What does it mean for a graph to bend upward or downward?
You will be able to: Relate concavity to changing first derivatives and signed second derivatives.
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.
What does it mean for a graph to bend upward or downward?
Walking downhill, you might find the path gradually leveling out. The slope is still negative, yet it is increasing toward zero. The graph can decrease and bend upward at the same time.
A useful starting point: What changes when an extremum candidate is a corner or an excluded endpoint? →
Words and symbols before equations
- Concave up
- The first derivative increases across an interval.
- Concave down
- The first derivative decreases across an interval.
- Second derivative f″
- The rate of change of f′.
- Bending
- How slopes change, rather than whether function heights are positive.
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
Where f″>0 throughout an interval, f′ increases and f is concave up. Where f″<0, f′ decreases and f is concave down. These claims concern intervals, not just a single plotted point.
For f=x³−3x, f″=6x. It is concave down for x<0 and concave up for x>0. The first derivative signs separately decide rising or falling.
On (0,1), this cubic is decreasing because f′<0, but concave up because f″>0. The negative slope is becoming less negative.
You may also infer concavity from a graph of f′: an increasing f′ graph means f is concave up even when that graph sits below zero. Read graph slope and graph height as different features.
A worked example, step by step
Describe increasing/decreasing and concavity for f=e^(−x).
- f′=−e^(−x), which is negative for every real x.
- Therefore f decreases everywhere.
- f″=e^(−x), which is positive everywhere.
- Therefore f is concave up everywhere: its negative slopes increase toward zero.
A negative first derivative does not imply concave down. Concavity depends on how the first derivative changes.
Can a function decrease while concave up?
Compare with an explanation
Yes; its slopes can be negative but becoming less negative.
Predict. Change one thing. Explain.
Compare x=−0.5 with x=0.5. Both have negative first derivatives; explain why their concavity differs.
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 questionFor f=x³−3x, describe the direction and concavity on each of (−2,−1), (−1,0), (0,1) and (1,2).
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Compare with the answer and four-point rubric
- 1 point: f′=3x²−3 and f″=6x.
- 1 point: On (−2,−1): increasing, concave down; on (−1,0): decreasing, concave down.
- 1 point: On (0,1): decreasing, concave up.
- 1 point: On (1,2): increasing, concave up; every claim follows the relevant derivative sign.
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 does concavity measure?
How slopes change with input.
RECALL 2What does the height of an f″ graph reveal?
The sign of the rate of change of f′.
RECALL 3What does the slope of an f′ graph reveal?
Concavity of f.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
What does it mean for a graph to bend upward or downward?
- f″>0 ⇒ concave up; f″<0 ⇒ concave down on the interval.
- Concavity can also be read from whether f′ increases or decreases.
Remember: A negative first derivative does not imply concave down. Concavity depends on how the first derivative changes.
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 BC Unit 5 · Objectives FUN-4.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.6, 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.
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