LEARN · EXPLAIN · REVISE

Choose a∈[0,1] to make the largest value of (x−a)² over x∈[−1,2] as small as possible. Prove your choice.

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 square Quadratic 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.

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 / Transfer#Your turn

Choose a∈[0,1] to make the largest value of (x−a)² over x∈[−1,2] as small as possible. Prove your choice.

M(a)=\max_{-1\le x\le2}(x-a)^2

a∈[0,1] and x∈[−1,2].

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

The worst input is one of the endpoints.

Hint 2

Balance distances a+1 and 2−a.

Worked solution

Express the competing endpoint costs.
$M(a)=\max((a+1)^2,(2-a)^2)$
Their distances sum to 3, so the larger is at least 1.5.
$M(a)\ge(3/2)^2=9/4$
Equality requires equal distances and an allowed centre.
$a+1=2-a\Rightarrow a=1/2$

a=1/2; smallest possible worst value 9/4.

Checks and common pitfalls: This minimises a maximum over a second variable; choosing the vertex output zero would answer a different question.

Reasoning checklist · self / teacher assessment

Express the competing endpoint costs.
Their distances sum to 3, so the larger is at least 1.5.
Equality requires equal distances and an allowed centre.

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

Teacher preparation and assessment

40-minute lesson

Minutes	Focus	Teaching and student action
5	Check admissibility	Use both interval diagnostics before calculating.
8	Build candidate table	Compare the two endpoints and the allowed vertex.
10	Change the domain	Pairs replace closed by open endpoints and explain attainment.
12	Balance worst cases	Groups draw the two endpoint-distance curves for the minimax task.
5	Exit comparison	Ask 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.
$L,h,U$
Balance endpoint distances.
$a+1=2-a$

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.