Problem 30 - Contemporary Precalculus

Question

Verify that \((f \circ g)(x)=x\) and \((g \circ f)(x)=x\) for every \(x\) $$f(x)=\sqrt[3]{x-1}, \quad g(x)=x^{3}+1$$

Step-by-Step Solution

Answer

Question: Verify that the function compositions \((f \circ g)(x)\) and \((g \circ f)(x)\) are equal to \(x\) for the given functions \(f(x)=\sqrt[3]{x-1}\) and \(g(x)=x^3+1\). Answer: We verified that \((f \circ g)(x)=x\) and \((g \circ f)(x)=x\) by plugging one function into the other, simplifying the expressions, and showing they are equal to \(x\).

Step 1: Find \((f \circ g)(x)\)

To find \((f \circ g)(x)\), we need to plug the function \(g(x)\) into the function \(f(x)\), so we have: \((f \circ g)(x)=f(g(x))\). Now substitute the given expressions for \(f(x)\) and \(g(x)\): \(f(g(x))=\sqrt[3]{(x^3+1)-1}=\sqrt[3]{x^3}\).

Step 2: Simplify \((f \circ g)(x)\)

Now simplify \((f \circ g)(x)\). We have the expression \((f \circ g)(x)=\sqrt[3]{x^3}\). The cube root of a number raised to the third power is simply the number itself, so \((f \circ g)(x) = x\).

Step 3: Find \((g \circ f)(x)\)

To find \((g \circ f)(x)\), we need to plug the function \(f(x)\) into the function \(g(x)\), so we have: \((g \circ f)(x)=g(f(x))\). Now substitute the given expressions for \(f(x)\) and \(g(x)\): \(g(f(x))=((\sqrt[3]{x-1})^3)+1\).

Step 4: Simplify \((g \circ f)(x)\)

Now simplify \((g \circ f)(x)\). We have the expression \((g \circ f)(x)=(\sqrt[3]{x-1})^3+1\). A number to the power of 3 and then taking the cube root will yield the original number, so we have \((g \circ f)(x)=x-1+1\). Thus, \((g \circ f)(x)=x\). In conclusion, we have verified that \((f \circ g)(x)=x\) and \((g \circ f)(x)=x\) for every x.

Key Concepts

Function Composition

Function composition is combining two functions to produce a new function. This means you apply one function to the result of another function. It is written as \((f \circ g)(x)\), which reads as "\(f\) composed with \(g\) at \(x\)." In function composition, order matters. Typically, you start with the inside function and then use the output as the input for the outside function.

• For the given functions \(f(x)=\sqrt[3]{x-1}\) and \(g(x)=x^3+1\), the composition \((f \circ g)(x)\) involves substituting \(g(x)\) into \(f(x)\).
• Similarly, \((g \circ f)(x)=g(f(x))\) requires substituting \(f(x)\) into \(g(x)\).

This exercise demonstrates how function composition can be used to check the inverse relationship between functions, as evidenced by their compositions yielding the original input \(x\). Understanding this is crucial in algebra and calculus because it allows verification of correct inverse pairs.

Cubic Functions

A cubic function is a polynomial function of degree three, generally expressed as \(a(x^3) + b(x^2) + cx + d\). These types of functions

• Have an \(x^3\) term as the highest power.
• Can cross the x-axis up to three times.
• Exhibit specific symmetries and inflection points.

In the exercise, \(g(x) = x^3 + 1\) is a cubic function.

• The main feature of this function is that it shifts the standard cubic curve upwards by one unit due to the "+1."
• This makes computation manageable when integrating or differentiating, while also simplifying the process of finding the inverse function.

Understanding cubic functions is important as they appear frequently in diverse areas of mathematics, modeling phenomena in physics, engineering, and statistics.

Simplifying Expressions

Once function composition is complete, simplification helps in identifying the essence of the expression. Simplifying involves reducing expressions into their simplest form to make them easier to understand and work with. In algebra, this often involves removing unnecessary factors or operations.

• For example, the expression \(f(g(x))=\sqrt[3]{x^3}\) was simplified to just \(x\) because the cube root and the cube operation cancel each other out.
• Similarly, \(g(f(x))=(\sqrt[3]{x-1})^3 + 1\) simplifies to \(x-1+1 = x\).

Simplification not only verifies steps but also confirms the accuracy of calculations. For instance, knowing how cube and cube root functions behave streamlines the validation of inverse functions. Simplification nurtures logical reasoning in mathematical problems, helping students to clearly see the core relationships between parts of an equation.