Problem 57 - A Graphical Approach to Precalculus with Limits

Question

Use \(f(x)\) and \(g(x)\) to find each composition. Identify is domain. (Use a calculator if necessary to find the domain.) \(\begin{array}{llll}\text { (a) }(f \circ g)(x) & \text { (b) }(g \circ f)(x) & \text { (c) }(f \circ f)(x)\end{array}\) $$f(x)=2 x+1, g(x)=4 x^{3}-5 x^{2}$$

Step-by-Step Solution

Answer

All compositions have domains of all real numbers.

Step 1: Understanding Composition

Composition of functions involves applying one function to the results of another. To find \((f \circ g)(x)\), substitute \(g(x)\) into \(f(x)\). Similarly, to find \((g \circ f)(x)\), substitute \(f(x)\) into \(g(x)\), and for \((f \circ f)(x)\), substitute \(f(x)\) back into itself.

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

To find \((f \circ g)(x)\), substitute \(g(x) = 4x^3 - 5x^2\) into \(f(x) = 2x + 1\). Thus, \((f \circ g)(x) = 2(4x^3 - 5x^2)+1\). Simplifying this gives \(8x^3 - 10x^2 + 1\).

Step 3: Determine Domain of \((f \circ g)(x)\)

The domain of \(g(x) = 4x^3 - 5x^2\) is all real numbers since it's a polynomial function. Because \(f(x)\) is also a polynomial, the domain of \((f \circ g)(x)\) is also all real numbers.

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

To find \((g \circ f)(x)\), substitute \(f(x) = 2x + 1\) into \(g(x) = 4x^3 - 5x^2\). Therefore, \((g \circ f)(x)\) is \(4(2x+1)^3 - 5(2x+1)^2\). Expanding this, we get: \(32x^3 + 48x^2 + 24x + 4 - 20x^2 - 20x - 5 = 32x^3 + 28x^2 + 4x - 1\).

Step 5: Determine Domain of \((g \circ f)(x)\)

The domain for \((g \circ f)(x)\) is also all real numbers, as both \(f(x)\) and \(g(x)\) are polynomials, which have domains consisting of all real numbers.

Step 6: Calculate \((f \circ f)(x)\)

To find \((f \circ f)(x)\), substitute \(f(x) = 2x + 1\) into itself. Thus, \((f \circ f)(x) = 2(2x+1) + 1 = 4x + 2 + 1 = 4x + 3\).

Step 7: Determine Domain of \((f \circ f)(x)\)

The domain of \(f(x)\) is all real numbers, so the domain of \((f \circ f)(x)\), being a linear polynomial of the form \(4x + 3\), is also all real numbers.

Key Concepts

Polynomial Functions

Polynomial functions are a fundamental aspect of algebra and precalculus. They are expressions consisting of variables raised to whole number powers and multiplied by coefficients. A simple example is \(f(x) = 2x + 1\), where 2x is the variable term, and 1 is the constant term.

Polynomials can take on various forms, such as quadratic (like \(ax^2 + bx + c\)) or cubic, as in \(g(x) = 4x^3 - 5x^2\). The highest power of the variable in a polynomial function determines its degree. For instance, \(g(x)\) is a third-degree polynomial because of the term \(4x^3\).

Polynomials are continuous functions, meaning they are smooth and unbroken. These functions do not have abrupt changes in direction or jumps. This property contributes to their domain being all real numbers, which we will explore next.

Domain of Functions

The domain of a function refers to all possible input values that the function can accept. For polynomial functions like \(f(x) = 2x + 1\) and \(g(x) = 4x^3 - 5x^2\), the domain is straightforward because polynomials are defined for all real numbers.

Unlike functions that might have restrictions—like square roots (since you cannot take the square root of a negative number) or fractions (since division by zero is undefined)—polynomials do not have such constraints. This makes their domain essentially unrestricted.

When performing function compositions, such as \((f \circ g)(x)\) or \((g \circ f)(x)\), it's essential to check the domains of individual functions first. Nevertheless, since both functions in our exercise are polynomials, we confirm that their domain remains all real numbers.

Function Operations

Function operations, including addition, subtraction, multiplication, and composition, allow us to create new functions from existing ones. These operations can be used to combine functions in various ways to solve complex problems.

In function composition, denoted by \((f \circ g)(x)\), we apply one function to the result of another. Effectively, it involves plugging \(g(x)\) into \(f(x)\).

For our specific exercise:

• \((f \circ g)(x)\) involves substituting \(g(x) = 4x^3 - 5x^2\) into \(f(x)\) to yield \(8x^3 - 10x^2 + 1\).
• \((g \circ f)(x)\) means substituting \(f(x) = 2x + 1\) into \(g(x)\) to result in a more expanded form, which simplifies to \(32x^3 + 28x^2 + 4x - 1\).
• \((f \circ f)(x)\) requires substituting \(f(x)\) into itself, resulting in \(4x + 3\).

Function operations such as these enrich our understanding of how functions interact and expand the toolkit available for analyzing mathematical relations.