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Which statement about |x| at zero is correct?

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 · 5.1 · PDF 64 / printed page 59

TOPIC 01

Rates and tangent meaning

Derive tangent slope from shrinking secants, then connect slope, point and local approximation.

What you will be able to explain

  • Derive tangent slope from shrinking secants, then connect slope, point and local approximation.
  • Distinguish average rate, instantaneous rate, height, normal and inverse rate.

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

Which statement about |x| at zero is correct?

f(x)=∣x∣f(x)=|x|
  • A derivative requires equal finite one-sided limits; an extremum itself does not require differentiability.
Skills and prerequisite lessons

Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

  1. Derivative is 0.
  2. Derivative is 1.
  3. Continuous, but not differentiable.
  4. Not continuous.
Working and explanation
BUILD THE REASONING
Hint 1
Compare left and right difference quotients.
Hint 2
Start from this intermediate relation.
lim⁡h→0−∣h∣/h=−1\lim_{h\to0^-}|h|/h=-1
Worked solution
  1. Compare left and right difference quotients.
  2. Establish and simplify this mathematical relation.
    lim⁡h→0−∣h∣/h=−1\lim_{h\to0^-}|h|/h=-1
  3. Establish and simplify this mathematical relation.
    lim⁡h→0+∣h∣/h=1\lim_{h\to0^+}|h|/h=1
  4. A corner differs from a jump.
Continuous, but not differentiable.
Checks and common pitfalls: A corner differs from a jump.
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
5PredictPredict slopes of secants to x² at a=1 from both sides.
8Build the ideaSample response: quotient 2+h tends to 2; h cannot initially be zero.
10Guided comparisonSample response: slope 4 at (2,4) gives y−4=4(x−2); normal slope is −1/4.
12Independent practice and discussionContrast the corner task with normal and local-error tasks; require a reason before calculation.
5Exit and next stepExit: explain why f(3)=9 but f′(3)=6, and when a normal is vertical.
Question sequence
  • Predict slopes of secants to x² at a=1 from both sides.
  • Sample response: slope 4 at (2,4) gives y−4=4(x−2); normal slope is −1/4.
Board development
  • From secant to tangent
    f′(a)=lim⁡h→0[f(a+h)−f(a)]/hf\prime(a)=\lim_{h\to0}[f(a+h)-f(a)]/h
Expected correct and incorrect responses
  • Correct sample: 6
  • Anticipated wrong response: 9 — Substitutes the function height for its derivative.
  • Anticipated wrong response: Derivative is 0. — Applies a differentiable interior-minimum theorem to a corner.
  • Confusing f(a) with f′(a).
  • Treating a corner as a horizontal tangent.
Follow-up questions
  • Follow-up: Differentiate the function, then evaluate its derivative. Then explain why 9 fails.
  • Follow-up: Compare left and right difference quotients. Then explain why Derivative is 0. fails.
  • Relate the vertical gap h² to the accuracy of the local approximation.
Remediation
  • Sample response: quotient 2+h tends to 2; h cannot initially be zero.
  • Return to the first example, identify the missing condition, then solve one fresh task without viewing its solution.
Exit check
  • Exit: explain why f(3)=9 but f′(3)=6, and when a normal is vertical.
Assessment checklist and sample responses
  • Representation and conditions
    For h≠0, [(1+h)²−1]/h=2+h; its limit is 2.
  • Connected mathematical reasoning
    At (2,4), f′(2)=4, so y−4=4(x−2).
  • Interpretation and boundary check
    The exact minus predicted value is h²≥0, even for negative h.
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Curriculum and source notes ↗