Problem 4 - Intermediate Algebra

Question

Determine \((f \circ g)(x)\) and \((g \circ f)(x)\) for each pair of functions. Also specify the domain of \((f \circ g)(x)\) and \((g \circ f)(x)\). (Objective 1\()\) \(f(x)=6 x-5\) and \(g(x)=-x+6\)

Step-by-Step Solution

Answer

\((f \circ g)(x) = -6x + 31\) and \((g \circ f)(x) = -6x + 11\) with domains \(\mathbb{R}\) for both.

Step 1 - Define the Composition (f ∘ g)(x)

The composition \((f \circ g)(x)\) is defined as \(f(g(x))\). This means we substitute \(g(x)\) into the function \(f(x)\).

Step 2 - Substitute g(x) into f(x) for (f ∘ g)(x)

Given \(f(x) = 6x - 5\) and \(g(x) = -x + 6\), substituting \(g(x)\) into \(f(x)\) gives: \[ f(g(x)) = f(-x + 6) = 6(-x + 6) - 5 \] Simplify this to find the expression for \((f \circ g)(x)\).

Step 3 - Simplify (f ∘ g)(x)

Simplify the function: \[ f(g(x)) = 6(-x + 6) - 5 = -6x + 36 - 5 = -6x + 31 \] Thus, \((f \circ g)(x) = -6x + 31\).

Step 4 - Define the Composition (g ∘ f)(x)

The composition \((g \circ f)(x)\) is defined as \(g(f(x))\). This means we substitute \(f(x)\) into the function \(g(x)\).

Step 5 - Substitute f(x) into g(x) for (g ∘ f)(x)

Given \(g(x) = -x + 6\) and \(f(x) = 6x - 5\), substituting \(f(x)\) into \(g(x)\) gives: \[ g(f(x)) = g(6x - 5) = -(6x - 5) + 6 \] Simplify this to find the expression for \((g \circ f)(x)\).

Step 6 - Simplify (g ∘ f)(x)

Simplify the function: \[ g(f(x)) = -(6x - 5) + 6 = -6x + 5 + 6 = -6x + 11 \] Thus, \((g \circ f)(x) = -6x + 11\).

Step 7 - Determine the Domain of (f ∘ g)(x)

The domain of \((f \circ g)(x)\) is the set of all \(x\) for which \(g(x)\) is defined and \(f(g(x))\) is also defined. Since both \(f(x)\) and \(g(x)\) are polynomials, they are defined for all real numbers \(x\). Hence, the domain of \((f \circ g)(x)\) is all real numbers \(\mathbb{R}\).

Step 8 - Determine the Domain of (g ∘ f)(x)

The domain of \((g \circ f)(x)\) is the set of all \(x\) for which \(f(x)\) is defined and \(g(f(x))\) is also defined. Similarly, since both \(f(x)\) and \(g(x)\) are polynomials, they are defined for all real numbers \(x\). Thus, the domain of \((g \circ f)(x)\) is all real numbers \(\mathbb{R}\).

Key Concepts

Domain of a Function

The domain of a function refers to all the possible input values, usually represented by the variable \(x\), for which the function is defined. In simpler terms, it is the set of all values you can put into a function without causing any glitches, like division by zero or negative numbers under a square root.
For polynomial functions, which are mathematical expressions involving variables raised to whole number powers and constants, the domain is quite straightforward. Polynomial functions are defined for all real numbers because they don't involve divisions by zero or square roots of negative numbers. This means you can choose any real number as an input and get a valid output.

• The domain of \(f(x)\) is all real numbers \(\mathbb{R}\).
• The domain of \(g(x)\) is all real numbers \(\mathbb{R}\).
  
• Therefore, the domain of \( (f \circ g)(x)\) and \( (g \circ f)(x) \) are also all real numbers \(\mathbb{R}\).
  

This simplicity makes working with polynomials a breeze when determining domains, especially important in the context of function composition.

Polynomial Functions

Polynomial functions are a backbone of algebra and mathematics as a whole. A polynomial is an expression made up of variables, coefficients, and non-negative integer exponents. The general form looks like this: \(a_nx^n + a_{n-1}x^{n-1} + \, ... \, + a_1x + a_0\).
Here are some characteristics of polynomial functions:

• Continuous and Smooth: They form curves that don't break or have sharp turns, allowing them to be differentiated and integrated easily.
• Degree of a Polynomial: This is determined by the highest exponent of the variable. For instance, in \(6x^3 + 3x^2 - 2x + 1\), the degree is 3.
• Coeffcients: Each \(a_i\) is a coefficient, which can be any real number.

In our context, both \(f(x) = 6x - 5\) and \(g(x) = -x + 6\) are linear polynomials (degree 1). When composed to create new functions \((f \circ g)(x)\) and \((g \circ f)(x)\), the resulting functions remain linear polynomials, confirming their domains as \(\mathbb{R}\).
Working with polynomials becomes particularly handy in intermediate algebra, where understanding the interplay between functions is critical.

Intermediate Algebra

Intermediate algebra is a crucial stepping stone in mathematics, positioned between basic arithmetic and more advanced subjects such as calculus. This level of algebra typically includes understanding functions, equations, inequalities, graphs, and in particular, polynomial functions and their properties.
Some skills developed in intermediate algebra include:

• Function Composition: This is the process of combining two functions to form a new one, denoted as \((f \circ g)(x)\). It involves substituting one function into another and critically analyzing their domains.
• Simplification: Skills such as expanding, factoring, and reducing expressions are honed.
• Understanding Structures: Recognizing and working within different algebraic structures, like polynomials, aids in solving complex problems.

In the given exercise, composing the functions \((f \circ g)(x) = -6x + 31\) and \((g \circ f)(x) = -6x + 11\) involves an application of these skills. The process requires substituting functions into one another and simplifying the results, a typical task in intermediate algebra. Understanding such compositions helps students see the beauty and utility of algebra in transforming and analyzing mathematical relationships.