Suppose that the range of g lies in the domain of f so that the composition fog is defined. If f and g are one-to-one, can anything be said about fog? Give reasons for your answer.

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Recall the definition of a one-to-one (injective) function: a function \( h \) is one-to-one if \( h(a) = h(b) \) implies \( a = b \).

Given that \( f \) and \( g \) are both one-to-one, consider the composition \( f \circ g \), defined by \( (f \circ g)(x) = f(g(x)) \).

To check if \( f \circ g \) is one-to-one, assume \( (f \circ g)(x_1) = (f \circ g)(x_2) \). This means \( f(g(x_1)) = f(g(x_2)) \).

Since \( f \) is one-to-one, \( f(g(x_1)) = f(g(x_2)) \) implies \( g(x_1) = g(x_2) \).

Because \( g \) is also one-to-one, \( g(x_1) = g(x_2) \) implies \( x_1 = x_2 \). Therefore, \( f \circ g \) is one-to-one.

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주요 개념

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Function Composition

Function composition involves applying one function to the result of another, denoted as (f ∘ g)(x) = f(g(x)). For composition to be defined, the range of g must lie within the domain of f. Understanding this ensures the combined function is valid and can be analyzed.

One-to-One (Injective) Functions

A function is one-to-one if it maps distinct inputs to distinct outputs, meaning no two different inputs share the same output. This property is crucial for invertibility and affects how compositions behave, especially regarding uniqueness of outputs.

Injectivity of Compositions

The composition of two one-to-one functions is also one-to-one. Since both f and g are injective, their composition f ∘ g preserves distinctness of inputs, ensuring that (f ∘ g)(x1) ≠ (f ∘ g)(x2) whenever x1 ≠ x2, thus making f ∘ g injective.