When are highest and lowest values guaranteed to exist?
You will be able to: Use the Extreme Value Theorem and distinguish attained extrema from limiting bounds.
BC foundation: Unit 5 shares these analytical differentiation objectives with AB. Check theorem hypotheses, justify derivative signs over intervals, and compare feasible candidates before claiming an optimum. State the local branch when analyzing an implicit relation.
When are highest and lowest values guaranteed to exist?
On a continuous temperature record from noon through 4 p.m., including both ends, the hottest and coldest recorded temperatures are actually reached. Removing an endpoint can change that guarantee.
A useful starting point: What do corners and equal endpoint heights change? →
Words and symbols before equations
- Absolute maximum
- A largest function value attained anywhere in the stated domain.
- Absolute minimum
- A smallest attained value on that domain.
- Extreme Value Theorem (EVT)
- A continuous function on a closed bounded interval attains both extrema.
- Attained
- Equal to f(x) for an allowed input, not only approached.
What this picture assumes
Original model; numerical readouts are rounded. Filled markers include endpoints; open markers exclude them. Finite traces do not by themselves prove the existence of extrema.
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.
- The polynomial is continuous on closed bounded [−1,2]. Minimum 0 at x=0; maximum 4 at x=2. EVT guarantees existence; the function confirms these values.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
EVT requires continuity throughout [a,b], with finite endpoints included. It guarantees an absolute maximum and minimum, but does not identify their locations.
For f=x² on [−1,2], the minimum 0 occurs at x=0 and maximum 4 at x=2. A global extremum can be at an endpoint or inside.
For f=x on (0,1), every value can be exceeded by another permitted value, and every value has a smaller permitted value. Neither 0 nor 1 is attained, so there is no absolute minimum or maximum.
If continuity fails, or the domain is not a closed bounded interval, extrema may still exist. For x² on all real numbers, a minimum exists at 0 even though EVT on a finite closed interval is not the reason.
A worked example, step by step
Does f(x)=1/x on [1,4] attain both extrema? Find them.
- The denominator never vanishes on [1,4], so f is continuous there.
- The finite interval includes both endpoints; EVT applies.
- Since f′=−1/x²<0, f decreases throughout.
- The maximum is 1 at x=1 and minimum is 1/4 at x=4.
A boundary value that is only approached is not an attained extremum. Failing EVT’s conditions does not prove extrema cannot exist.
Does EVT require differentiability?
Compare with an explanation
No. Continuity suffices, so a continuous graph with a corner can qualify.
Predict. Change one thing. Explain.
Compare x² on [−1,2] with x on (0,1). Read the closed and open endpoint markers and state which extreme values are actually attained.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
The polynomial is continuous on closed bounded [−1,2]. Minimum 0 at x=0; maximum 4 at x=2. EVT guarantees existence; the function confirms these values.
Original model; numerical readouts are rounded. Filled markers include endpoints; open markers exclude them. Finite traces do not by themselves prove the existence of extrema.
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 both absolute extrema of abs(x−1) on [−1,3] and explain why they are guaranteed despite the corner.
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Compare with the answer and four-point rubric
- 1 point: The absolute-value function is continuous throughout this closed bounded interval.
- 1 point: EVT guarantees both extrema; differentiability is not required.
- 1 point: The minimum is 0 at x=1.
- 1 point: The maximum is 2, attained at both x=−1 and x=3.
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 1Does EVT locate extrema?
No; it guarantees existence.
RECALL 2May an absolute extremum occur at an endpoint?
Yes, if that endpoint is in the domain.
RECALL 3Is a corner a problem for EVT?
Not if the function is continuous there.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
When are highest and lowest values guaranteed to exist?
- Continuous on a closed bounded [a,b] ⇒ both absolute extrema attained.
- State the value and the input where it occurs.
Remember: A boundary value that is only approached is not an attained extremum. Failing EVT’s conditions does not prove extrema cannot exist.
Conditions: Original model; numerical readouts are rounded. Filled markers include endpoints; open markers exclude them. Finite traces do not by themselves prove the existence of extrema.
Refresh Kid · AP Calculus BC Unit 5 · Objectives FUN-1.C · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.2, FUN-1.C. 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 BC 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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