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For x³+x, what is the tangent slope at its inflection?

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 domainsMonotonicity 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.

01 / Foundation#Your turn

For x³+x, what is the tangent slope at its inflection?

f=x3+xf=x^3+x
  • 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.

  1. 1
  2. 0
  3. 3
  4. Undefined
Working and explanation
BUILD THE REASONING
Hint 1
Use f″ to locate, f′ to find slope.
Hint 2
Start from this intermediate relation.
f′′=6x  ⟹  x=0f''=6x\implies x=0
Worked solution
  1. Use f″ to locate, f′ to find slope.
  2. Establish and simplify this mathematical relation.
    f′′=6x  ⟹  x=0f''=6x\implies x=0
  3. Establish and simplify this mathematical relation.
    f′(0)=1f'(0)=1
  4. An inflection need not have zero slope.
1
Checks and common pitfalls: An inflection need not have zero slope.
Think first. Reveal a hint when the class is ready.
Focus on one question
Teacher preparation and assessment
40-minute lesson
MinutesFocusTeaching and student action
5PredictCompare the shapes of x³ and x⁴ at zero before differentiating.
8Build the ideaSample response: 6x changes sign but 12x² does not.
10Guided comparisonSample response: x³+x has inflection zero and tangent slope one.
12Independent practice and discussionUse derivative graphs to explain the two parameter and tangent-comparison transfer tasks.
5Exit and next stepExit: 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: concave upf''>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.
No sign-in. Work stays in this browser. Export before clearing browser data. Written reasoning is assessed with a checklist.
Curriculum and source notes ↗