Functions, inverse branches and composition
| English | Español |
|---|---|
| bijection/baɪˈdʒekʃn/ | bijection |
| involution/ɪnvəˈluːʃn/ | involution |
A decision before an answer
- A reversible encoding must recover exactly one input. A formula alone cannot tell you whether recovery is possible: its allowed inputs and outputs matter.
- Your goal: Distinguish injectivity and surjectivity using the stated domain and codomain.
Read the relationship
- A function assigns exactly one output to each allowed input. Injective means equal outputs force equal inputs; surjective means every element of the stated codomain is reached. The map x↦x² from R to [0,∞) is surjective but not injective. From [0,∞) to [0,∞) it is both, hence bijective. From [0,∞) to R it remains injective but fails surjectivity. Keep domain, image and codomain separate when testing each claim.
- Choose and verify an inverse branch by composing in both directions.
For f:[2,∞)→[−5,∞), f(x)=(x−2)²−5, what is f⁻¹(y)?
The input restriction x≥2 forces x−2≥0, so select the positive square-root branch.
Use the defining rule
- An inverse function reverses a bijection. Solve y=f(x) for x, then use the original domain to choose a branch. For f(x)=(x−2)²−5 on x≥2, the inverse is 2+√(y+5) on y≥−5. The minus branch would return inputs outside the chosen domain. Check f⁻¹(f(x))=x for allowed x and f(f⁻¹(y))=y for allowed y; one unchecked composition can conceal a domain error.
- Trace repeated function composition while preserving the domain.
For g(x)=1/(1−x) on R excluding 0 and 1, what is g⁸(3)?
The orbit is 3, −1/2, 2/3, then 3 again. Eight is congruent to two modulo three.
Check the conditions
- The inverse graph reflects the original graph across y=x; it does not take reciprocals of output values. A self-inverse function, or involution, satisfies f(f(x))=x wherever the composition is defined. Both −x on R and 1/x on R excluding zero are involutions. A strictly increasing involution on an interval must be the identity: if f(x)>x, increasingness gives f(f(x))>f(x)>x, and the analogous argument rules out f(x)<x.
- Trace repeated function composition while preserving the domain.
For f(x)=(x−2)²−5 with x≥2, f(5)=4 and f⁻¹(4)=2+√9=5. The unrestricted quadratic would have two inputs, −1 and 5, giving output 4. For g(x)=1/(1−x), the orbit 3→−1/2→2/3→3 has period three, so g⁸(3)=2/3; the count concerns compositions, not eighth powers.
For h(x)=−x on R, the value of h(h(7)) is ____.
Negation twice returns the original input; this map is an involution.
Apply the task format
- Iteration fⁿ means repeated composition, not the power (f(x))ⁿ. Calculate the first few compositions and check their domains before looking for a period. For f(x)=1/(1−x) on R excluding 0 and 1, f²(x)=(x−1)/x and f³(x)=x. The image stays in the same allowed domain. Therefore reduce an iteration count modulo three. A displayed formula equal to x after cancellation does not restore forbidden inputs.
- Trace repeated function composition while preserving the domain.
Changing a codomain changes surjectivity. Reflection across y=x is an inverse graph; reciprocating y-values is a different operation. Do not cancel away excluded inputs.
Which answer fits this case?
Distinguish injectivity and surjectivity using the stated domain and codomain
The function x↦x² from [0,∞) to R is surjective.
No negative element of the codomain is reached. The image is only [0,∞).
Keep the distinctions
- bijection 双射 — A function that is both injective and surjective between its specified sets.
- involution 对合 — A function whose composition with itself is the identity on its domain.
- Distinguish injectivity and surjectivity using the stated domain and codomain.
- Choose and verify an inverse branch by composing in both directions.
- Trace repeated function composition while preserving the domain.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.