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Find the minimum on the interval.

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 / Foundation#Worked example

Find the minimum on the interval.

f=x^2-4x+5,\quad x\in[0,3]

All inputs are real unless a smaller domain is stated.

Skills and prerequisite lessons

Scope: JM01 / JM02

Working and explanation

BUILD THE REASONING

Hint 1

Complete the square and locate its centre.

Hint 2

x=2 is inside [0,3].

Worked solution

Use vertex form.
$f=(x-2)^2+1$
The allowed vertex attains the lower bound.
$f(2)=1$

1 at x=2.

Checks and common pitfalls: The interval includes the global minimiser.

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.