Critical points and extrema

Peaks and valleys

Extrema (singular: extremum) are the maximum and minimum values of a function.

Critical points

Local extrema can only occur at critical points: where f′(x) = 0 (horizontal tangent) or f′(x) is undefined (a corner or cusp).

Not every critical point is an extremum (y = x³ at 0 is flat but neither).

Classifying critical points

First derivative test: check the sign of f′ on each side.

• • to − → local max
• − to + → local min

Second derivative test: at f′(a) = 0,

• f″(a) < 0 → local max
• f″(a) > 0 → local min
• f″(a) = 0 → inconclusive

Example

f(x) = 2x³ − 3x² − 12x + 4

• f′(x) = 6x² − 6x − 12 = 6(x − 2)(x + 1) → x = 2, x = −1
• f″(x) = 12x − 6
• f″(−1) = −18 < 0 → local max: f(−1) = −2 − 3 + 12 + 4 = 11
• f″(2) = 18 > 0 → local min: f(2) = 16 − 12 − 24 + 4 = −16

Absolute extrema on a closed interval

On a closed interval [a, b], a continuous function always has an absolute max and min (Extreme Value Theorem). Use the closed interval method:

1. Find critical points inside (a, b).
2. Evaluate f at each critical point and at both endpoints.
3. The largest value is the absolute max; the smallest is the absolute min.

Example: f(x) = 2x³ − 3x² − 12x + 4 on [−2, 3]

Absolute max 11 at x = −1; absolute min −16 at x = 2.

Why endpoints matter

On [0, 1], f(x) = x has no critical points, but it still has a max (1) and min (0), both at endpoints.

Applications

• Maximum height of a projectile
• Minimum cost of production
• Maximum profit
• Optimal dimensions (see optimization)

Worked examples

Example 1: Critical points of f(x) = x² − 8x + 3? 2x − 8 = 0 → x = 4 (local and absolute min −13).

Example 2: f(x) = x³ − 27x. Local extrema? x = ±3; f(−3) = 54 (max), f(3) = −54 (min).

Example 3: Absolute max of f(x) = x² on [−1, 3]? Values: 1, 0, 9 → 9.

Common mistakes

• Forgetting endpoints on closed intervals.
• Assuming every critical point is an extremum.
• Mixing up local and absolute extrema.

Try it yourself

1. Critical points of f(x) = x³ − 12x.
2. Classify them.
3. Absolute max and min of f(x) = x³ − 12x on [−3, 5].
4. Can a local max be lower than a local min? Explain.

Answers: 1) x = ±2 2) local max at −2 (16), local min at 2 (−16) 3) max 65 at x = 5, min −16 at x = 2 4) Yes; on a wavy graph, a small hill can be lower than a valley elsewhere.

Quick recap

• Extrema occur at critical points (f′ = 0 or undefined).
• Use the first or second derivative test to classify a local maximum or local minimum.
• Closed interval method: check critical points and endpoints for the absolute maximum and minimum.