How do you find the absolute extrema on a closed interval?
You will be able to: Compare function values at every interior critical number and both included endpoints.
How do you find the absolute extrema on a closed interval?
The tallest hill inside a park may still be lower than the elevation at the park gate. Searching only for flat tangents would miss that endpoint.
A useful starting point: How do changing slope signs prove a local high or low? →
Words and symbols before equations
- Candidate list
- All interior critical inputs plus both included endpoints.
- Candidates test
- Compare f-values on that complete list for a continuous function on a closed interval.
- Location
- The input where an extremum occurs.
- Extreme value
- The output f at that location.
What this picture assumes
Original model; numerical readouts are rounded. f=x³−3x on [−2,b]. Include both endpoints and all interior critical numbers. Compare original function values and report all tied locations.
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.
- Minimum -2 at x=-2 and 1; maximum 18 at x=3. Endpoints are included. All interior critical numbers (−1 and 1) were evaluated; compare function values, not derivatives.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
First confirm continuity on the closed bounded interval. EVT guarantees the extrema exist. An absolute extremum can occur at an endpoint or an interior critical point, so build the complete candidate list.
For f=x³−3x on [−2,3], f′=3(x−1)(x+1). Both critical numbers −1 and 1 lie inside. Include −2 and 3 as endpoints.
Evaluate f(−2)=−2, f(−1)=2, f(1)=−2, f(3)=18. The largest is 18 at x=3; the smallest is −2 at both x=−2 and x=1.
Compare f, not f′, and report all tied locations. A local maximum need not be the absolute maximum on the selected interval.
A worked example, step by step
Find absolute extrema of f=x²−4x+1 on [0,5].
- The polynomial is continuous on the closed interval.
- f′=2x−4=0 gives interior critical number 2.
- Evaluate f(0)=1, f(2)=−3 and f(5)=6.
- Absolute minimum is −3 at 2; absolute maximum is 6 at 5.
A derivative equal to zero identifies a location to inspect; it does not supply the output to compare. Include endpoints and ties.
Can two inputs tie for an absolute minimum?
Compare with an explanation
Yes. Both x=−2 and x=1 give −2 in the displayed cubic on [−2,3].
Predict. Change one thing. Explain.
Change the right endpoint from 2 to 3. Watch how the candidate table and winning locations change while the derivative formula stays the same.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Minimum -2 at x=-2 and 1; maximum 18 at x=3. Endpoints are included. All interior critical numbers (−1 and 1) were evaluated; compare function values, not derivatives.
| Candidate x | f(x) | Result |
|---|---|---|
| -2 | -2 | absolute minimum |
| -1 | 2 | does not win |
| 1 | -2 | absolute minimum |
| 3 | 18 | absolute maximum |
Original model; numerical readouts are rounded. f=x³−3x on [−2,b]. Include both endpoints and all interior critical numbers. Compare original function values and report all tied locations.
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 questionFind all absolute extrema of f=x³−3x on [−2,2].
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Compare with the answer and four-point rubric
- 1 point: f is continuous; critical numbers inside are −1 and 1.
- 1 point: Candidates are −2,−1,1,2.
- 1 point: The corresponding values are −2,2,−2,2.
- 1 point: Absolute maximum 2 occurs at −1 and 2; absolute minimum −2 occurs at −2 and 1.
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 1What guarantees existence before comparison?
Continuity on the closed bounded interval.
RECALL 2Which inputs join critical numbers?
Both included endpoints.
RECALL 3What must you do with tied outputs?
Report all locations that attain the extreme.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do you find the absolute extrema on a closed interval?
- For continuous f on [a,b]: evaluate f at a, b and all interior critical numbers.
- Largest and smallest candidate outputs determine absolute extrema.
Remember: A derivative equal to zero identifies a location to inspect; it does not supply the output to compare. Include endpoints and ties.
Conditions: Original model; numerical readouts are rounded. f=x³−3x on [−2,b]. Include both endpoints and all interior critical numbers. Compare original function values and report all tied locations.
Refresh Kid · AP Calculus AB Unit 5 · Objectives FUN-4.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.5, 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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