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Find the inflection point of x³.

Read the idea, work independently, then explain what changed.

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高二選擇性必修第二册(A版).pdf · 5.3 · PDF 89 / printed page 84

Revisit first: Derivative rules and domains Monotonicity and constrained extrema

TOPIC 01

Curvature, inflections and derivative graphs

Verify a concavity change at a point on the curve, distinguishing inflections from stationary points.

What you will be able to explain

Verify a concavity change at a point on the curve, distinguishing inflections from stationary points.
Read second-derivative signs and parameter degeneracies.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

Examples 1–3: study a full demonstration; finish a solution from its first step; then choose a method independently before opening the solution. The remaining exercises may continue after class.

01 / Foundation#Worked example

Find the inflection point of x³.

f=x^3

Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.

Skills and prerequisite lessons

Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

Working and explanation

BUILD THE REASONING

Hint 1

Require a change of concavity at a point on the curve.

Hint 2

Start from this intermediate relation.

f''=6x

Worked solution

Require a change of concavity at a point on the curve.
Establish and simplify this mathematical relation.
$f''=6x$
Establish and simplify this mathematical relation.
$x<0:f\prime\prime<0;\ x>0:f\prime\prime>0$
A zero second derivative alone does not prove inflection.

(0,0), since f″=6x changes sign.

Checks and common pitfalls: A zero second derivative alone does not prove inflection.

Reasoning checklist · self / teacher assessment

Require a change of concavity at a point on the curve.
Give the intermediate relations and explain the conclusion with its restrictions.
A zero second derivative alone does not prove inflection.

Think first. Reveal a hint when the class is ready.

Teacher preparation and assessment

40-minute lesson

Minutes	Focus	Teaching and student action
5	Predict	Compare the shapes of x³ and x⁴ at zero before differentiating.
8	Build the idea	Sample response: 6x changes sign but 12x² does not.
10	Guided comparison	Sample response: x³+x has inflection zero and tangent slope one.
12	Independent practice and discussion	Use derivative graphs to explain the two parameter and tangent-comparison transfer tasks.
5	Exit and next step	Exit: test membership in the domain, curvature change, and the separate tangent slope.

Question sequence

Compare the shapes of x³ and x⁴ at zero before differentiating.
Sample response: x³+x has inflection zero and tangent slope one.

Board development

Curvature is change in slope
$f''>0:\ \text{concave up}$

Expected correct and incorrect responses

Correct sample: x³
Anticipated wrong response: x² — Confuses a minimum with a change in concavity.
Anticipated wrong response: 0 — Assumes an inflection must have a horizontal tangent.
Confusing an inflection with a horizontal tangent.
Ignoring a hole in the domain.

Follow-up questions

Follow-up: Check second-derivative signs. Then explain why x² fails.
Follow-up: Use f″ to locate, f′ to find slope. Then explain why 0 fails.
Explain why the disappearance of two inflections does not eliminate the minimum at zero.

Remediation

Sample response: 6x changes sign but 12x² does not.
Return to the first example, identify the missing condition, then solve one fresh task without viewing its solution.

Exit check

Exit: test membership in the domain, curvature change, and the separate tangent slope.

Assessment checklist and sample responses

Representation and conditions
A candidate must be on the curve and concavity must change across it.
Connected mathematical reasoning
For x³−6x²+3x, f″=6(x−2) changes sign and f(2)=−10.
Interpretation and boundary check
At k=0, x⁴ has f″=12x²≥0 on both sides, so no inflection.