When does the second derivative classify a stationary point?
You will be able to: Apply the second derivative test and recognize when it is inconclusive.
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 does the second derivative classify a stationary point?
A bowl-shaped graph with a horizontal tangent at its bottom has a local low. But upward bending without a horizontal tangent can just be a sloped part of the bowl.
A useful starting point: Why does f″=0 not automatically mean an inflection point? →
Words and symbols before equations
- Stationary candidate
- An input c with f′(c)=0.
- Second derivative test
- Uses the sign of f″(c) at a stationary point under suitable smoothness.
- Inconclusive result
- The test cannot decide; another argument is needed.
- Global justification
- Evidence covering the whole allowed interval, beyond a local classification.
What this picture assumes
Original model; numerical readouts are rounded. All have f′(0)=0 and continuous second derivatives. If f″(0)=0 the second derivative test is inconclusive; actual extrema need another argument.
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.
- For x², f′(0)=0 and f″(0)=2. Second derivative test: local minimum. Actual classification: local minimum. The second derivative is continuous near zero, satisfying the stated smoothness condition.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
For a function with continuous second derivative near c and f′(c)=0, f″(c)>0 gives a local minimum; f″(c)<0 gives a local maximum. These are sufficient conditions used here.
If f″(c)=0, the test is inconclusive. The functions x⁴, −x⁴ and x³ all have first and second derivatives zero at 0, but give a minimum, maximum and neither, respectively.
Use the first derivative sign test or direct value comparison in an inconclusive case. Do not report “no extremum” merely because this test failed.
A local result alone does not generally prove a global one. For a continuous function on an interval with exactly one interior critical point, a local extremum there is also the corresponding absolute extremum; verify the interval and uniqueness rather than assuming them.
A worked example, step by step
Classify the critical points of f=x³−3x using the second derivative test.
- f′=3x²−3=0 gives c=−1 and c=1.
- f″=6x is continuous everywhere.
- f″(−1)=−6<0 gives a local maximum at (−1,2).
- f″(1)=6>0 gives a local minimum at (1,−2); on all real numbers neither is absolute because the cubic is unbounded.
Check f′(c)=0 before using f″(c). A zero second derivative means inconclusive, not “neither.”
If f′(c)=5 and f″(c)>0, does this test give a minimum?
Compare with an explanation
No. The stationary-point hypothesis f′(c)=0 is missing.
Predict. Change one thing. Explain.
Compare x², −x², x⁴ and x³ at zero. Predict the second-derivative verdict, then compare it with the actual graph and first-derivative signs.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
For x², f′(0)=0 and f″(0)=2. Second derivative test: local minimum. Actual classification: local minimum. The second derivative is continuous near zero, satisfying the stated smoothness condition.
Original model; numerical readouts are rounded. All have f′(0)=0 and continuous second derivatives. If f″(0)=0 the second derivative test is inconclusive; actual extrema need another argument.
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 questionFor f=x⁴, apply the second derivative test at zero, then use a different method to settle its classification.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: f′=4x³ and f″=12x²; both are zero at 0.
- 1 point: The second derivative test is inconclusive.
- 1 point: The first derivative is negative before zero and positive after.
- 1 point: Therefore zero is a local minimum; x⁴≥0 also proves it is an absolute minimum.
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 condition must come before the f″ sign test?
The first derivative must be zero at the candidate.
RECALL 2What does f″=0 tell this test?
Nothing conclusive about the extremum.
RECALL 3What can replace an inconclusive test?
A first-derivative sign change or direct value comparison.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
When does the second derivative classify a stationary point?
- At a stationary point with continuous f″ nearby: f″>0 gives local min; f″<0 gives local max.
- If f″=0, use another test.
Remember: Check f′(c)=0 before using f″(c). A zero second derivative means inconclusive, not “neither.”
Conditions: Original model; numerical readouts are rounded. All have f′(0)=0 and continuous second derivatives. If f″(0)=0 the second derivative test is inconclusive; actual extrema need another argument.
Refresh Kid · AP Calculus BC Unit 5 · Objectives FUN-4.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.7, 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 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.
Want to work through this with a tutor?
Bring your question about When does the second derivative classify a stationary point? Your explanation and answers remain free to access.