Problem 31 - Precalculus

Question

Let \(f\) and \(g\) be invertible functions. With the help of your classmates show that \((f \circ g)\) is one-to-one, hence invertible, and that \((f \circ g)^{-1}(x)=\left(g^{-1} \circ f^{-1}\right)(x)\).

Step-by-Step Solution

Answer

The composition \((f \circ g)\) is one-to-one and invertible, with inverse \((g^{-1} \circ f^{-1})\).

Step 1: Understanding One-to-One Functions

A function is one-to-one (injective) if different inputs produce different outputs. Mathematically, a function \(h\) is one-to-one if \(h(a) = h(b)\) implies \(a = b\). For \((f \circ g)(x) = f(g(x))\) to be one-to-one, both \(f\) and \(g\) must be one-to-one.

Step 2: Prove \(g\) is Injective

To prove \((f \circ g)\) is one-to-one, start by noting that \(g\) is injective. If \(g(x_1) = g(x_2)\), then since \(g\) is one-to-one, it must be true that \(x_1 = x_2\). Thus, \(g\) passes the injective property.

Step 3: Prove \(f\) is Injective

Since \(f\) is also one-to-one, if \(f(y_1) = f(y_2)\), then it follows that \(y_1 = y_2\). This ensures that distinct inputs in \(f\) result in distinct outputs.

Step 4: Show \((f \circ g)\) is Injective

Consider \((f \circ g)(x_1) = (f \circ g)(x_2)\), which expands to \(f(g(x_1)) = f(g(x_2))\). Since \(f\) is injective, \(g(x_1) = g(x_2)\). Because \(g\) is injective, it follows that \(x_1 = x_2\). Thus, \((f \circ g)\) is one-to-one.

Step 5: Show Inverse of \((f \circ g)\)

Given that \((f \circ g)\) is one-to-one, it is invertible. The inverse can be found by applying the inverses in reverse order. We have \((f \circ g)^{-1}(x)=(g^{-1} \circ f^{-1})(x)\). This means you first apply \(f^{-1}\), then \(g^{-1}\).

Step 6: Verify the Inverse

Check that \((f \circ g)\circ (g^{-1} \circ f^{-1}) = I\) and \((g^{-1} \circ f^{-1}) \circ (f \circ g) = I\), where \(I\) is the identity function. If \((f \circ g)((g^{-1} \circ f^{-1})(x)) = x\) and \((g^{-1} \circ f^{-1})((f \circ g)(x)) = x\), both above conditions are verified, confirming the inverse.

Key Concepts

Injective Functions

Injective functions are a fundamental concept in mathematics, especially within the study of function inverses and compositions. A function is called injective (or one-to-one) if it maps distinct inputs to distinct outputs. This property is formally expressed as: if a function \( h \) is injective and \( h(a) = h(b) \), then it must be true that \( a = b \). This means each input corresponds to a unique output.

Why is the injective property so crucial? In the context of function inverses and compositions, injectivity ensures that a function can be "reversed" through an inverse function. Without this property, an inverse cannot exist because the process of "undoing" the function isn't uniquely determined.

In our initial problem, demonstrating that both functions \( f \) and \( g \) are injective is key in proving that their composition \((f \circ g)\) is also one-to-one. Each function must pass the injective test to ensure that their combination does too.

Composition of Functions

The composition of functions is an operation that takes two functions \( f \) and \( g \) and produces a new function \( f \circ g \). This composed function is defined as \( (f \circ g)(x) = f(g(x)) \). Composition is akin to processing an input through multiple stages, with each function acting on the output of the previous one.

In essence, the composition \( f \circ g \) applies \( g \) to the input \( x \), then applies \( f \) to the result \( g(x) \). The critical aspect is understanding how the two functions interact. When both \( f \) and \( g \) are injective, it ensures that \( (f \circ g) \) remains one-to-one. This means that mapping through \( g \) first, followed by \( f \), doesn’t collapse distinct inputs into a single output, preserving the integrity of the injective property.

Function composition is fundamental in various mathematical concepts, solidifying the relationship between multiple transformations and how they affect inputs and outputs through sequential application.

Inverse Functions

An inverse function is a function that reverses the effect of the original function. For a function \( f \) to have an inverse, it must be both injective (one-to-one) and surjective (onto). The inverse of a function \( f \) is denoted as \( f^{-1} \), and it satisfies the condition \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \).

In the exercise, we look into the inverse of a composition of functions, \( (f \circ g) \), where the inverse is given by \( (f \circ g)^{-1} = g^{-1} \circ f^{-1} \). This formula implies that the composition's inverse is found by reversing the order of application and taking the inverse of each component.

Thus, to "undo" the effect of \( (f \circ g) \), we first "undo" \( f \) with \( f^{-1} \) and then "undo" \( g \) with \( g^{-1} \). This ensures the input returns to its original state, demonstrating the deeply interlinked nature of functions and their inverses.

One-to-One Functions

One-to-one functions, also known as injective functions, are functions where each element of the domain maps to a distinct element of the codomain. This concept is critical as it ensures the function can be reversed using an inverse function.

Mathematically, a function \( f \) is one-to-one if, for every \( x_1 \) and \( x_2 \) in the domain where \( f(x_1) = f(x_2) \), it implies \( x_1 = x_2 \). Since no two different domain elements have the same image in the codomain, this property supports the existence of an inverse function.

In practical terms, one-to-one functions are significant because their inverse can be used to "undo" the function's effect. This principle forms the backbone of many logical arguments and problem-solving scenarios in mathematics, and especially plays a crucial role in our exercise concerning function compositions and inverses.