AP Calculus AB and BC

Candidates Test vs First Derivative Test

The candidates test compares function values at critical points and endpoints to find the absolute maximum and minimum on a closed interval. The first derivative test reads sign changes of the derivative to label relative extrema, and it never looks at endpoints.

Candidates test

Use when: The question asks for the absolute largest or smallest value of a continuous function on a closed interval, so the endpoints are live contenders.

First derivative test

Use when: You need to label one critical point as a relative maximum or minimum, on any interval at all, including one that has no endpoints.

Side by side

	Candidates test	First derivative test
What it finds	Absolute extrema across the whole interval	Relative extrema at one critical number
What you compute	The values $f$ takes at each candidate	The sign of $f'$ on each side of $c$
Endpoints	Listed as candidates alongside the critical numbers	Never examined, since only one side exists there
Interval required	Closed and bounded, written $[a,b]$	Any interval, open or unbounded
What you write down	A number, and the place where it occurs	A label: relative maximum, relative minimum, or neither

The candidates test is a search rather than a classification. The Extreme Value Theorem promises that a continuous function on $[a,b]$ attains an absolute maximum and an absolute minimum, and those can only occur where $f' = 0$ , where $f'$ fails to exist, or at one of the two endpoints. Collect that short list, evaluate $f$ at every entry, then read the largest and smallest numbers straight off. No sign chart appears anywhere in the method.

The first derivative test handles one point at a time. Around a critical number $c$ where $f$ is continuous, a sign change in $f'$ from positive to negative marks a relative maximum, negative to positive marks a relative minimum, and no change marks neither. It says nothing about how that point compares with anywhere else on the interval. Take $f(x) = x^{3} - 3x$ on $[-2,3]$ , where $f'(x) = 3(x^{2} - 1)$ gives critical numbers at $x = \pm 1$ .

f(-2) = -2, \qquad f(-1) = 2, \qquad f(1) = -2, \qquad f(3) = 18

The endpoint maximum the sign chart cannot see

Run the first derivative test on that function and it reports a relative maximum at $x = -1$ , where $f = 2$ . A student who stops there answers $2$ , and the real absolute maximum is $18$ at the endpoint $x = 3$ , roughly nine times larger. This is the standard way marks are lost on closed interval optimisation: the sign of $f'$ is a local instrument, and endpoints are invisible to it. Once the interval is closed, compare values.

Frequently asked questions

Do I have to do the first derivative test to justify an absolute maximum?

The comparison of values is itself the justification on a closed interval, provided you show the full candidate list. Adding a sign chart is extra work that answers a different question.

Can the absolute maximum happen at an endpoint?

Yes, and it often does, especially when the function is increasing across most of the interval. That is precisely the case the first derivative test cannot detect on its own.

What do I use when the interval is open?

The candidates test does not apply, because nothing guarantees an absolute extremum exists. Classify the critical points with the first or second derivative test, then argue from the behaviour of $f$ near the open ends.

In the CED: Unit 5: Analytical Applications