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Where is the minimum attained?

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

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高一必修 第一册(A版).pdf · 3.2 · PDF 83 / printed page 76

Revisit first: Vertex form and completing the squareQuadratic signs and solution sets

TOPIC 01

Quadratic extrema on restricted intervals

Compare admissible vertex and endpoints, including open and moving intervals.

What you will be able to explain

  • Find extrema by comparing only valid critical candidates.
  • Distinguish attained extrema, unattained bounds and minimax design.

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

Where is the minimum attained?

f=(x−3)2,x∈[0,1]f=(x-3)^2,\quad x\in[0,1]
  • All inputs are real unless a smaller domain is stated.
Skills and prerequisite lessons

Scope: JM01 / JM02

  1. x=3
  2. x=1
  3. x=0
  4. No minimum
Working and explanation
BUILD THE REASONING
Hint 1
Mark the permitted interval.
Hint 2
Use the permitted input nearest 3.
Worked solution
  1. Compare endpoint distances.
    ∣1−3∣=2<3=∣0−3∣|1-3|=2<3=|0-3|
  2. The vertex is not in the domain.
    x=1x=1
B: x=1.
Checks and common pitfalls: An unrestricted vertex rule cannot override the domain.
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
5Check admissibilityUse both interval diagnostics before calculating.
8Build candidate tableCompare the two endpoints and the allowed vertex.
10Change the domainPairs replace closed by open endpoints and explain attainment.
12Balance worst casesGroups draw the two endpoint-distance curves for the minimax task.
5Exit comparisonAsk which candidates remain under integer inputs.
Question sequence
  • Which candidates are forbidden before any calculation?
  • Are we minimising f or the largest value of f?
Board development
  • Use an admissible candidate table.
    L,h,UL,h,U
  • Balance endpoint distances.
    a+1=2−aa+1=2-a
Expected correct and incorrect responses
  • “Vertex outside, so use the nearest endpoint” is justified for a>0.
  • “The minimum is zero” on (0,1) confuses attainment with a bound.
Follow-up questions
  • How does the candidate role reverse if a<0?
Remediation
  • Shade the domain before plotting the vertex; circle excluded candidates.
Exit check
  • For −(x−1)² on [2,4], find both extrema. Expected maximum −1, minimum −9.
Assessment checklist and sample responses
  • 2 marks: candidate validity.
    Full: endpoint and interior checks; partial: global vertex without domain.
  • 2 marks: minimax proof.
    Full: distances sum 3, maximum at least 1.5, equality at a=0.5; partial: guessed midpoint only.
No sign-in. Work stays in this browser. Export before clearing browser data. Written reasoning is assessed with a checklist.
Curriculum and source notes ↗