Problem 59 - Calculus Single Variable

Question

Prove that if \(f:(a, b) \rightarrow(c, d)\) and \(g:(c, d) \rightarrow(\alpha, \beta)\) are invertible, then \(g\) o \(f\) is invertible.

Step-by-Step Solution

Answer

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

Step 1: Understand Invertibility

A function is invertible if there exists a function such that the composition of the two functions in both orders gives back the identity function on the respective domains. That means for a function \( f: (a, b) \rightarrow (c, d) \), there exists a function \( f^{-1}: (c, d) \rightarrow (a, b) \) such that \( f(f^{-1}(x)) = x \) for \( x \in (a, b) \) and \( f^{-1}(f(x)) = x \) for \( x \in (c, d) \). Similarly, function \( g: (c, d) \rightarrow (\alpha, \beta) \) has an inverse \( g^{-1}: (\alpha, \beta) \rightarrow (c, d) \).

Step 2: Express the Composition

The composition of functions \( g \circ f \) is defined by \((g \circ f)(x) = g(f(x))\) for all \( x \in (a, b) \). Our task is to show that there exists a function \((g \circ f)^{-1}\) that makes this composition invertible.

Step 3: Find the Inverse of the Composition

Assume the invertibility of \( f \) and \( g \), with their respective inverses \( f^{-1} \) and \( g^{-1} \). We hypothesize that \( (g \circ f)^{-1} = f^{-1} \circ g^{-1} \).

Step 4: Verify the Inverse

To confirm that \((g \circ f)^{-1} = f^{-1} \circ g^{-1}\) is the correct inverse, we calculate the following compositions: 1. \((g \circ f)\left((f^{-1} \circ g^{-1})(x)\right) = g(f(f^{-1}(g^{-1}(x)))) = g(g^{-1}(x)) = x\) for all \( x \in (\alpha, \beta)\). 2. \((f^{-1} \circ g^{-1})((g \circ f)(x)) = f^{-1}(g^{-1}(g(f(x)))) = f^{-1}(f(x)) = x\) for all \( x \in (a, b)\). The identity condition is satisfied in both cases.

Step 5: Conclusion

Since \( (g \circ f)\left((f^{-1} \circ g^{-1})(x)\right) = x \) and \( (f^{-1} \circ g^{-1})((g \circ f)(x)) = x \) for all relevant \( x \), the function \( g \circ f \) is invertible, and its inverse is \( f^{-1} \circ g^{-1} \).

Key Concepts

Function Composition

Function composition is a way to combine two functions to form a new function. If you have a function \(f\) that takes an input \(x\) and returns an output \(f(x)\), and another function \(g\) that takes the output of \(f\) and provides a result as \(g(f(x))\), then the composition of these functions is denoted as \((g \circ f)(x) = g(f(x))\).

Imagine a factory process: raw material goes through Function \(f\), becomes an intermediate product, which then goes through Function \(g\) to become the finished good. Just like those factory stages, function composition involves output from one process becoming input for the next.

When composing functions, it's crucial to ensure that the output range of \(f\) matches the input domain of \(g\), ensuring the process flows smoothly without errors. When this happens, the composition \(g \circ f\) is seamless and fully functional.

Function Inverses

A function's inverse reverses the effect of the original function. For a function \(f\), its inverse is denoted as \(f^{-1}\), and it satisfies the property that \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\). This means that applying \(f\) and then \(f^{-1}\), or vice versa, brings you back to your starting point.

Think of function inverses as retracing your steps in a maze – every move you made initially through the function \(f\), you directly undo with \(f^{-1}\). Similarly, for another function \(g\), if it has an inverse \(g^{-1}\), we can 'backtrack' the effect of \(g\) as well.

In the context of composition, if both functions \(f\) and \(g\) are invertible, meaning they each have their own inverses, then the composition \(g \circ f\) itself can be reversed. The inverse of this composition is found via \((g \circ f)^{-1} = f^{-1} \circ g^{-1}\), effectively reversing \(g\) first and \(f\) second.

Identity Function

The identity function is essentially the "do nothing" function. For any set \(A\), the identity function on \(A\) is denoted by \(I_A(x) = x\) for all \(x\) in \(A\). This function returns the input unchanged, serving as a kind of neutral "element" in the context of function composition and inverses.

In simpler terms, if \(x\) represents a number put into the function, the identity function simply echoes \(x\) back out with no modifications. The identity function plays a pivotal role in verifying inverses because it confirms correctness when composing a function with its inverse. For example, if \(f\) is a function, then \(f(f^{-1}(x)) = I(x) = x\) and \(f^{-1}(f(x)) = I(x) = x\). These conditions reflect the idea that applying a function and its inverse should leave you back at the original starting point – just like the identity function does.

Understanding identity functions is fundamental when discussing invertibility, as they provide the benchmark against which proper inverses are measured.