T.6 · Funciones, ramas inversas y composición
GRE Subject Test · GRE Mathematics Tema 19 3:00 Narración en inglés · Subtítulos en inglés + 中文 quemados en pantalla
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This unit distinguishes injectivity and surjectivity using the stated domain and codomain, chooses and verifies an inverse branch by composing both ways, and traces repeated composition while preserving the domain.
本单元用给定的定义域和陪域区分单射与满射，通过双向复合来选择并验证反函数分支，并在保持定义域的前提下追踪反复复合。
Injective means equal outputs force equal inputs; surjective means every element of the stated codomain is reached.
单射指输出相等则输入相等；满射指给定陪域的每个元素都被取到。
x squared from R to the nonnegatives is surjective but not injective; on the nonnegatives it is bijective; into all of R it is injective but not surjective.
x² 从 R 到非负实数是满射但不是单射；在非负实数上是双射；映到整个 R 时是单射但不是满射。
An inverse reverses a bijection.
反函数是双射的逆。
Solve y equals f of x for x, then use the original domain to choose the branch: for x at least 2, the inverse is 2 plus root y plus 5.
从 y = f(x) 解出 x，再用原定义域选择分支：对 x ≥ 2，反函数是 2 + √(y + 5)。
The minus branch would return inputs outside the domain.
负号分支会返回定义域以外的输入。
Check both compositions.
要检验两个方向的复合。
The inverse graph reflects the original across y equals x; it does not take reciprocals.
反函数的图像是原图像关于 y = x 的反射；不是取倒数。
A self-inverse function satisfies f of f of x equals x; a strictly increasing one must be the identity.
自反函数满足 f(f(x)) = x；严格递增的自反函数只能是恒等函数。
Iteration means repeated composition, not a power: for 1 over 1 minus x, three steps return x.
迭代指反复复合，不是乘方：对 1/(1 − x)，三次复合回到 x。
f of 5 is 4, and f inverse of 4 is 5: reflecting across y equals x swaps the point.
f(5) = 4，f⁻¹(4) = 5：关于 y = x 反射把点对调。
The unrestricted quadratic would also send minus 1 to 4.
不加限制的二次函数也会把 −1 映到 4。
For g, the orbit 3 to minus a half to 2 thirds has period three, so eight steps land where two do: 2 thirds.
对 g，轨道 3 → −1/2 → 2/3 的周期是 3，所以八步和两步落在同一处：2/3。
Changing a codomain changes surjectivity.
改变陪域会改变满射性。
Reflection across y equals x is an inverse graph; reciprocating y-values is a different operation.
关于 y = x 的反射是反函数图像；把 y 值取倒数是另一种运算。
Do not cancel away excluded inputs.
不要通过约分消去被排除的输入。
The image is 2 to infinity.
像是 [2, ∞)。
Solving with x at least 1 selects 1 plus root y minus 2, mapping 2 to infinity onto 1 to infinity.
在 x ≥ 1 下求解选出 1 + √(y − 2)，把 [2, ∞) 映到 [1, ∞)。
Both compositions check: the first because x minus 1 is nonnegative.
两个复合都成立：第一个是因为 x − 1 非负。
The negative root belongs to a different restriction.
负根属于另一种限制。
The first stage requires x not 1; the second requires h of x not 1, which excludes 0.
第一步要求 x ≠ 1；第二步要求 h(x) ≠ 1，排除了 0。
The third adds nothing, and h cubed is x on R without 0 and 1.
第三步不再增加限制，h³(x) = x 在 R 去掉 0 和 1 上成立。
The identity formula is defined more widely, but adding 0 or 1 would be an extension, not the composite.
恒等公式定义得更广，但加入 0 或 1 是延拓，而不是这个复合函数。
The orbit 3 to minus a half to 2 thirds has period three.
轨道 3 → −1/2 → 2/3 的周期是 3。
Check yourself.
自我检测。
x squared from R is not injective.
从 R 出发的 x² 不是单射。
The inverse branch is 2 plus root y plus 5.
反函数分支是 2 + √(y + 5)。
An inverse reflects; it does not reciprocate.
反函数是反射，不是取倒数。
g to the 8 of 3 is 2 thirds.
g⁸(3) = 2/3。
And the triple composite lives on R without 0 and 1.
三重复合定义在 R 去掉 0 和 1 上。