Problem 66 - 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. $$ \begin{array}{llll} \text { (a) }(f \circ g)(x) & \text { (b) }(g \circ f)(x) & \text { (c) }(f \circ f)(x) \end{array} $$ $$ f(x)=2 x+1, \quad g(x)=4 x^{3}-5 x^{2} $$

Step-by-Step Solution

Answer

(a) \((f \circ g)(x) = 8x^3 - 10x^2 + 1\), domain: all real numbers. (b) \((g \circ f)(x) = 32x^3 + 28x^2 + 4x - 1\), domain: all real numbers. (c) \((f \circ f)(x) = 4x + 3\), domain: all real numbers.

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

To compute the composition \((f \circ g)(x)\), we need to substitute \(g(x)\) into \(f(x)\). The function \(g(x) = 4x^3 - 5x^2\) becomes the input for \(f(x) = 2x + 1\). Therefore, \((f \circ g)(x) = f(g(x)) = 2(4x^3 - 5x^2) + 1 = 8x^3 - 10x^2 + 1\).

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

The domain of \((f \circ g)(x)\) is determined by the domain of \(g(x)\) because \(g(x)\) must be defined for \(f\). Both functions \(f\) and \(g\) are polynomials, which are defined for all real numbers. Therefore, the domain of \((f \circ g)(x)\) is all real numbers, \((-\infty, \infty)\).

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

For \((g \circ f)(x)\), substitute \(f(x)\) into \(g(x)\). \(f(x) = 2x + 1\) becomes the input for \(g(x) = 4x^3 - 5x^2\). Thus, \((g \circ f)(x) = g(f(x)) = 4(2x + 1)^3 - 5(2x + 1)^2\). Simplify: \((2x + 1)^3 = 8x^3 + 12x^2 + 6x + 1\) and \((2x + 1)^2 = 4x^2 + 4x + 1\). Therefore, \((g \circ f)(x) = 4(8x^3 + 12x^2 + 6x + 1) - 5(4x^2 + 4x + 1)\). Simplifying yields \(32x^3 + 48x^2 + 24x + 4 - 20x^2 - 20x - 5 = 32x^3 + 28x^2 + 4x - 1\).

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

The domain of \((g \circ f)(x)\) is determined by the domain of \(f(x)\), which is all real numbers, \((-\infty, \infty)\). Since \(g(x)\) is also a polynomial, it is defined wherever \(f(x)\) is defined, so the domain is all real numbers, \((-\infty, \infty)\).

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

To compute \((f \circ f)(x)\), substitute \(f(x)\) into itself. \(f(x) = 2x + 1\) is substituted into \(f(x) = 2x + 1\), so \((f \circ f)(x) = f(f(x)) = 2(2x + 1) + 1 = 4x + 2 + 1 = 4x + 3\).

Step 6: Identify Domain of \((f \circ f)(x)\)

Polynomials are defined for all real numbers, and the composition \((f \circ f)(x)\) is found by repeatedly applying \(f\), which is a polynomial. Hence, the domain of \((f \circ f)(x)\) is all real numbers, \((-\infty, \infty)\).

Key Concepts

Domain Determination

Knowing the domain of a function is vital. The domain tells us the input values for which the function is defined. Elements outside this set are not processed by the function. Let's dive deeper into how we determine this for function compositions.

For functions like - polynomials, the domain is straightforward. Both \(f(x) = 2x + 1\) and \(g(x) = 4x^3 - 5x^2\) are polynomial functions.Therefore, they are defined for all real numbers, written as \((-\infty, \infty)\).

When composing functions, we must ensure that the inputs of the inner function (say \(g(x)\)) are within the valid range of the outer function (here, \(f(x)\)). This means the domain of \((f \circ g)(x)\) is determined by the domain of \(g(x)\), and it remains \((-\infty, \infty)\).

The same principle applies to \((g \circ f)(x)\). The domain is determined by \(f(x)\), again resulting in \((-\infty, \infty)\). Finally, for \((f \circ f)(x)\), since it's just \(f(x)\) composed with itself, the domain stays \((-\infty, \infty)\).

Polynomial Functions

Polynomial functions are fundamental in algebra and appear in many math problems. Understanding them is crucial for tackling function compositions.

A polynomial function consists of sums and products of variables and coefficients. The simplest example is \(f(x) = 2x + 1\), which is linear with a constant and a single variable term. Another example is \(g(x) = 4x^3 - 5x^2\), a cubic polynomial.

The degree of the polynomial is the highest power of the variable. For \(g(x)\), the degree is 3, while \(f(x)\) has a degree of 1.Polyomials are continuous and defined for all real numbers, which simplifies domain determination.

They also have certain determined properties:- They are smooth and continuous- They don't have any gaps, jumps, or holes- They increase or decrease in predictably shaped curves

Algebraic Manipulation

Algebraic manipulation is key when composing more elaborate functions. It involves expanding, simplifying, and rearranging algebraic expressions so that they are easier to work with or solve.

Consider finding \((g \circ f)(x)\) : The task required expanding \((2x + 1)^3\) and \((2x + 1)^2\) as part of the process. The expanded form ensures all parts of the calculation are visible and combinable.

The next step is combining like terms — simplifying \(4(8x^3 + 12x^2 + 6x + 1) - 5(4x^2 + 4x + 1)\). Through careful addition and subtraction, a cleaner, simpler function emerges: \(32x^3 + 28x^2 + 4x - 1\).

The technique simplifies compositions by breaking them into smaller, more manageable problems. With practice, one can manipulate these expressions efficiently, significantly expanding the range of function work even when complex.