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For f=x²−4x, find the local minimum value.

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

TOPIC 01

Monotonicity and constrained extrema

Use derivative sign changes, endpoints and feasible domains to distinguish local extrema from global optima.

What you will be able to explain

Use derivative sign changes, endpoints and feasible domains to distinguish local extrema from global optima.
Handle parameter boundaries and non-attained bounds.

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#Your turn

For f=x²−4x, find the local minimum value.

f=x^2-4x

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

Solve the first derivative and confirm positive curvature.

Hint 2

Start from this intermediate relation.

f'=2x-4=0\implies x=2;\ f''=2

Worked solution

Solve the first derivative and confirm positive curvature.
Establish and simplify this mathematical relation.
$f'=2x-4=0\implies x=2;\ f''=2$
Establish and simplify this mathematical relation.
$f(2)=-4$
The point’s x-coordinate 2 differs from its minimum value −4.

The requested value is -4.

Checks and common pitfalls: The point’s x-coordinate 2 differs from its minimum value −4.

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	Predict what changes in x³−kx as k crosses zero.
8	Build the idea	Sample response: k>0 gives two sign-changing roots; k=0 is a stationary non-extremum.
10	Guided comparison	Sample response: for [−2,1], compare −2,2,−2 and choose maximum 2.
12	Independent practice and discussion	Students contrast endpoint, corner and stationary cases before solving the two transfer tasks.
5	Exit and next step	Exit: name all candidates and explain why a supremum on an open interval need not be a maximum.

Question sequence

Predict what changes in x³−kx as k crosses zero.
Sample response: for [−2,1], compare −2,2,−2 and choose maximum 2.

Board development

Candidates need classification
$f'<0\to f'>0:\ \text{local minimum}$

Expected correct and incorrect responses

Correct sample: Stationary but no extremum
Anticipated wrong response: Local minimum — Calls every stationary point a minimum without checking signs.
Anticipated wrong response: 0 — Checks only stationary points and omits interval endpoints.
Calling every stationary point an extremum.
Reporting a limit at an excluded endpoint as an attained maximum.

Follow-up questions

Follow-up: Compare derivative signs on both sides. Then explain why Local minimum fails.
Follow-up: Compare both endpoints and the stationary point. Then explain why 0 fails.
Change the visible interval and explain why a global optimum also needs endpoint comparisons.

Remediation

Sample response: k>0 gives two sign-changing roots; k=0 is a stationary non-extremum.
Return to the first example, identify the missing condition, then solve one fresh task without viewing its solution.

Exit check

Exit: name all candidates and explain why a supremum on an open interval need not be a maximum.

Assessment checklist and sample responses

Representation and conditions
On a closed interval I must compare included endpoints with stationary points inside it.
Connected mathematical reasoning
For x³−3x the derivative signs are +,−,+, so the local maximum is (−1,2).
Interpretation and boundary check
On (0,1), 2x−x²<1 but tends to 1: supremum 1, no attained maximum.