Why Polynomial Extrema Require a Comparison
A polynomial is continuous and differentiable everywhere, so its absolute extrema on a closed interval can be found by a systematic process. The derivative identifies possible locations of an interior extremum, but it does not tell us which candidate has the largest or smallest function value. That requires comparing values, including the values at both endpoints.
This distinction matters because a polynomial can have several interior critical numbers, and an endpoint can be the absolute winner even when the function has an interior local maximum or minimum. The goal is to identify every candidate that could attain an absolute extremum and then compare the candidates’ function values.
The Candidate List for a Polynomial
Every polynomial is continuous on \([a,b]\), so the Extreme Value Theorem guarantees that it attains both an absolute maximum and an absolute minimum there. Since a polynomial is also differentiable at every interior point, any interior point where it attains an absolute extremum must satisfy \(p'(c)=0\). The endpoints must be considered separately: they can be absolute extrema whether or not the derivative vanishes there.
Only roots of the derivative inside the interval are included as interior candidates. A root outside the interval cannot be an extremum point on \([a,b]\). A root at an endpoint is already covered by evaluating that endpoint.
Why the Candidate Comparison Works
The candidate rule follows from two facts: a polynomial attains its extrema on a closed interval, and at an interior extremum of a differentiable function the derivative is zero. The short argument for the second fact explains why critical numbers belong on the list.
A polynomial is continuous on the closed, bounded interval \([a,b]\). By the Extreme Value Theorem, it attains an absolute maximum and an absolute minimum on that interval.
Suppose \(c\in(a,b)\) is an absolute maximum. For sufficiently small positive \(h\), \(p(c+h)-p(c)\le 0\), so \(\frac{p(c+h)-p(c)}{h}\le 0\). For sufficiently small negative \(h\), \(p(c+h)-p(c)\le 0\), but now \(h<0\), so \(\frac{p(c+h)-p(c)}{h}\ge 0\). Because the derivative exists, the two one-sided limits are the same number. That number must be both nonpositive and nonnegative, hence \(p'(c)=0\). For an absolute minimum the inequalities reverse, and the same conclusion follows.
An absolute extremum occurs at an endpoint or at an interior point where \(p'(c)=0\). Thus at least one point attaining each absolute extremum is in the candidate set. Comparing the polynomial’s values on that set therefore gives the actual absolute maximum and minimum.
This argument establishes why the list is complete; it does not say that every candidate is an extremum. A critical number might be neither an absolute maximum nor an absolute minimum on the interval. It might even be a local extremum whose value is beaten by an endpoint.
If \(p'\) is the zero polynomial, then \(p\) is constant, and the rule of comparing a finite list of critical numbers is not needed: every point of the interval has the same function value. For a nonconstant polynomial, \(p'\) has only finitely many real roots, so the candidate comparison involves finitely many values.
A Reliable Calculation Procedure
For a nonconstant polynomial, the calculation can be organized into four steps. Keep the candidate points and their function values distinct: the derivative helps find points, while evaluating \(p\) determines the winners.
- Differentiate the polynomial to obtain \(p'(x)\).
- Solve \(p'(x)=0\), then retain only roots in the open interval \((a,b)\).
- Evaluate \(p\) at each retained root and at both endpoints \(a\) and \(b\).
- Compare the values. Report every candidate point attaining the greatest value and every candidate point attaining the least value.
When a value occurs at more than one candidate, all such points are absolute extremum points. Do not report only one of them. Exact fractions and integers are especially useful for comparisons because they make ties and inequalities visible without rounding.
Worked Example: A Cubic with Tied Endpoint and Interior Winners
Find the absolute maximum and minimum of \(p(x)=x^3-3x\) on \([-2,2]\).
Step 1: Differentiate.
Thus \(p'(x)=0\) at \(x=-1\) and \(x=1\). Both lie in \((-2,2)\), so the complete candidate list is \(-2,-1,1,2\).
Step 2: Evaluate the polynomial at every candidate.
The greatest candidate value is \(2\), attained at \(x=-1\) and \(x=2\). The least candidate value is \(-2\), attained at \(x=-2\) and \(x=1\).
Answer and check: The absolute maximum is \(2\), at \(x=-1\) and \(x=2\); the absolute minimum is \(-2\), at \(x=-2\) and \(x=1\). Substituting the reported points gives \(p(-1)=p(2)=2\) and \(p(-2)=p(1)=-2\), matching the comparison table. The derivative factors as \(3(x-1)(x+1)\), confirming that the two interior critical numbers were found.
