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Finding maxima and minima

f′(x) = 0 only finds candidates — a sign change decides max or min.

Appears onAP Calc AB

The common mistake

Calling every point where f′(x) = 0 a maximum without checking whether f′ actually changes sign there.

Instead: Find the critical points, then test the sign of f′ on each side. + to − is a local maximum, − to + is a local minimum, and no sign change is neither. On a closed interval, also check the endpoints.

Worked example with Guide me

Classify the critical points of f(x) = x³ − 3x.

  1. Sugar · What's f′(x), and where is it zero?
    Student · f′(x) = 3x² − 3, zero at x = 1 and x = −1.
  2. Sugar · Test x = 0, between them. Is f′ positive or negative?
    Student · f′(0) = −3, negative.
  3. Sugar · And at x = 2 or x = −2?
    Student · f′ = 9, positive on both sides outside.

Answer: x = −1 is a local max (+ to −); x = 1 is a local min (− to +).

Try 3 practice questions

  1. 1. Where does f(x) = x² − 6x have a local minimum?

  2. 2. Does f(x) = x³ have a max or min at x = 0?

  3. 3. Find the local maximum x-value of f(x) = −x² + 4x.

Quick answers

What is the most common mistake with finding maxima and minima?
Calling every point where f′(x) = 0 a maximum without checking whether f′ actually changes sign there.
How do I get finding maxima and minima right?
Find the critical points, then test the sign of f′ on each side. + to − is a local maximum, − to + is a local minimum, and no sign change is neither. On a closed interval, also check the endpoints.
Can you show a worked example?
Classify the critical points of f(x) = x³ − 3x. x = −1 is a local max (+ to −); x = 1 is a local min (− to +).

Related concepts

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