Problem 57 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The composition \( f \circ g \) is one-to-one since both \( f \) and \( g \) are one-to-one functions.

Step 1: Understand One-to-One Functions

A function is one-to-one if every element of the range is mapped to by exactly one element of the domain. This means for a function \( f \), if \( f(a) = f(b) \), then \( a = b \). It ensures that different inputs produce different outputs.

Step 2: Composition of Functions

The composition \( f \circ g \) is defined if the range of \( g \) is contained within the domain of \( f \). Mathematically, this means \( (f \circ g)(x) = f(g(x)) \).

Step 3: Analyzing One-to-One Composition

To determine if \( f \circ g \) is one-to-one, we need to check that if \( (f \circ g)(a) = (f \circ g)(b) \), then \( a = b \). For \( g \): \( g(a) = g(b) \) implies \( a = b \) due to \( g \) being one-to-one. For \( f \): if \( f(g(a)) = f(g(b)) \), then \( g(a) = g(b) \) (because \( f \) is one-to-one), thus \( a = b \).

Step 4: Conclusion

Since both \( g \) and \( f \) are one-to-one, the composition \( f \circ g \) is also one-to-one. Therefore, the composition of two one-to-one functions remains one-to-one.

Key Concepts

One-to-One Functions

One-to-one functions, or injective functions, have a special property. Every element from the range is paired with exactly one unique element from the domain. This means that no two different inputs will produce the same output. Consider this as a situation where each key opens one and only one lock. This ensures if two outputs are equal, their corresponding inputs must also be equal.
Mathematically for a function \ \( f \ \), it holds: \ \( f(a) = f(b) \ \implies a = b \ \). In simpler terms, if the function outputs the same result for two different inputs, those inputs must actually be the same. This property is crucial when dealing with function composition because it helps us retain distinct outputs for distinct inputs across the composed functions.

Domain and Range

The domain and range are fundamental concepts across function analysis. The domain refers to all possible input values a function can accept, while the range is all possible outputs it can produce. When dealing with function composition, understanding these sets becomes even more critical.
For the composition \ \( f \circ g \ \) to be defined, the range of \ \( g \ \) (outputs from \ \( g \ \)) must fit within the domain of \ \( f \ \) (inputs to \ \( f \ \)). Imagine \ \( g \ \) picking fruits and \ \( f \ \) sorting them - if \ \( g \ \) collects something \ \( f \ \) can't register, the process halts.
In the specific question, because the range of \ \( g \ \) lies within the domain of \ \( f \ \), we can seamlessly apply \ \( f \ \) to the outputs of \ \( g \ \), hence the function composition \ \( f \circ g \ \) is valid.

Mathematical Proof

Mathematical proof is a process that involves logical reasoning to demonstrate the truth of a statement. When we talk about proving that the composition of two one-to-one functions \ \( f \circ g \ \) is itself one-to-one, we aim to establish this via logical steps.
We start by assuming two equal outputs from the composed function: \ \( (f \circ g)(a) = (f \circ g)(b) \ \). Unfolding the composition, \ \( f(g(a)) = f(g(b)) \ \). Since \ \( f \ \) is one-to-one, it means \ \( g(a) = g(b) \ \). Applying the property of \ \( g \ \) being one-to-one gives \ \( a = b \ \).
So, we've logically arrived at the conclusion that different inputs for the composed function result in different outputs, establishing that \ \( f \circ g \ \) is indeed one-to-one just like its component functions. This step-by-step verification solidifies our understanding and application of function composition principles.