A Complete Process for Closed-Interval Extrema
An absolute-extrema problem asks for the greatest and least function values on a specified interval. The calculation can involve different kinds of functions, but the logic is consistent: first check that the function is continuous on the whole closed interval, then locate endpoints and interior points where an extremum could occur, and finally compare the function values.
The comparison is straightforward when there are finitely many candidates. The method itself does not guarantee a finite candidate set, however. Continuity guarantees that absolute extrema exist on a closed bounded interval; it does not guarantee that there are only finitely many critical numbers.
For a continuous function on a closed bounded interval \([a,b]\), the Extreme Value Theorem guarantees that both absolute extrema exist. To locate them, include the endpoints \(a\) and \(b\), along with interior critical numbers: points \(c\in(a,b)\) where \(f'(c)=0\) or where \(f'(c)\) does not exist, provided \(f(c)\) is defined.
Why Interior Extrema Are Candidates
The reason for checking interior critical numbers is Fermat’s theorem: if \(f\) has a local maximum or minimum at an interior point \(c\), and \(f\) is differentiable at \(c\), then \(f'(c)=0\). If \(f\) is not differentiable there, the point is still a candidate. An absolute extremum on \([a,b]\) must therefore occur at an endpoint or at an interior critical number.
For completeness, suppose \(c\) is an interior local maximum and \(f'(c)\) exists. For sufficiently small \(h>0\), \(f(c+h)-f(c)\le 0\), so the right-hand difference quotient is at most zero. For sufficiently small \(h<0\), \(f(c+h)-f(c)\le 0\), and division by \(h<0\) makes the left-hand difference quotient at least zero. Since both limits equal \(f'(c)\), it follows that \(f'(c)\le0\) and \(f'(c)\ge0\); hence \(f'(c)=0\). Reversing the inequalities proves the same conclusion for a local minimum.
A Reliable Workflow
Confirm that the function is defined and continuous throughout the closed interval. For rational functions, check that the denominator is never zero there. For piecewise functions, check the values at joins.
Find interior points where the derivative is zero or does not exist, including relevant piecewise joins.
Find the function value at every candidate. Use exact values when possible.
The greatest candidate value is the absolute maximum and the least is the absolute minimum. Report every point where each value occurs.
Worked Polynomial Example
Worked Example: A Cubic on a Closed Interval
Find the absolute maximum and minimum of \(f(x)=x^3-3x\) on \([-2,2]\).
Continuity: A polynomial is continuous for every real \(x\), so it is continuous on \([-2,2]\). The Extreme Value Theorem guarantees both extrema exist.
Interior candidates: Differentiate and solve:
Thus \(f'(x)=0\) at \(x=-1\) and \(x=1\), both interior points. The full candidate list is \(-2,-1,1,2\).
Evaluate:
Compare and check: The greatest value is \(2\), occurring at \(x=-1\) and \(x=2\). The least value is \(-2\), occurring at \(x=-2\) and \(x=1\). Substituting those points into \(x^3-3x\) gives exactly the listed values, confirming both extrema and their locations.
Worked Rational Example
Worked Example: Checking the Denominator Before Differentiating
Find the absolute extrema of \(f(x)=\frac{x+1}{x^2+1}\) on \([0,2]\).
Continuity: The denominator \(x^2+1\ge1>0\) for every real \(x\). Thus the rational function is defined and continuous throughout \([0,2]\), and the Extreme Value Theorem applies.
Interior candidates: By the quotient rule,
The denominator of \(f'(x)\) is positive. Setting its numerator to zero gives \(x^2+2x-1=0\), so \(x=-1\pm\sqrt{2}\). Only \(c=-1+\sqrt{2}\) lies in \((0,2)\). The candidates are \(0,c,2\).
Evaluate:
At \(c=\sqrt{2}-1\), the numerator is \(\sqrt{2}\), and the denominator is \((\sqrt{2}-1)^2+1=4-2\sqrt{2}\). Therefore
Compare and check: Since \(\frac{\sqrt{2}+1}{2}>1>\frac35\), the absolute maximum is \(\frac{\sqrt{2}+1}{2}\) at \(x=\sqrt{2}-1\), and the absolute minimum is \(\frac35\) at \(x=2\). Substitution into the original function gives \(f(c)=\frac{\sqrt{2}+1}{2}\) and \(f(2)=\frac35\), as calculated.
