Problem 53 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose that both \(f\) and \(g\) have inverses and that \(h(x)=(f \circ g)(x)=f(g(x)) .\) Show that \(h\) has an inverse given by \(h^{-1}=g^{-1} \circ f^{-1}\).

Step-by-Step Solution

Answer

The inverse of \(h\) is \(h^{-1} = g^{-1} \circ f^{-1}\).

Step 1: Understand Function Composition

We are given that \(h(x) = (f \circ g)(x) = f(g(x))\). This means that \(h\) is the composite function where first \(g(x)\) is applied and then \(f\) is applied to the result of \(g(x)\).

Step 2: Identify the Components of the Inverse

The task is to show that the inverse of \(h\) is \((g^{-1} \circ f^{-1})(x) = g^{-1}(f^{-1}(x))\). This means we first apply \(f^{-1}\) and then \(g^{-1}\) to \(x\).

Step 3: Apply \(h\) and \(h^{-1}\) to \(x\)

To show \(h^{-1}\) is \(g^{-1} \circ f^{-1}\), we must verify that applying \(h\) followed by \(h^{-1}\) yields the original \(x\). Similarly, applying \(h^{-1}\) followed by \(h\) should also yield \(x\).

Step 4: Calculate \(h(h^{-1}(x))\)

Compute \(h(h^{-1}(x)) = h((g^{-1} \circ f^{-1})(x)) = h(g^{-1}(f^{-1}(x)))\). This expression becomes \(f(g(g^{-1}(f^{-1}(x)))) = f(f^{-1}(x))\). Since \(f\) and \(f^{-1}\) are inverses, \(f(f^{-1}(x)) = x\).

Step 5: Calculate \(h^{-1}(h(x))\)

Now compute \(h^{-1}(h(x)) = (g^{-1} \circ f^{-1})(f(g(x)))\). This becomes \(g^{-1}(f^{-1}(f(g(x)))) = g^{-1}(g(x))\). Since \(g\) and \(g^{-1}\) are inverses, \(g^{-1}(g(x)) = x\).

Step 6: Conclude that \(h^{-1} = g^{-1} \circ f^{-1}\)

The calculations show that both \(h(h^{-1}(x)) = x\) and \(h^{-1}(h(x)) = x\). Therefore, \(h^{-1}\) is indeed \(g^{-1} \circ f^{-1}\), confirming the inverse function property in both directions.

Key Concepts

Function Composition

Function composition occurs when you apply one function to the result of another function. It's like a sequence of actions: you do one thing first, and then you follow it up with another. In mathematical terms, if you have two functions, say \(f\) and \(g\), their composition is written as \((f \circ g)(x) = f(g(x))\). This means you begin with the function \(g\), applying it to \(x\), and then take that resulting output and input it into the function \(f\).

Understanding the order in function composition is important, as switching it can lead to different results. So, always remember in the notation \((f \circ g)(x)\), \(g(x)\) is computed first, followed by \(f\).

In practical terms, function composition is like following recipes: preparatory steps must be completed in the right order to achieve the desired result, just like solving mathematical expressions using function compositions.

Inverse Functions

Inverse functions are fascinating because they "undo" each other's action. If you have a function \(f\), and you apply it to an input, the inverse function \(f^{-1}\) will return you to start. Mathematically, if you apply \(f\) to \(x\) and obtain a result, applying \(f^{-1}\) to that result will get you back to \(x\).

For this inverse relationship to work, both functions must be applied correctly. It's known that for the function \(f\), you require that \(f(f^{-1}(x)) = x\) for all \(x\) in the domain of \(f^{-1}\). Likewise, \(f^{-1}(f(x)) = x\) must hold for all \(x\) in the domain of \(f\).

Think of inverse functions like a forward-backward motion: anytime you take two steps forward with one function, the inverse takes you two steps back, bringing you back to your original position. This elegant dance between a function and its inverse is foundational in understanding how to "reverse" operations in calculus.

Composite Functions

Composite functions can be visualized as nesting functions inside each other, where one function's output becomes another's input. For example, a composite function \(h(x) = (f \circ g)(x) = f(g(x))\) takes \(x\), passes it through \(g\), and the result is then processed by \(f\).

When we talk about the inverse of a composite function, the process is slightly nuanced. You need to reverse the order. If \(h(x) = f(g(x))\), then \(h^{-1}(x) = g^{-1}(f^{-1}(x))\). This essentially means that when finding the inverse of a composite function, the inverse operations are applied in the reverse order of the original functions. Consider using two locks with different keys: the order of unlocking is vital. First unlock with the key for \(f^{-1}\), then with the key for \(g^{-1}\).

Comprehending composite functions and their inverses requires practice and a clear grasp of how each function alters its input. With time, these concepts become intuitive, simplifying the handling of complex functional relationships.