How do the graphs of f, f′ and f″ fit together?
You will be able to: Translate graph heights and slopes between a function and its derivatives.
How do the graphs of f, f′ and f″ fit together?
Three graphs can describe the same process at different levels: an amount, how fast it changes, and how fast that rate changes. Their heights answer different questions at the same input.
A useful starting point: How do you build a graph from derivative evidence? →
Words and symbols before equations
- f graph
- Shows function value against input.
- f′ graph
- Its height is the slope of f.
- f″ graph
- Its height is the slope of f′.
- Shared input
- The same horizontal coordinate compared across all panels.
What this picture assumes
Original model; numerical readouts are rounded. f=x³−3x+C. Both derivatives are independent of C. Each panel has its own vertical units and scale; screen heights are not directly comparable.
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
For f=x³−3x+C, changing C raises or lowers f but does not change f′=3x²−3 or f″=6x. Match slopes rather than assuming the graphs share vertical positions.
Where the f′ graph crosses from positive to negative, f has a local maximum, provided the continuity and side-sign conditions hold. A peak of f′ instead signals a possible inflection of f, if f′ changes from increasing to decreasing.
The f″ graph crossing zero indicates a change in the slope trend of f′; verify its side signs and f’s continuity before asserting inflection. Merely touching zero is insufficient.
Each plot uses its own labeled vertical scale. A large-looking height in one panel cannot be compared directly with another panel’s screen height; use the numerical axes and units.
| Graph | Height means | Slope means |
|---|---|---|
| f | Function value | First derivative |
| f′ | First derivative | Second derivative |
| f″ | Second derivative | Change in second derivative |
A worked example, step by step
At x=0.5 for f=x³−3x, interpret f′ and f″.
- f′(0.5)=3(0.25)−3=−2.25.
- So f is decreasing locally at this input.
- f″(0.5)=6(0.5)=3.
- So f′ is increasing locally and f bends upward; the original value is a separate quantity f(0.5)=−1.375.
A maximum of f′ is not automatically a maximum of f. The labels on each graph determine the meaning.
If f′ has a negative minimum, is f necessarily at a minimum there?
Compare with an explanation
No. Negative f′ means f is decreasing; the minimum of f′ concerns its slope trend.
Predict. Change one thing. Explain.
Move x, then change the vertical shift C. Track what changes in the f panel and what stays fixed in the derivative panels.
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+C. Both derivatives are independent of C. Each panel has its own vertical units and scale; screen heights are not directly comparable.
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 questionAt an input c, suppose f′(c)=2 and f″(c)=−1, with these signs persisting nearby. State four justified graph interpretations.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: The original function increases nearby.
- 1 point: Its local slope at c is 2 output units per input unit.
- 1 point: The first derivative decreases nearby, so f is concave down.
- 1 point: The local rate of change of that slope at c is −1 output units per input unit squared.
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 is the slope of the f′ graph?
f″.
RECALL 2Can f and f′ have different signs?
Yes; function height and slope are different.
RECALL 3What happens to derivative graphs under a vertical shift?
They stay unchanged.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do the graphs of f, f′ and f″ fit together?
- Height of f′ = slope of f; height of f″ = slope of f′.
- Vertical shifts of f leave both derivative graphs unchanged.
Remember: A maximum of f′ is not automatically a maximum of f. The labels on each graph determine the meaning.
Conditions: Original model; numerical readouts are rounded. f=x³−3x+C. Both derivatives are independent of C. Each panel has its own vertical units and scale; screen heights are not directly comparable.
Refresh Kid · AP Calculus AB Unit 5 · Objectives FUN-4.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.9, 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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