Why a Candidate List Needs a Proof
When looking for the absolute maximum or minimum of a function on a closed interval, a common strategy is to inspect the endpoints and the interior critical numbers. But why should that list be complete? Without a reason, it is only a plausible collection of places to check.
The guarantee comes from combining two results with different jobs. The Extreme Value Theorem guarantees that a continuous function on a closed interval actually attains an absolute maximum and minimum. Fermat’s theorem then restricts where an interior extremum can occur if the function is differentiable there. Together, they show that an absolute extremum cannot be missed by the candidate list.
What Counts as an Interior Critical Number?
For this result, an interior critical number is a number \(c\in(a,b)\) such that \(f'(c)=0\) or \(f'(c)\) does not exist. Since \(f\) is defined on \([a,b]\), the second possibility refers to failure of differentiability at an interior point, not failure of the function to have a value there.
If \(f\) is differentiable at every point of \((a,b)\), the only interior critical numbers are those where \(f'(c)=0\). In that important case, the candidate list consists of the two endpoints together with the interior solutions of \(f'(x)=0\).
The endpoints are included separately. The derivative test used below is about interior points, where values of the function can be considered on both sides of the point. An endpoint has no such two-sided neighborhood within the interval.
The Two Results Behind Completeness
Continuity and the closed, bounded interval are essential hypotheses of this theorem. Its conclusion is existence: it says that the largest and smallest function values are reached somewhere in the interval. On its own, it does not say where those points are.
A local maximum or minimum compares \(f(c)\) with values sufficiently close to \(c\). An absolute maximum or minimum compares \(f(c)\) with values throughout the interval. An absolute extremum at an interior point is automatically a local extremum: every sufficiently nearby point in the interval is among the points being compared.
Why Fermat’s Theorem Holds
Suppose \(f\) has a local maximum at an interior point \(c\) and is differentiable there. By the definition of derivative,
For all sufficiently small \(h\), both positive and negative, \(c+h\) remains in the domain near \(c\). At a local maximum, \(f(c+h)\leq f(c)\). When \(h>0\), dividing by \(h\) preserves the inequality, so the difference quotient is at most zero. When \(h<0\), dividing by \(h\) reverses the inequality, so the quotient is at least zero. Therefore, if the two-sided limit exists, it must be both at most zero and at least zero. It is zero.
At a local minimum the inequalities reverse: the quotient is at least zero for positive \(h\) and at most zero for negative \(h\). Again, if its two-sided limit exists, that limit must be zero. This proves Fermat’s theorem for both kinds of interior extremum.
Differentiability matters here because it ensures the two-sided derivative limit exists. If the derivative does not exist, Fermat’s theorem does not conclude that it is zero. That is why interior points where the derivative does not exist are included in the critical-number definition.
Proof That the Closed-Interval List Is Complete
Assume \(f\) is continuous on \([a,b]\). By the Extreme Value Theorem, choose a point \(x_{\max}\in[a,b]\) where \(f\) attains its absolute maximum, and a point \(x_{\min}\in[a,b]\) where it attains its absolute minimum. Consider either one of these extremizing points, denoted by \(x_*\).
Continuity on the closed, bounded interval ensures that the absolute extremum is attained at some \(x_*\in[a,b]\).
If \(x_*=a\) or \(x_*=b\), it is already on the candidate list.
If \(x_*\in(a,b)\), an absolute extremum there is also a local extremum.
If \(f'(x_*)\) exists, Fermat’s theorem gives \(f'(x_*)=0\). If it does not exist, \(x_*\) is an interior critical number by definition.
Thus every point where an absolute maximum or minimum is attained belongs to the endpoints-and-interior-critical-numbers list. This is a completeness statement: no absolute extremum can occur outside the list. It does not claim that every point on the list is an extremum.
Worked Examples: Finding the Complete Candidate Locations
Worked Example: A Quadratic on a Closed Interval
Let \(f(x)=x^2-4x+1\) on \([0,5]\). The polynomial is continuous on the closed, bounded interval, so the Extreme Value Theorem guarantees that it attains both absolute extrema. Differentiate:
Solve \(f'(x)=0\):
Since \(0<2<5\), this is an interior critical number. The complete candidate list is therefore \(0,2,5\). To check the calculation, \(f'(2)=2(2)-4=0\), and \(0\) and \(5\) are the interval endpoints. The theorem guarantees that any absolute extremum is attained at one of these listed points; forming the list alone does not decide which value is largest or smallest.
