for the interior critical numbers and endpoints of a function on \([a,b]\).
