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How does a sign change in the first derivative classify a critical point as a local maximum or minimum?

Topic 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema: classify critical points using sign changes in the first derivative.

A focused answer to AP Calculus AB Topic 5.4, using sign changes of the first derivative to classify critical points as relative maxima, relative minima, or neither, with worked sign-chart classifications and the required justification.

Generated by Claude Opus 4.88 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this topic is asking
  2. The test
  3. A worked classification
  4. When the derivative does not change sign
  5. Why the First Derivative Test is robust
  6. Reading the test from a derivative graph
  7. Stating the conclusion in words

What this topic is asking

The College Board (Topic 5.4) asks you to classify critical points as relative maxima, relative minima, or neither, using how the sign of ff' changes across each critical point. This is the First Derivative Test.

The test

A worked classification

When the derivative does not change sign

A critical point need not be an extremum. For f(x)=x3f(x) = x^3, the derivative f(x)=3x2f'(x) = 3x^2 is zero at x=0x = 0 but is positive on both sides, so there is no sign change and x=0x = 0 is neither a max nor a min: it is a horizontal point of inflection. This is the standard counterexample, and the exam tests it precisely to see whether students assume that f(c)=0f'(c) = 0 automatically means an extremum. It does not; the sign change is what classifies the point.

Why the First Derivative Test is robust

The First Derivative Test works even where the Second Derivative Test fails or is awkward. It applies at critical points where ff' is undefined (cusps and corners), where the second derivative test cannot be used because there is no useful ff''. It also needs no second derivative computation, which can be heavy. On free-response questions the marks require an explicit statement of the sign change: write "ff' changes from positive to negative at x=cx = c, so ff has a relative maximum there." A bare conclusion without the sign-change reason does not earn the justification credit.

Reading the test from a derivative graph

A frequent exam format gives only the graph of ff' and asks for the relative extrema of ff. The First Derivative Test reads directly off that graph: a relative maximum of ff occurs where the graph of ff' crosses the axis from above to below (positive to negative), and a relative minimum where it crosses from below to above (negative to positive). A point where the graph of ff' merely touches the axis without crossing, like the bottom of a parabola sitting on the xx-axis, gives no sign change and so no extremum of ff. This is the graphical face of the same test, and it is worth practicing because the exam often hides the formula and supplies only the picture. The key is to track where ff' changes sign, not where ff' is large or small, since the magnitude of the derivative says nothing about extrema; only its sign change does.

Stating the conclusion in words

Because this topic is graded on justification, the wording matters as much as the answer. A complete statement names the critical point, the direction of the sign change, and the resulting classification in one sentence: "At x=2x = 2, ff' changes from negative to positive, so ff has a relative minimum at x=2x = 2." Writing the sign change as a fact you read from the chart or graph, then drawing the conclusion, earns both the analysis and the reasoning. Vague phrasing such as "the graph turns around here" does not, because it does not tie the conclusion to the sign of the derivative. Practice the precise sentence so it becomes automatic under exam time pressure.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)1 marksSection I (multiple choice, no calculator). If f(x)=(x2)(x+3)f'(x) = (x - 2)(x + 3), then ff has a relative minimum at (A) x=3x = -3 (B) x=2x = 2 (C) x=0x = 0 (D) both x=3x = -3 and x=2x = 2
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The correct answer is (B), x=2x = 2.

f(x)=(x2)(x+3)f'(x) = (x-2)(x+3) is zero at x=3x = -3 and x=2x = 2. At x=3x = -3, ff' changes from ++ to - (relative max). At x=2x = 2, ff' changes from - to ++ (relative min).

AP 2024 (style)4 marksSection II (free response, no calculator). Let f(x)=x48x2f(x) = x^4 - 8x^2. (a) Find f(x)f'(x) and its critical points. (b) Use the First Derivative Test to classify each critical point as a relative max, relative min, or neither, with justification.
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A 4-point classification question.

(a) (1 point) f(x)=4x316x=4x(x24)=4x(x2)(x+2)f'(x) = 4x^3 - 16x = 4x(x^2 - 4) = 4x(x-2)(x+2), critical points x=2,0,2x = -2, 0, 2.
(b) (3 points) Sign of ff': on (,2)(-\infty,-2) negative; (2,0)(-2,0) positive; (0,2)(0,2) negative; (2,)(2,\infty) positive. At x=2x = -2: - to ++, relative minimum. At x=0x = 0: ++ to -, relative maximum. At x=2x = 2: - to ++, relative minimum. Each conclusion is justified by the sign change of ff'.

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