Worked Example: A Cubic with Two Interior Critical Numbers
Let \(f(x)=x^3-3x\) on \([-2,2]\). This polynomial is continuous on the interval, so its absolute extrema exist. Its derivative is
Setting the derivative equal to zero gives
Both solutions are in \((-2,2)\). The endpoints are \(-2\) and \(2\), so the complete candidate list is \(-2,-1,1,2\). Check the derivative at each interior candidate:
The endpoints are included because they bound the interval; the interior points are included because their derivatives vanish. By the completeness result, every point attaining an absolute extremum must be on this list.
Worked Example: No Interior Critical Numbers
Let \(f(x)=x\) on \([1,4]\). The function is continuous on the closed interval, so it attains an absolute maximum and minimum. Its derivative is
The equation \(f'(x)=0\) becomes \(1=0\), which has no solutions. Thus there are no interior critical numbers, and the complete candidate list consists only of the endpoints \(1\) and \(4\). The completeness theorem says that the points attaining the absolute extrema must be among these endpoints. This example shows why endpoints cannot be omitted even when the derivative never vanishes.
Worked Example: Several Interior Candidates
Let \(f(x)=(x^2-1)^2\) on \([-2,2]\). It is a polynomial, hence continuous on the interval. Differentiate using the chain rule:
The derivative is zero when one of the factors is zero:
All three numbers lie in \((-2,2)\). Including the endpoints gives the complete candidate list \(-2,-1,0,1,2\). Substitution into the derivative checks the interior candidates:
The endpoints \(-2\) and \(2\) are included by their membership in the interval boundary. The theorem ensures every absolute extremum is attained at a listed point; it does not assert that every listed point is an extremum.
What the Candidate List Does—and Does Not—Say
The result gives a necessary location condition, not a classification of every candidate. An endpoint may or may not attain an absolute extremum. An interior critical number may or may not do so. The theorem says only that the search can be restricted to this list under its hypotheses.
It is also important not to confuse the roles of the two theorems. The Extreme Value Theorem supplies existence, while Fermat’s theorem restricts the location of a differentiable interior extremum. Neither result, by itself, identifies which candidate gives the absolute maximum or minimum.
A Common Mistake: Using the Extreme Value Theorem as a Location Test
A tempting but incorrect argument is: “The function is continuous on a closed interval, so the Extreme Value Theorem says its extrema occur where \(f'(x)=0\).” That conclusion does not follow from the Extreme Value Theorem. For instance, for \(f(x)=x\) on \([0,1]\), the function is continuous and \(f'(x)=1\), so there is no point where \(f'(x)=0\). The endpoints are nevertheless on the candidate list, and the theorem guarantees the extrema are attained somewhere in the interval.
The correction is to keep the two results separate. First, use the Extreme Value Theorem to know an extremum is attained. Then, if an attaining point is interior and differentiability holds there, use Fermat’s theorem to conclude the derivative is zero. Endpoints remain candidates regardless of whether the derivative vanishes.
A second error is to treat the candidate list as a promise that every listed point is an extremum. The proof establishes the implication from “absolute extremum” to “listed candidate.” It does not establish the reverse implication. Candidate points must be examined further when the goal is to determine which values are actually absolute extrema.
Why the Hypotheses Matter
Continuity on the whole closed interval is what permits the Extreme Value Theorem to guarantee attainment. If continuity is absent, or if the interval is not closed and bounded, an absolute extremum need not be attained. In that case, there may be no point to locate using the candidate argument.
Differentiability has a more local role. At an interior extremum where the derivative exists, Fermat’s theorem forces that derivative to be zero. If the derivative fails to exist at an interior point, the zero-derivative conclusion is unavailable; that point is still included as a critical number. This is why the general candidate list includes both zero-derivative points and interior points of nondifferentiability.
Check Your Understanding
For each question, distinguish carefully between existence of an extremum and the location of a point where it is attained.
- What does the Extreme Value Theorem guarantee for a continuous function on \([a,b]\)?
- Why does an absolute extremum at an interior point also qualify as a local extremum?
- State the conclusion of Fermat’s theorem and the differentiability condition it requires.
- If \(f\) is continuous on \([a,b]\) and differentiable throughout \((a,b)\), what points belong to the complete candidate list?
- Does membership in the candidate list prove that a point is an absolute extremum? Explain the direction of the implication that the theorem establishes.
