Why Absolute Extrema Occur at Candidate Points

Why Search for Candidates?

Finding the absolute highest or lowest value of a function on an interval can seem to require checking every input. Calculus gives a reason to narrow the search. For a continuous function on a closed interval, absolute extrema are guaranteed to exist, and any such extremum must occur at an endpoint or at an interior point where the derivative is zero or does not exist.

This result does not say that every candidate is an absolute maximum or minimum. It says that an absolute extremum cannot be overlooked if the endpoints and all interior critical numbers are checked. Comparing the function values at those candidates then identifies the absolute extrema.

The Guarantee: The Extreme Value Theorem

The theorem guarantees that the highest and lowest function values are actually reached, not merely approached. Its hypotheses matter: the function must be continuous on the whole interval, and the interval must be closed and bounded. When either condition fails, the guarantee may fail, as the earlier material on the Extreme Value Theorem illustrates.

The theorem alone does not tell us where the extrema occur. The endpoints are natural candidates, but interior points must also be considered. The derivative explains why only certain interior points need to be included.

Why an Interior Extremum Has Derivative Zero

Suppose \(f\) has an absolute maximum at an interior point \(c\), where \(a<c<b\), and suppose \(f\) is differentiable at \(c\). Because \(c\) is interior, inputs slightly to its left and right are still in the interval. Since \(f(c)\) is a maximum, nearby function values cannot be greater than \(f(c)\).

For a small positive \(h\), the difference quotient is therefore nonpositive:

For a small negative \(h\), the numerator is still nonpositive, but the denominator is negative. Thus the difference quotient is nonnegative:

Differentiability means that these difference quotients approach the same limit as \(h\) approaches zero from either side. The limit must be both nonpositive and nonnegative, so it must be zero. Therefore \(f'(c)=0\). For an interior absolute minimum, the inequalities reverse: the quotient is nonnegative for positive \(h\) and nonpositive for negative \(h\). Its two-sided limit is again zero.

This argument is sometimes called Fermat’s theorem. It is a necessary condition, not a guarantee: a point where \(f'(c)=0\) need not be an extremum. For instance, a graph can flatten at a point and continue increasing through it.

Why Include Points Where the Derivative Does Not Exist?

The derivative argument applies only when the function is differentiable at the interior extremum. A continuous function can have a corner, cusp, or other nondifferentiable point and still attain an interior maximum or minimum there. Such a point cannot be ruled out by setting the derivative equal to zero.

A critical number of \(f\) is an interior number \(c\) in the domain of \(f\) such that either \(f'(c)=0\) or \(f'(c)\) does not exist. For the candidate test on \([a,b]\), we consider critical numbers in the open interval \((a,b)\). The endpoints are included separately whether or not a derivative exists there.

The Candidate Test

The Extreme Value Theorem supplies an absolute maximum and minimum. The derivative argument shows that if either occurs in the interior at a differentiable point, that point has derivative zero. If it occurs at an interior point where the derivative does not exist, the point is critical by definition. The only remaining locations are the endpoints. Together, these facts give the candidate test.

The test is a way to reduce the search, not a claim that every candidate is an extremum. A critical number might produce neither the greatest nor the least value. Also, the candidate set is not necessarily finite. If there are infinitely many critical numbers, the theorem still identifies where extrema must occur, but comparing a short finite list is not enough; further analysis may be needed.

Worked Examples

Common Pitfall: Looking Only for \(f'(x)=0\)

A frequent error is to assume every interior extremum must occur where the derivative equals zero. That conclusion ignores points where the function is not differentiable. Consider \(g(x)=|x-1|\) on \([0,3]\). A mistaken approach says, “The derivative is \(-1\) to the left and \(1\) to the right, so it is never zero; therefore there are no interior candidates.” This leaves out \(x=1\), where the derivative does not exist. But \(g(1)=0\), less than both endpoint values \(g(0)=1\) and \(g(3)=2\), so \(x=1\) is the absolute minimum.

The correction is to include both types of interior critical number: points where the derivative is zero and points where it does not exist. Another mistake is to omit endpoints because the derivative test concerns interior points. For example, on \([1,3]\), \(f(x)=x^2\) has \(f'(x)=2x\), which is never zero in the interval. Yet its absolute minimum is \(f(1)=1\), and its absolute maximum is \(f(3)=9\). Both extrema occur at endpoints.

Why the Candidate Test Matters

The test turns a global question—how does the function compare across an entire interval?—into a structured search. Continuity on a closed interval ensures that extrema exist, while the derivative condition restricts where interior extrema can occur. These are two distinct roles: the Extreme Value Theorem provides existence; the candidate result narrows the possible locations.

On the AP exam, make the assumptions and the candidates visible in your reasoning. State that continuity applies on the closed interval, include both endpoints, include interior critical numbers whether the derivative is zero or undefined, and compare function values. Finding a critical number alone does not establish an absolute extremum; the comparison is essential.