Why Start with a Candidate List?
When looking for absolute extrema on a closed interval, it is useful to reduce the search to a finite or otherwise manageable set of points. The key idea is that possible locations come from two sources: the interval's endpoints and critical numbers in its interior.
This gives a systematic first step. Identify the interval, include both endpoints, and find every interior point where the derivative is zero or does not exist. The resulting points are candidates for absolute extrema. A candidate is not automatically a maximum or minimum; the list identifies where further investigation is needed.
The method applies to a function that is continuous on a closed interval \([a,b]\), with \(a<b\). Continuity matters because the Extreme Value Theorem guarantees that such a function attains both an absolute maximum and an absolute minimum on the interval. The derivative helps organize the search for where those values can occur.
The candidate set is formed by adjoining the endpoints to the interior critical numbers:
Use only critical numbers that lie strictly between \(a\) and \(b\) in the second part of the set. The endpoints are included separately, whether or not they are critical numbers.
Why These Sources Are Relevant
The endpoint part of the list is necessary because an absolute extremum can occur at an endpoint. The derivative condition is relevant to possible extrema inside the interval: at an interior local extremum where \(f\) is differentiable, the derivative must be zero.
Here is the basic reason for that derivative condition. Suppose \(c\) is an interior local maximum and \(f\) is differentiable at \(c\). For sufficiently small positive \(h\), the local maximum property gives \(f(c+h)-f(c)\leq 0\). Dividing by \(h>0\) preserves the inequality, so the right-hand difference quotients are at most zero. For sufficiently small negative \(h\), the same local maximum property gives \(f(c+h)-f(c)\leq 0\), but dividing by \(h<0\) reverses the inequality: the left-hand difference quotients are at least zero. Since the derivative exists, both one-sided limits agree. Their common value must therefore be both at most and at least zero, hence it is zero.
For a local minimum, the inequalities reverse: right-hand difference quotients are at least zero and left-hand difference quotients are at most zero. Differentiability again forces the common derivative to be zero. This explains why stationary points belong on the list. If the derivative does not exist at an interior point, the zero-derivative argument does not apply there, so the definition of critical number includes such points as well.
This reasoning explains the kinds of points the candidate search must include. It does not say that each listed point is an extremum. Nor does it replace the need to check the hypotheses: if the function is not continuous on the closed interval, the Extreme Value Theorem may not apply.
A Practical Procedure
For a continuous function on \([a,b]\), use the following process. Keep the interval fixed throughout: a critical number outside \((a,b)\) is not an interior candidate for this interval.
Write down \(a\) and \(b\) before solving any derivative equations. Both belong in the candidate list.
Compute \(f'(x)\) wherever the derivative exists in the interior of the interval.
Solve \(f'(x)=0\), and check for points in \((a,b)\) where \(f'(x)\) does not exist.
Retain only critical numbers strictly between \(a\) and \(b\), then combine them with the endpoints.
In this tutorial, the goal is to generate the points, not to decide which point gives the largest or smallest function value. Keeping those tasks separate helps prevent an incomplete search: first establish the whole candidate list, then use the function values to make the global comparison.
Worked Examples
Worked Example: A Cubic on a Symmetric Interval
Generate candidates for \(f(x)=x^3-3x\) on \([-2,2]\). The function is a polynomial, so it is continuous on the entire interval.
The endpoints are \(-2\) and \(2\). Differentiate:
Set the derivative equal to zero:
Both solutions lie in the interior, since \(-2<-1<2\) and \(-2<1<2\). The derivative exists everywhere, so there are no additional interior points where it fails to exist. The candidate list is therefore \(-2,-1,1,2\).
Check the stationary points by substitution into the derivative: \(f'(-1)=3(-1)^2-3=3-3=0\), and \(f'(1)=3(1)^2-3=3-3=0\). The endpoints are included as endpoints, not because their derivatives must vanish. This verifies the list's sources and interval membership.
Worked Example: A Critical Number Outside the Interval
Generate candidates for \(g(x)=x^3-6x\) on \([0,2]\). As a polynomial, \(g\) is continuous on \([0,2]\). Record the endpoints first: \(0\) and \(2\).
