Problem 27 - Thomas Calculus in SI Units

Question

Gives a formula for a function \(y=f(x) .\) In each case, find \(f^{-1}(x)\) and identify the domain and range of \(f^{-1} .\) As a check, show that \(f\left(f^{-1}(x)\right)=f^{-1}(f(x))=x\). $$f(x)=x^{3}+1$$

Step-by-Step Solution

Answer

The inverse function is \(f^{-1}(x) = \sqrt[3]{x - 1}\) with domain and range as all real numbers.

Step 1: Understand the Function

We start by looking at the given function, which is \(f(x) = x^3 + 1\). This is a cubic function that shifts the basic \(x^3\) graph up by 1 unit.

Step 2: Define the Inverse Function

To find the inverse function, \(f^{-1}(x)\), we need to express \(x\) in terms of \(y\) from the given equation \(y = x^3 + 1\). Swap \(x\) and \(y\), so the equation becomes \(x = y^3 + 1\). Solving for \(y\) gives \(y = \sqrt[3]{x - 1}\), which means \(f^{-1}(x) = \sqrt[3]{x - 1}\).

Step 3: Determine the Domain and Range of the Inverse Function

The domain of \(f(x)\) is all real numbers, since you can cube any real number and add 1. Consequently, the range of \(f(x)\) is also all real numbers. For the inverse, \(f^{-1}(x)\), the domain is also all real numbers because any real number can be used inside a cube root. Therefore, the range is all real numbers as well.

Step 4: Verify Inverse Function by Composition

We need to verify that \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\).1. Compute \(f(f^{-1}(x))\): \[f(f^{-1}(x)) = f\left(\sqrt[3]{x - 1}\right) = \left(\sqrt[3]{x - 1}\right)^3 + 1 = (x - 1) + 1 = x\]2. Compute \(f^{-1}(f(x))\): \[f^{-1}(f(x)) = f^{-1}(x^3 + 1) = \sqrt[3]{x^3 + 1 - 1} = \sqrt[3]{x^3} = x\]Since both compositions result in \(x\), the inverse function is verified.

Key Concepts

Cubic Functions

Cubic functions are a type of polynomial function where the highest degree of the variable is three. These functions look like this:

• General form: \(f(x) = ax^3 + bx^2 + cx + d\)
• Specific function we're working with: \(f(x) = x^3 + 1\)

This specific function takes the graph of \(x^3\) and shifts it up by 1 unit. Cubic functions are known for their smooth, continuous curves and can have up to three real roots. They can also display various types of symmetry, sometimes being symmetric about a point called inflection point.
Another interesting property is that cubic functions cover all real numbers. This means for every \(x\), there is a unique \(y\), ensuring the function is reversible, allowing us to find an inverse.

Domain and Range

Understanding the domain and range is crucial when working with functions and their inverses. Let's define these terms:

• Domain: All possible input values (\(x\)-values) for which the function is defined.
• Range: All possible output values (\(y\)-values) that the function can produce.

For the function \(f(x) = x^3 + 1\), the domain is all real numbers since you can cube any real number, and adding one doesn't impose restrictions.
The range is also all real numbers because no matter which \(x\) is chosen, \(y\) will cover every possible real number.
The inverse function \(f^{-1}(x) = \sqrt[3]{x - 1}\) also maintains the same domain and range: all real numbers. This symmetry of domain and range across the original and inverse functions is common for cubic and other polynomial functions of odd degree.

Function Composition

Function composition involves applying one function to the results of another. It essentially allows for the combination of two functions into a single operation. When verifying inverse functions, function composition is used to check if one function truly undoes another.

• For a function \(f\) and its inverse \(f^{-1}\), we test: \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\).

In the given exercise, we verified \(f(f^{-1}(x))\) through:\[f(f^{-1}(x)) = f\left(\sqrt[3]{x - 1}\right) = \left(\sqrt[3]{x - 1}\right)^3 + 1 = x - 1 + 1 = x\]And similarly checked \(f^{-1}(f(x))\) as:\[f^{-1}(f(x)) = f^{-1}(x^3 + 1) = \sqrt[3]{x^3 + 1 - 1} = \sqrt[3]{x^3} = x\]These compositions confirm our inverse function is correct, essentially asserting that \(f\) and \(f^{-1}\) "undo" each other's effects, returning the original input \(x\).