How do changing slope signs prove a local high or low?
You will be able to: Classify a continuous critical point using derivative signs on either side.
How do changing slope signs prove a local high or low?
As a path climbs and then descends, the transition is a hilltop. In a derivative sign chart that same transition is positive to negative.
A useful starting point: How does a derivative sign chart reveal where a function rises? →
Words and symbols before equations
- First derivative test
- Classifies a continuous critical point using neighboring derivative signs.
- Positive-to-negative
- Increasing before and decreasing after: a local maximum.
- Negative-to-positive
- Decreasing before and increasing after: a local minimum.
- No sign change
- The monotonic direction persists across the point.
What this picture assumes
Original model; numerical readouts are rounded. f=x³−3x; f′=3x²−3; f″=6x. Algebra proves the interval sign chart. Separate plot axes have independently labeled scales. x and f are dimensionless.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- x=0.5: f=-1.375, f′=-2.25, f″=3. Negative slope; concave up at this input. C=0 affects only f’s heights. The plotted window is finite; exact interval signs come from the polynomial factors.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
Assume f is continuous at c and differentiable on nearby intervals to either side. If f′ changes from positive to negative, f rises into f(c) then falls away: c is a local maximum.
A negative-to-positive change proves a local minimum. Equal nonzero signs on both sides give no extremum at c. Merely knowing f′(c)=0 is not enough.
For f=x³−3x, the derivative signs +,−,+ give a local maximum (−1,2) and local minimum (1,−2). Report both input locations and function values.
The test also handles a corner such as abs(x): it is continuous at zero with side derivative signs −,+, so zero is a minimum despite f′(0) being undefined.
A worked example, step by step
Use derivative signs to classify zero for f=x⁴ and g=x³.
- For f, f′=4x³ is negative before zero and positive after.
- Thus f has a local minimum at zero.
- For g, g′=3x² is positive on both sides.
- Thus g has no local extremum at zero even though g′(0)=0.
Writing “f′=0, so maximum” omits the needed sign-change evidence.
Can a local minimum have an undefined derivative?
Compare with an explanation
Yes, a continuous corner such as abs(x) at zero qualifies under the side-sign test.
Predict. Change one thing. Explain.
At each marked critical input, say the derivative sign just before and just after. Turn that sign change into a complete extrema justification.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
x=0.5: f=-1.375, f′=-2.25, f″=3. Negative slope; concave up at this input. C=0 affects only f’s heights. The plotted window is finite; exact interval signs come from the polynomial factors.
| Interval | f′ sign | f behavior | f″ sign | Concavity |
|---|---|---|---|---|
| (−∞,−1) | + | increasing | − | down |
| (−1,0) | − | decreasing | − | down |
| (0,1) | − | decreasing | + | up |
| (1,∞) | + | increasing | + | up |
Original model; numerical readouts are rounded. f=x³−3x; f′=3x²−3; f″=6x. Algebra proves the interval sign chart. Separate plot axes have independently labeled scales. x and f are dimensionless.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the relevant derivative signs, theorem conditions, domain or geometric constraint. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA continuous f has f′>0 on (−2,0), f′<0 on (0,3) and f′>0 on (3,5). Classify the critical inputs 0 and 3 and state what their heights require.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: At zero the derivative changes positive to negative.
- 1 point: Therefore f has a local maximum at 0.
- 1 point: At 3 the derivative changes negative to positive, so f has a local minimum.
- 1 point: The numerical values f(0) and f(3) require additional function-value information.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Which sign change gives a hilltop?
Positive to negative.
RECALL 2Which gives a valley?
Negative to positive.
RECALL 3Does a sign chart supply exact function heights?
Not by itself.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do changing slope signs prove a local high or low?
- f′: +→− gives local maximum; −→+ gives local minimum.
- Check continuity at the candidate and signs on neighboring intervals.
Remember: Writing “f′=0, so maximum” omits the needed sign-change evidence.
Conditions: Original model; numerical readouts are rounded. f=x³−3x; f′=3x²−3; f″=6x. Algebra proves the interval sign chart. Separate plot axes have independently labeled scales. x and f are dimensionless.
Refresh Kid · AP Calculus AB Unit 5 · Objectives FUN-4.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.4, FUN-4.A. CED effective Fall 2020, with Fall 2026 clarifications, checked September 17, 2026. Unit 5 has twelve official topics. Focused lesson titles and questions are original Refresh Kid teaching material.
Existence theorems require their stated hypotheses. Critical numbers must belong to the function’s domain; interior local extrema and included endpoints are handled explicitly. Extrema and inflection candidates need justification, not just a zero derivative. Sparse derivative samples do not establish interval-wide signs. Optimization includes the constraint, feasible domain and global comparison. Implicit-curve conclusions specify the branch and distinguish finite slopes from vertical tangents.
The Organic Chemistry Tutor Mean Value Theorem and Optimization Problems video titles, creator and relevant descriptions were checked; full videos were not reviewed. The free optimization video mentions additional paid material, which is not required here. Khan Academy’s unit destination was checked; its JavaScript lesson contents were not fully readable by the research tool. OpenStax Sections 4.3, 4.4, 4.5 and 4.7 were consulted for conceptual cross-checking. No creator scripts, questions, diagrams or artwork were copied. Refresh Kid is not affiliated with or endorsed by these providers.
GitHub’s 3D website collection and its camera-control example informed optional spatial inspection. The open-box model is original geometry using existing self-hosted Three.js with its MIT license. A 12 cm square sheet loses four equal corner squares. The remaining net folds into an open box; the geometry uses the same lengths as the labeled 2D net. Fold angle is a construction view, not an independent design variable; displayed volume refers to the fully upright box. Camera rotation changes no mathematical values. Complete 2D diagrams and explanations remain available without WebGL.
Independent teacher review and observation of students remain pending. Technical checks do not certify mathematical accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus AB questions and scoring guides. The archive spans multiple units; no entire exam question is assigned here.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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