Problem 15 - College Algebra

Question

Use each pair of functions to find \(f(g(x))\) and \(g(f(x))\). Simplify your answers. $$ f(x)=\sqrt[3]{x}, g(x)=\frac{x+1}{x^{3}} $$

Step-by-Step Solution

Answer

\( f(g(x)) = \frac{\sqrt[3]{x+1}}{x} \) and \( g(f(x)) = \frac{\sqrt[3]{x} + 1}{x} \).

Step 1: Find f(g(x))

To find \( f(g(x)) \), substitute \( g(x) \) into the function \( f(x) \): \[ f(g(x)) = f\left(\frac{x+1}{x^3}\right) \] Since \( f(x) = \sqrt[3]{x} \), we have: \[ f\left(\frac{x+1}{x^3}\right) = \sqrt[3]{\frac{x+1}{x^3}} \] This represents the cube root of the expression \( \frac{x+1}{x^3} \).

Step 2: Simplify f(g(x))

Simplify the expression: \[ \sqrt[3]{\frac{x+1}{x^3}} = \frac{\sqrt[3]{x+1}}{\sqrt[3]{x^3}} \] Using the property of radicals \( \sqrt[3]{x^3} = x \), we get: \[ \frac{\sqrt[3]{x+1}}{x} \] So, \( f(g(x)) = \frac{\sqrt[3]{x+1}}{x} \).

Step 3: Find g(f(x))

To find \( g(f(x)) \), substitute \( f(x) \) into the function \( g(x) \): \[ g(f(x)) = g(\sqrt[3]{x}) \] Since \( g(x) = \frac{x+1}{x^3} \), replace every \( x \) in \( g(x) \) with \( \sqrt[3]{x} \): \[ g(\sqrt[3]{x}) = \frac{\sqrt[3]{x} + 1}{(\sqrt[3]{x})^3} \]

Step 4: Simplify g(f(x))

Simplify the expression \((\sqrt[3]{x})^3 = x\): \[ g(\sqrt[3]{x}) = \frac{\sqrt[3]{x} + 1}{x} \] Thus, \( g(f(x)) = \frac{\sqrt[3]{x} + 1}{x} \).

Key Concepts

Function Operations

Function operations allow you to combine functions in different ways to form new functions. One common type of operation is function composition, where one function is applied to the results of another. In this exercise, we work with compositions such as \( f(g(x)) \) and \( g(f(x)) \).
When we want to find \( f(g(x)) \), we substitute the entire function \( g(x) \) into the function \( f(x) \). This means that wherever there is an \( x \) in \( f(x) \), we replace it with \( g(x) \).

In the given problem, \( f(x) = \sqrt[3]{x} \) and \( g(x) = \frac{x+1}{x^3} \).

• To find \( f(g(x)) \), we substitute \( g(x) \) into \( f(x) \), resulting in \( f\left(\frac{x+1}{x^3}\right) = \sqrt[3]{\frac{x+1}{x^3}} \).
• To find \( g(f(x)) \), we do the reverse—substitute \( f(x) \) into \( g(x) \). Thus, \( g(\sqrt[3]{x}) = \frac{\sqrt[3]{x} + 1}{(\sqrt[3]{x})^3} \).

With an understanding of these operations, you can handle even more complex function combinations in the future!

Algebraic Functions

Algebraic functions involve polynomials, rational expressions, or radicals. In this exercise, both \( f(x) \) and \( g(x) \) are algebraic functions. These functions are commonly used in various branches of mathematics due to their straightforward structure and properties.
Algebraic functions can be composed to create new functions, as demonstrated here.

\( f(x) = \sqrt[3]{x} \) is a radical function, as it contains a cube root. \( g(x) = \frac{x+1}{x^3} \) is a rational function, given the fraction involving a polynomial in the numerator and a power of \( x \) in the denominator.

• Radical functions are important for illustrating how functions can be defined on specific domains, such as non-negative numbers for even roots.
• Rational functions show how division of polynomials can create expressions with distinct domains due to denominators.

In working through these functions, you will encounter unique algebraic properties that can aid you in solving various mathematical problems.

Simplification of Expressions

Simplification makes complex expressions easier to work with and understand. In function operations involving composition, simplification helps us see the underlying structure of the resultant expressions. By breaking down and simplifying \( f(g(x)) \) and \( g(f(x)) \), we obtain clearer forms that hold the same value.
For \( f(g(x)) = \sqrt[3]{\frac{x+1}{x^3}} \), simplification involves using properties of radicals and knowing that \( \sqrt[3]{x^3} = x \). This results in \( \frac{\sqrt[3]{x+1}}{x} \).
Similarly, \( g(f(x)) = \frac{\sqrt[3]{x} + 1}{x} \) simplifies directly because \( (\sqrt[3]{x})^3 \) simplifies back to \( x \).

• Simplification often involves recognizing patterns and rules like the power rule for exponents and roots.
• Good simplification can make future steps in algebra more manageable by reducing the likelihood of errors.

With these simplifications, the functions are easier to analyze and use in further mathematical explorations.