Problem 61 - College Algebra with Modeling and Visualization

Question

Exercises \(57-72:\) Use the given \(f(x)\) and \(g(x)\) to find each of the following. Identify its domain. $$ \begin{array}{llll} \text { (a) }(f \circ g)(x) & \text { (b) }(g \circ f)(x) & \text { (c) }(f \circ f)(x) \end{array} $$ $$ f(x)=2-3 x, \quad g(x)=x^{3} $$

Step-by-Step Solution

Answer

(a) \(2 - 3x^3\), \(x \in \mathbb{R}\); (b) \((2 - 3x)^3\), \(x \in \mathbb{R}\); (c) \(9x - 4\), \(x \in \mathbb{R}\).

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

The function \((f \circ g)(x)\) means we apply \(g(x)\) and then \(f(x)\). First, compute \(g(x) = x^3\). Now, substitute \(g(x)\) into \(f(x)\): \[ f(g(x)) = f(x^3) = 2 - 3(x^3). \]Thus, \((f \circ g)(x) = 2 - 3x^3.\)

Step 2: Determine the Domain of \((f \circ g)(x)\)

For \(f(g(x))\) to be defined, \(g(x)\) and then \(f(y)\) (where \(y = g(x)\)) must both be defined. Since \(g(x) = x^3\) and \(f(x) = 2 - 3x\), there are no restrictions such as division by zero or square roots of negative numbers. Therefore, the domain is all real numbers: \((f \circ g)(x): x \in \mathbb{R}\).

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

The function \((g \circ f)(x)\) means we apply \(f(x)\) and then \(g(x)\). First, compute \(f(x) = 2 - 3x\). Now, substitute \(f(x)\) into \(g(x)\):\[ g(f(x)) = g(2 - 3x) = (2 - 3x)^3. \]Thus, \((g \circ f)(x) = (2 - 3x)^3.\)

Step 4: Determine the Domain of \((g \circ f)(x)\)

For \(g(f(x))\) to be defined, \(f(x)\) and then \(g(y)\) (where \(y = f(x)\)) must both be defined. With \(f(x) = 2 - 3x\) and \(g(x) = x^3\), there are no restrictions. Thus, the domain is all real numbers: \((g \circ f)(x): x \in \mathbb{R}\).

Step 5: Find \((f \circ f)(x)\)

The function \((f \circ f)(x)\) means we apply \(f(x)\) to itself. So, compute:\[ f(f(x)) = f(2 - 3x) = 2 - 3(2 - 3x). \]Distribute the \(-3\):\[ 2 - 6 + 9x = 9x - 4. \]Thus, \((f \circ f)(x) = 9x - 4.\)

Step 6: Determine the Domain of \((f \circ f)(x)\)

For \(f(f(x))\) to be defined, \(f(x)\) must first be calculated, and then passed into \(f\) again, with no restrictions. As previously noted, \(f(x) = 2 - 3x\) imposes no domain restrictions. Therefore, the domain is all real numbers: \((f \circ f)(x): x \in \mathbb{R}\).

Key Concepts

Domain of Functions

The domain of a function is a fundamental concept, representing all the possible input values (often represented as \(x\)) for which the function is defined. In simpler terms, it's the complete set of values that you can plug into a function without encountering any mathematical problems like division by zero or taking the square root of a negative number. When dealing with functions such as \(f(x) = 2 - 3x\) and \(g(x) = x^3\), both are defined for all real numbers.

• For \(f(x) = 2 - 3x\), there's no restriction on \(x\). There's no division, no logarithms, or square roots requiring consideration of non-negative numbers.
• Similarly, \(g(x) = x^3\) is defined for all real numbers again because a cube function can accept any real number without complications.

When you perform operations like function compositions, such as \((f \circ g)(x)\) or \((g \circ f)(x)\), you need to ensure that each function in the composition is defined at each step. For the functions given, there's no additional restriction, maintaining the domain as all real numbers, \(x \in \mathbb{R}\). Therefore, the composed functions also inherit this freedom. Understanding the domain allows us to understand where our functions are valid and how they behave across different values.

Polynomial Functions

Polynomial functions are among the simplest and most important types of functions in algebra. They are formed by combining constants and variables with the operations of addition, subtraction, and multiplication, using only non-negative integer exponents. A basic representation of a polynomial is: \[ P(x) = a_nx^n + a_{n-1}x^{n-1} + \, \ldots \, + a_1x + a_0 \]where the coefficients \(a_i\) are real numbers and \(n\) is a non-negative integer.

• The function \(g(x) = x^3\) is a simple polynomial known as a cubic polynomial due to its highest exponent being 3.
• The function \(f(x) = 2 - 3x\) is a linear polynomial, as it is of the first degree (the highest exponent of \(x\) is 1).

Polynomial functions, like the ones given in the exercise, can be combined in various ways, such as addition, subtraction, multiplication, and composition. Each operation yields another polynomial. When these operations are performed, the resulting function maintains as a polynomial, and the domain typically remains all real numbers. This attribute makes polynomial functions particularly straightforward and handy in numerous mathematical applications, including modeling real-world phenomena.

Algebraic Expressions

Algebraic expressions are combinations formed from variables, numbers, and operations (such as addition, subtraction, multiplication, and division). They serve as the building blocks for more complex mathematical constructs like equations and functions.Learning to manipulate and understand these expressions is crucial for solving algebraic problems. When dealing with function composition, such as \((f \circ g)(x)\), you're essentially substituting one function into another as seen in the exercise:

• In \((f \circ g)(x) = 2 - 3x^3\), the expression \(x^3\) from \(g(x)\) was placed into \(f(x)\), replacing each occurrence of \(x\) in the polynomial.
• For \((g \circ f)(x) = (2 - 3x)^3\), the expression \(2 - 3x\) from \(f(x)\) was placed into \(g(x)\), effectively cubing the entire expression.
• For combined functions \(f(f(x)) = 9x - 4\), you substitute \(f(x)\) back into \(f\), showing the self-recursion property.

The ability to simplify these expressions correctly requires attention to detail and an understanding of algebraic laws and properties. Students need to practice mastering the fundamentals of expression manipulation to handle more advanced math topics competently. Such understanding allows for the simplification of complex problems into manageable steps, ensuring clarity and correctness in solutions.