Answer: (h∘g)∘f = h∘(g∘f) always, while g∘f = f∘g may fail.
- A Composition is undefined unless the functions commute
- B (h∘g)∘f = h∘(g∘f) always, while g∘f = f∘g may fail
- C g∘f = f∘g always, while grouping matters
- D Both grouping and order are irrelevant to the result
Correct answer: B. (h∘g)∘f = h∘(g∘f) always, while g∘f = f∘g may fail
Explanation: Regrouping never changes the result, but swapping the order generally does - f(x) = x + 1 and g(x) = x² already differ.
Concept context
Types of relations, equivalence classes, one-one and onto functions, composition, and invertible functions