How does a derivative sign chart reveal where a function rises?
You will be able to: Use derivative signs on connected intervals of the domain to justify monotonic behavior.
How does a derivative sign chart reveal where a function rises?
A graph can sit below the horizontal axis while climbing. Its height is negative, but its slope is positive. Rising and falling describe changes in height, not the sign of the height.
A useful starting point: Why is a critical point only a candidate for an extremum? →
Words and symbols before equations
- Increasing
- Larger inputs give larger outputs on the interval.
- Decreasing
- Larger inputs give smaller outputs.
- Sign chart
- An interval-by-interval record of positive, negative or zero values.
- Domain break
- A missing input that separates intervals and cannot be crossed in a conclusion.
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
If f′>0 throughout an interval, MVT implies f increases there; if f′<0, it decreases. Split the domain at zeros of f′ and places where the derivative or function fails.
For f=x³−3x, f′=3(x−1)(x+1). The factors give signs +,−,+ on (−∞,−1), (−1,1), (1,∞). Hence f increases, decreases, then increases.
Testing one point per interval works here because the factored continuous derivative has no additional zeros or discontinuities inside those intervals. A few arbitrary samples alone cannot prove an interval-wide sign.
A function can be strictly increasing with a zero derivative at an isolated point: x³ illustrates this. Also, 1/x decreases on each of (−∞,0) and (0,∞), but not on the disconnected union as a single monotonic interval.
A worked example, step by step
Determine increasing and decreasing intervals of f=x²−6x+1.
- Differentiate: f′=2x−6.
- Solve 2x−6=0 to split at x=3.
- The derivative is negative for x<3 and positive for x>3.
- Thus f decreases on (−∞,3) and increases on (3,∞); the reasoning uses f′, not f’s height.
The sign of f is not the sign of f′. Never join intervals across a missing domain point.
Can f<0 and f′>0 hold at the same time?
Compare with an explanation
Yes. A negative output can be increasing toward zero.
Predict. Change one thing. Explain.
Move the input across −1 and 1. Compare the original function height with the first derivative’s sign and the fixed sign chart.
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 questionGiven f′(x)=(x−2)(x+1) for a differentiable f, determine all increasing/decreasing intervals and justify the signs.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: The only zeros are −1 and 2.
- 1 point: Both factors are negative for x<−1, giving a positive product.
- 1 point: Between −1 and 2 the product is negative; above 2 it is positive.
- 1 point: f increases on (−∞,−1) and (2,∞), and decreases on (−1,2).
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 derivative height tell you?
The slope of the original function at that input.
RECALL 2What makes a sign-chart test point sufficient?
A reason the derivative has constant sign throughout that subinterval.
RECALL 3May you cross a domain gap?
No; keep conclusions on connected intervals.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How does a derivative sign chart reveal where a function rises?
- f′>0 throughout an interval ⇒ f increases there.
- f′<0 throughout an interval ⇒ f decreases there.
Remember: The sign of f is not the sign of f′. Never join intervals across a missing domain point.
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 AB Unit 5 · Objectives FUN-4.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.3, 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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