Every test for classifying a critical point
A critical point exists — now is it a max, a min, or neither, and is it the biggest one on the whole interval?
Find the critical points first — nothing below applies anywhere else. To classify one as a LOCAL max/min, the First Derivative Test always works (checks a sign change in f') and the Second Derivative Test is usually faster when f'' is easy to compute, but it's inconclusive exactly when f''(c)=0, in which case fall back to the first derivative test. To find the ABSOLUTE max/min on a closed interval, skip classification entirely and use the closed-interval candidates test instead — it only needs candidate values, not sign changes.
Find the candidates
Where f'(c)=0 or f'(c) doesn't exist — the only places a local extremum can occur.
Classify a local extremum
Always applicable — watches for a sign change in f' across c.
Faster when it applies; inconclusive when f''(c)=0.
Find an absolute extremum on a closed interval
If f is continuous on a closed interval [a, b], then f attains both an absolute maximum value and an absolute minimum value on [a, b] -- that is, there exist c, d in [a, b] with f(c) \geq f(x) and f(d) \leq f(x) for every x in [a, b].
Guarantees an absolute max and min exist on [a,b] before you go looking for them.
Show proof
Both hypotheses are load-bearing, and dropping either one produces a function with no maximum, no minimum, or neither. f(x) = x on the OPEN interval (0, 1) is continuous but has no largest value -- for any candidate close to 1, a larger one is still available, since 1 itself is never reached. f(x) = 1/x on the closed interval [-1, 1], with f(0) left undefined (or defined arbitrarily), fails continuity at a single interior point and its values run off to \pm\infty on either side -- no maximum or minimum exists at all. A closed, bounded interval together with continuity is exactly the pair of hypotheses that rules out both failure modes: closedness keeps the interval's own endpoints in play as candidates, and continuity keeps the function's values from escaping to infinity or leaving a gap where a max or min would otherwise sit. A full proof (typically via the Bolzano–Weierstrass theorem, showing a maximizing sequence has a convergent subsequence whose limit is itself in [a,b]) belongs to real analysis; here it's enough to know EVT is what guarantees the "candidates test" in this chapter's next section always has a winner to find.