Worked Radical Example
Worked Example: A Radical with an Endpoint Derivative Issue
Find the absolute maximum and minimum of \(f(x)=\sqrt{x+1}-x\) on \([-1,3]\).
Continuity: The radicand \(x+1\) is nonnegative throughout the interval. The square-root function and the function \(-x\) are continuous on their relevant domains, so their sum is continuous on \([-1,3]\). Both extrema exist.
Interior candidates: For \(x>-1\),
Solving \(f'(x)=0\) gives \(\frac{1}{2\sqrt{x+1}}=1\), hence \(\sqrt{x+1}=\frac12\), \(x+1=\frac14\), and \(x=-\frac34\). The derivative does not exist at \(x=-1\), but that point is an endpoint and is already included. The candidate list is \(-1,-\frac34,3\).
Evaluate:
Compare and check: The absolute maximum is \(\frac54\) at \(x=-\frac34\), and the absolute minimum is \(-1\) at \(x=3\). Substitution into \(\sqrt{x+1}-x\) reproduces these values. The derivative issue at the left endpoint does not remove it from consideration: endpoints must be checked whether or not the derivative exists there.
Worked Piecewise Example
Worked Example: A Nondifferentiable Join
Let \(f\) be defined on \([-2,2]\) by
Find its absolute extrema.
Continuity: Each formula is continuous on its piece. At the join, the left formula gives \(f(0)=0+2=2\). The limit from the right is \(2-0=2\), so the two pieces meet continuously. Thus \(f\) is continuous on the whole closed interval and both extrema exist.
Interior candidates: On the left piece, \(f'(x)=1\); on the right piece, \(f'(x)=-1\). Neither derivative is zero. At \(x=0\), the one-sided derivatives are \(1\) and \(-1\), so \(f'(0)\) does not exist. The join is an interior critical number. The candidates are \(-2,0,2\).
Evaluate and compare:
The absolute maximum is \(2\) at \(x=0\). The absolute minimum is \(0\), attained at both endpoints. Substituting each candidate into its applicable piece confirms the values.
Common Mistake: Skipping the Join or an Endpoint
A common error is to differentiate the smooth pieces, find no points where their derivatives are zero, and conclude that there are no interior candidates. In the piecewise example, that reasoning would miss \(x=0\): the function is continuous there, but its derivative does not exist. The correction is to check each interior join and include it when differentiability fails.
A related error is to report only the point of an extremum rather than its value, or to report just one location when a value occurs more than once. In the polynomial example, the minimum value \(-2\) occurs at two points, \(x=-2\) and \(x=1\). State both the extremum value and every point where it is attained.
Finite Lists and General Candidate Sets
In the four examples, there are finitely many candidates, so evaluating and comparing a short list completes each problem. Do not turn that feature of these examples into a general claim. For example, continuous functions can have infinitely many interior critical numbers on a closed interval. In that situation the Extreme Value Theorem still guarantees that the absolute extrema exist, and the candidate principle still says where to look, but the candidates cannot necessarily be handled by a finite table of values.
The practical lesson is to distinguish the logic of the method from the size of the list it produces. First establish continuity and identify all relevant locations. Then use a finite comparison when the candidate set is finite; otherwise, additional analysis may be needed to determine the extrema without treating the set as a finite list.
Check Your Understanding
For each question, explain the reasoning as well as giving any requested calculation.
- Why must the denominator of a rational function be checked on the entire closed interval before using the Extreme Value Theorem?
- For a continuous piecewise function, why can an interior join be a candidate even when the derivative on each individual piece is nonzero?
- Find the candidates for \(g(x)=x^2-4x\) on \([0,3]\), and calculate the function value at each one.
- What does the Extreme Value Theorem guarantee, and what does it not guarantee about the number of critical numbers?
- If an absolute minimum value occurs at two different points, what should a complete answer report?
