Critical points, inflection points and the derivative tests

By Ashraf Bhura, Co-Founder, STEM-Tutor · Updated October 2026

The first and second derivatives tell you where a function rises and falls, where it has peaks and troughs, and where it changes the way it bends. These ideas appear on every AP Calculus exam, and the justifications matter as much as the answers.

Critical points

A critical point of f is a value x = c in the domain where f'(c) = 0 or f'(c) does not exist. Every relative maximum or minimum of a function happens at a critical point, but not every critical point is a maximum or minimum. For example, f(x) = x³ has f'(0) = 0, but the graph keeps rising through x = 0.

Increasing and decreasing

If f'(x) > 0 on an interval, f is increasing there. If f'(x) < 0, f is decreasing. To find these intervals, find the critical points, then test the sign of f' in each interval between them, using a sign chart.

The first derivative test

At a critical point c:

• If f' changes from positive to negative, f has a relative maximum at c.
• If f' changes from negative to positive, f has a relative minimum at c.
• If f' does not change sign, there is no relative extremum at c.

Concavity and the second derivative

If f''(x) > 0, the graph is concave up: it bends upwards like a cup, and its slopes are increasing. If f''(x) < 0, it is concave down, like a cap, with slopes decreasing.

Inflection points

A point of inflection is where the graph changes concavity: from concave up to concave down or the other way round. That happens where f'' changes sign. A common error is to say f''(c) = 0 is enough. It is not: f(x) = x⁴ has f''(0) = 0, but the graph is concave up on both sides, so there is no inflection point at 0. Equivalently, an inflection point of f is where f' has a relative maximum or minimum.

The second derivative test

At a critical point where f'(c) = 0:

• If f''(c) > 0, the graph is concave up there, so f has a relative minimum.
• If f''(c) < 0, it is concave down, so f has a relative maximum.
• If f''(c) = 0, the test is inconclusive; use the first derivative test instead.

Worked examples

Example 1: a cubic

Find and classify the critical points of f(x) = x³ − 3x² − 9x + 5, and find any inflection point.

• f'(x) = 3x² − 6x − 9 = 3(x − 3)(x + 1), so the critical points are x = −1 and x = 3.
• f''(x) = 6x − 6. f''(−1) = −12 < 0, so a relative maximum at x = −1. f''(3) = 12 > 0, so a relative minimum at x = 3.
• f'' = 0 at x = 1 and changes sign there (negative before, positive after), so there is an inflection point at x = 1.

Example 2: a derivative that does not exist

Find the critical points of f(x) = x^2/3.

• f'(x) = (2/3)x^−1/3, which is never 0 but does not exist at x = 0.
• x = 0 is a critical point. f' is negative to the left and positive to the right, so f has a relative minimum at x = 0, at a sharp point called a cusp.

Example 3: the test fails

Use both tests on f(x) = x⁴ at x = 0.

• f'(x) = 4x³ = 0 at x = 0. f''(0) = 0, so the second derivative test is inconclusive.
• First derivative test: f' < 0 for x < 0 and f' > 0 for x > 0, so there is a relative minimum at x = 0.

Example 4: reading the graph of f'

The graph of f' crosses the x-axis from above to below at x = 2, and has a minimum point at x = 5. What does this tell you about f?

• f' changes from positive to negative at x = 2, so f has a relative maximum at x = 2.
• f' has a minimum at x = 5, so f'' changes from negative to positive there: f has an inflection point at x = 5.
• Questions that give the graph of f' instead of f are very common on AP free-response sections.

Absolute extrema on a closed interval

To find the absolute maximum and minimum of a continuous function on [a, b], use the candidates test: evaluate f at every critical point in the interval and at both endpoints. The largest value is the absolute maximum; the smallest is the absolute minimum. Do not forget the endpoints; the absolute extremum is often there.

Example 5: the candidates test

Find the absolute maximum and minimum of f(x) = x³ − 3x² − 9x + 5 on [−2, 4].

• Critical points inside the interval: x = −1 and x = 3.
• f(−2) = 3, f(−1) = 10, f(3) = −22, f(4) = −15.
• Absolute maximum 10 at x = −1; absolute minimum −22 at x = 3.

Writing full-credit justifications

AP readers want the reason tied to the derivative, using the function's name and the point. Strong answers look like this:

• “f has a relative maximum at x = 2 because f' changes from positive to negative at x = 2.”
• “The graph of f has a point of inflection at x = 5 because f'' changes sign at x = 5” or “because f' changes from decreasing to increasing at x = 5”.

Avoid vague references like “the slope changes” or “it goes up then down” without saying which function.

Common mistakes

• Saying f''(c) = 0 guarantees an inflection point.
• Forgetting critical points where f' does not exist.
• Leaving out the endpoints in the candidates test.
• Confusing which graph is given: f, f' or f''.
• Justifying with “the slope changes” instead of naming f' or f''.

Practice questions

Try each one before opening the answer.

Frequently asked questions

Are these topics on AP Calculus AB?

Yes. Analytical applications of differentiation, including the first and second derivative tests, concavity and the candidates test, are Unit 5 of both AB and BC.

Are they in A level Maths?

A level Maths covers stationary points, the second derivative test and points of inflection, using the same ideas and similar notation.

Key takeaways

• Critical points: f' = 0 or f' undefined.
• First derivative test: sign change of f' decides max or min.
• Second derivative test: f'' > 0 min, f'' < 0 max, f'' = 0 inconclusive.
• Inflection points need a sign change of f''.
• Absolute extrema: check critical points and endpoints.

Related guides

• The mean value theorem
• Product and quotient rules
• AP Calculus AB review