First derivative test

Let c be a critical point of a function continuous there. If f′ changes sign from + to − as x passes c, then c is a local maximum. If it changes from − to +, c is a local minimum.

If f′ keeps the same sign on both sides of c, c is neither; it is a point of inflection.

Worked example: f(x) = x³ − 3x + 3. f′ = 3(x − 1)(x + 1). Around −1 the sign goes + to −, so f(−1) = 5 is a local maximum; around 1 it goes − to +, so f(1) = 1 is a local minimum.

Worked example: f(x) = 2x³ − 6x² + 6x + 5. f′ = 6(x − 1)² ≥ 0 does not change sign at 1, so there is no local extremum; x = 1 is a point of inflection.

The test works even where f′ does not exist. f(x) = 3 + |x| has slope −1 on the left and +1 on the right of 0, so f(0) = 3 is a local minimum.

A small sign table (intervals across, sign of each factor down) makes the test quick and keeps the factors' signs honest.