Several Critical Numbers and an Endpoint Maximum
A polynomial derivative may have several roots inside the interval. The candidate rule does not require deciding in advance which of these roots gives a local maximum or minimum. Include every interior root, evaluate the original polynomial at each one, and then compare those values with both endpoints.
Worked Example: Three Interior Critical Numbers
Find the absolute extrema of \(q(x)=x^4-2x^2\) on \([-2,\frac32]\).
Step 1: Differentiate and find the critical numbers.
The derivative is zero at \(-1,0,1\). Each of these numbers lies in \((-2,\frac32)\), so the candidates are \(-2,-1,0,1,\frac32\).
Step 2: Evaluate at the candidates.
The greatest value is \(8\), at \(x=-2\). The least value is \(-1\), at \(x=-1\) and \(x=1\).
Answer and check: The absolute maximum is \(8\) at the left endpoint \(x=-2\), and the absolute minimum is \(-1\) at \(x=-1\) and \(x=1\). Substitution gives \(q(-2)=8\) and \(q(-1)=q(1)=-1\). The derivative \(4x(x-1)(x+1)\) vanishes at all three listed interior points, confirming the critical-number calculations.
Keep the Interval in View
Solving \(p'(x)=0\) for all real \(x\) is only part of the task. The specified interval determines which roots can be interior candidates. It also determines the endpoints that must be evaluated. A root outside the interval is irrelevant to the extrema on that interval, even though it is a genuine root of the derivative.
Worked Example: Excluding a Critical Number Outside the Interval
Find the absolute maximum and minimum of \(r(x)=(x^2-1)^2\) on \([-2,\frac12]\).
Step 1: Differentiate and filter the roots.
The derivative is zero at \(-1,0,1\). Since \(1\notin[-2,\frac12]\), it is not a candidate for this interval. The candidates are \(-2,-1,0,\frac12\).
Step 2: Evaluate at the candidates.
The greatest value is \(9\), at \(x=-2\), and the least value is \(0\), at \(x=-1\).
Answer and check: The absolute maximum is \(9\) at \(x=-2\), and the absolute minimum is \(0\) at \(x=-1\). Direct substitution gives \(r(-2)=9\) and \(r(-1)=0\), as reported. The derivative factors as \(4x(x-1)(x+1)\); its root \(1\) was correctly excluded because it is outside the interval.
A Common Error: Omitting the Endpoints
A frequent mistake is to find the critical numbers, evaluate the polynomial only there, and declare the largest and smallest of those values to be the absolute extrema. This comparison is incomplete because an endpoint can have a more extreme value than every interior critical number.
For example, in the preceding example, comparing only the interior critical values would compare \(r(-1)=0\) and \(r(0)=1\). One might then incorrectly claim that the absolute maximum is \(1\) at \(x=0\). But the omitted endpoint value is \(r(-2)=9\), which is larger than \(1\). Including both endpoints corrects the conclusion: the absolute maximum is \(9\) at \(x=-2\), while the absolute minimum remains \(0\) at \(x=-1\).
The correction is procedural: write down both endpoints before evaluating, and do not finalize either extremum until every candidate value has been compared. A second useful check is to verify that each reported extremum point lies in the interval and that its function value agrees with direct substitution.
What the Results Tell You
For polynomial absolute extrema on a closed interval, the derivative gives a finite list of possible interior locations, while the interval endpoints complete the list. The calculation is not finished until the original polynomial has been evaluated at every candidate and the resulting values have been compared.
This method also clarifies the difference between a critical number and an absolute extremum point. A critical number is a point where the derivative is zero; it is a candidate, not a guarantee of a global maximum or minimum. A candidate can lose to another critical point or to an endpoint, and more than one point can attain the same absolute value.
Check Your Understanding
For each question, keep track of both candidate points and polynomial values.
- For a nonconstant polynomial on \([a,b]\), which points belong in the candidate list for absolute extrema?
- Why must both endpoints be checked even if the polynomial has interior critical numbers?
- If \(p'(x)=0\) at a point outside \([a,b]\), should that point be included in the candidate list for extrema on \([a,b]\)? Explain.
- A candidate value is attained at two different points and is the greatest value in the list. How should the absolute maximum be reported?
- For \(q(x)=x^4-2x^2\) on \([-2,\frac32]\), which candidate gives the absolute maximum, and what is that maximum value?
