Chapter 3 · Functions
Choose a valid monotonic interval for equal-exponent comparisons and prove square–cube comparisons by factoring their difference.
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Give your reason before checking.
| x | x² | x³ | x² − x³ |
|---|---|---|---|
| 0.5 | 0.25 | 0.125 | 0.125 |
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Chapter 3 · Functions
Class ______ Name __________________ Date __________
Predict the order of x² and x³ for x = 0.8, 1 and 2. Decide whether one rule can cover all positive inputs.
Record x² and x³ at x = 0.5, 0.8, 1 and 2. Group your observations into 0 < x < 1, x = 1 and x > 1.
Factor x² − x³ and record the sign of x² and 1 − x in each of the three regions.
Compare √2.1 with √2.4, and 1/2.1 with 1/2.4 without decimal approximations. Name the function and interval used each time.
Compare (3/4)² and (3/4)³, then (5/4)² and (5/4)³ without evaluating the powers. Explain why the conclusions reverse.
For 0 < x < 1, compare x² and x³ and justify the answer for every such x. State what happens at x = 1 and for x > 1.
Chapter 3 · Functions · Teacher reference
3.3.2(第12課時) · 12_運用冪函數性質比較大小.docx
5 min predict square–cube order; 12 min collect and group evidence; 8 min prove the sign pattern; 12 min paper comparisons; 3 min exit check.
Predict the order of x² and x³ for x = 0.8, 1 and 2. Decide whether one rule can cover all positive inputs.
For the same increasing function on a valid interval, input order is preserved; for a decreasing function it is reversed. Thus √2.1 < √2.4, while 1/2.1 > 1/2.4 on the positive interval. Comparing different exponents needs a different argument: x² − x³ = x²(1 − x). For positive x, it is positive when 0 < x < 1, zero at 1, and negative when x > 1. Therefore x² > x³, x² = x³, and x² < x³ respectively. Numerical trials locate the pattern; the sign argument proves it throughout the stated regions.
Compare (3/4)² and (3/4)³, then (5/4)² and (5/4)³ without evaluating the powers. Explain why the conclusions reverse.
For 0 < x < 1, compare x² and x³ and justify the answer for every such x. State what happens at x = 1 and for x > 1.
x² − x³ = x²(1 − x) > 0, so x² > x³. At x = 1 they are equal; for x > 1 the difference is negative, so x² < x³.