Differentiate and factor:
The derivative is zero at \(x=\sqrt{2}\) and \(x=-\sqrt{2}\). Since \(1<\sqrt{2}<2\), the number \(\sqrt{2}\) lies in \((0,2)\). Since \(-\sqrt{2}<0\), the number \(-\sqrt{2}\) lies outside the interval and is excluded.
The candidate list is \(0,\sqrt{2},2\). Verify the derivative at the retained critical number: \(g'(\sqrt{2})=3(\sqrt{2})^2-6=3\cdot 2-6=0\). Thus it satisfies the stationary-point condition and lies in the interval's interior. The endpoints remain on the list independently of this calculation.
Worked Example: A Rational Function with an Interior Critical Number
Generate candidates for \(h(x)=\dfrac{x+1}{x^2+1}\) on \([-1,2]\). The denominator satisfies \(x^2+1\geq 1>0\) for every real \(x\), so the function is defined and continuous throughout this interval. The endpoints are \(-1\) and \(2\).
Using the quotient rule, with numerator \(x+1\) and denominator \(x^2+1\), gives:
The denominator is positive, so \(h'(x)=0\) exactly when the numerator is zero:
The quadratic formula gives \(x=\dfrac{-2\pm\sqrt{4+4}}{2}=-1\pm\sqrt{2}\). Since \(1<\sqrt{2}<2\), we have \(0<-1+\sqrt{2}<1\), so \(-1+\sqrt{2}\) lies inside \([-1,2]\). Also, \(-1-\sqrt{2}<-1\), so that root is outside the interval.
The candidate list is \(-1,-1+\sqrt{2},2\). To check the retained critical number, substitution gives \((-1+\sqrt{2})^2+2(-1+\sqrt{2})-1=0\), so the derivative numerator vanishes there. Its denominator is positive, so the derivative is indeed zero. The other root has been excluded because it does not lie in the interval.
Worked Example: Trigonometric Critical Numbers on a Closed Interval
Generate candidates for \(p(x)=\sin x\) on \([0,2\pi]\). The sine function is continuous on this interval. Record the endpoints \(0\) and \(2\pi\).
Differentiate:
Interior critical numbers satisfy \(\cos x=0\). On \([0,2\pi]\), the solutions are \(x=\dfrac{\pi}{2}\) and \(x=\dfrac{3\pi}{2}\); both lie strictly between \(0\) and \(2\pi\). The derivative exists throughout, so there are no points to add from derivative failure.
The candidate list is \(0,\dfrac{\pi}{2},\dfrac{3\pi}{2},2\pi\). Check the stationary points: \(p'(\frac{\pi}{2})=\cos(\frac{\pi}{2})=0\), and \(p'(\frac{3\pi}{2})=\cos(\frac{3\pi}{2})=0\). Both satisfy the derivative equation, and the endpoints have been included separately.
Common Mistakes
A frequent error is to solve \(f'(x)=0\) and treat its solutions as the entire candidate list. This omits endpoints, which may be where an absolute extremum occurs. For example, take \(q(x)=x^2\) on \([0,1]\). Its derivative is \(q'(x)=2x\), so solving \(2x=0\) gives only \(x=0\). The incorrect list \(\{0\}\) leaves out the endpoint \(1\). The corrected list is \(\{0,1\}\): \(0\) is an endpoint and a stationary point, while \(1\) is included as the other endpoint.
A second error is retaining every solution of \(f'(x)=0\), even when it lies outside the specified interval. In the example with \(g(x)=x^3-6x\) on \([0,2]\), \(-\sqrt{2}\) solves the derivative equation but is not an interior point of that interval. Only \(\sqrt{2}\) belongs among its interior critical numbers.
Finally, do not assume that every candidate is an absolute extremum. The list identifies points that must be considered; candidate status alone does not determine what value the function takes there or whether that value is globally largest or smallest.
Check Your Understanding
For each question, focus on building the candidate list rather than deciding which listed point gives an absolute extremum.
- For \(f(x)=x^2-4x\) on \([0,3]\), what are the endpoints and interior critical numbers?
- Why must an endpoint be included even when the derivative there is not zero?
- For \(g(x)=x^3-3x\) on \([0,1]\), which solutions of \(g'(x)=0\) lie inside the interval?
- What two conditions make an interior point a critical number?
- Does membership in the candidate list by itself prove that a point is an absolute maximum or minimum? Explain briefly.
