Problem 28 - A Graphical Approach to Precalculus with Limits

Question

For each pair of fimetions, (a) find ( \(f+g)(x),(f-g)(x),\) and \((f g)(x) ;\) (b) give the domains of the functions in part (a); (c) find \(\frac{t}{x}\) and give its domain; (d) find \(f \circ g\) and give is domain: and (e) find \(g \circ f\) and give its domain. Do not use a calculator. $$f(x)=\sqrt[3]{6-3 x}, g(x)=2 x^{3}+1$$

Step-by-Step Solution

Answer

Domains: All real numbers except \(-\sqrt[3]{1/2}\) for \(\frac{f}{g}(x)\); others are all real numbers.

Step 1: Define Function Operations

Given the functions \( f(x) = \sqrt[3]{6-3x} \) and \( g(x) = 2x^3 + 1 \), we need to find several operations between them. Begin by identifying \((f+g)(x)\), \((f-g)(x)\), and \((fg)(x)\): 1. \((f+g)(x) = \sqrt[3]{6-3x} + (2x^3 + 1) = \sqrt[3]{6-3x} + 2x^3 + 1\)2. \((f-g)(x) = \sqrt[3]{6-3x} - (2x^3 + 1) = \sqrt[3]{6-3x} - 2x^3 - 1\)3. \((fg)(x) = \sqrt[3]{6-3x} \times (2x^3 + 1) = \sqrt[3]{6-3x} \cdot (2x^3 + 1)\).

Step 2: Determine Domains for Operations

Next, we determine the domains for these operations. Since \(\sqrt[3]{6-3x}\) is defined for all real numbers and \(2x^3+1\) is also defined for all real numbers, the domains for \((f+g)(x)\), \((f-g)(x)\), and \((fg)(x)\) are all real numbers, \( \mathbb{R} \).

Step 3: Find Division of Functions

We must now find \( \frac{f}{g}(x) \) which is \( \frac{\sqrt[3]{6-3x}}{2x^3 + 1} \). Since division by zero is undefined, its domain excludes values of \(x\) that make the denominator zero. Therefore, solve \(2x^3 + 1 = 0\): \[ 2x^3 + 1 = 0 \implies 2x^3 = -1 \implies x^3 = -\frac{1}{2} \implies x = -\sqrt[3]{\frac{1}{2}} \approx -0.7937 \]Thus, the domain is all real numbers except \(-\sqrt[3]{\frac{1}{2}}\).

Step 4: Compute Composition (f ◦ g)

To find \( f \circ g \), substitute \( g(x) \) into \( f(x) \): \[ (f \circ g)(x) = f(g(x)) = \sqrt[3]{6 - 3(2x^3 + 1)} = \sqrt[3]{6 - 6x^3 - 3} = \sqrt[3]{3 - 6x^3} \]Here \( f \circ g \) is defined for all real numbers since a cube root is defined for all real inputs.

Step 5: Compute Composition (g ◦ f)

To find \( g \circ f \), substitute \( f(x) \) into \( g(x) \):\[ (g \circ f)(x) = g(f(x)) = 2(\sqrt[3]{6-3x})^3 + 1 \]Since \((\sqrt[3]{6-3x})^3 = 6 - 3x\), \[ (g \circ f)(x) = 2(6 - 3x) + 1 = 12 - 6x + 1 = 13 - 6x \]No restrictions on the cube root or resultant expression, so the domain is all real numbers.

Key Concepts

Function Composition

Function composition is a technique to combine two or more functions, creating a new function. Let’s consider two functions, say \( f(x) \) and \( g(x) \). The composition of these functions, denoted as \( f \circ g \), is defined as \( f(g(x)) \). This means you first apply \( g(x) \), and then the result is used as the input for \( f(x) \).

In this exercise, we are given \( f(x) = \sqrt[3]{6-3x} \) and \( g(x) = 2x^3 + 1 \). For \( f \circ g \), we substitute \( g(x) \) into \( f \), resulting in \( f(g(x)) = \sqrt[3]{3 - 6x^3} \).

Similarly, when calculating \( g \circ f \), you start with \( f(x) \) and plug it into \( g \), giving us \( g(f(x)) = 13 - 6x \). Both compositions are straightforward due to the application of the cube root and polynomial operations.

Domain of Functions

The domain of a function is the complete set of possible values of the independent variable, typically \( x \), which makes the function work without any mathematical issues, such as division by zero or taking the square root of a negative number.

In the original exercise, calculating the domain for operations such as \((f + g)(x)\), \((f - g)(x)\), and \((fg)(x)\) is straightforward because each component function is defined for all real numbers. This makes the domain of these operations to be all real numbers \( \mathbb{R} \).

However, when considering \( \frac{f}{g}(x) \), we must exclude any \( x \) that makes the denominator zero, which involves solving \( 2x^3 + 1 = 0 \). The solution is \( x = -\sqrt[3]{\frac{1}{2}} \), so the domain of the division is all real numbers except this specific value.

Function Division

Function division involves creating a new function by dividing one function by another, symbolically written as \( \frac{f}{g}(x) \). The key challenge is ensuring the denominator never becomes zero, as division by zero is undefined.

In our problem, the expression \( \frac{\sqrt[3]{6-3x}}{2x^3 + 1} \) requires careful consideration of when \( 2x^3 + 1 = 0 \). Solving this gives \( x = -\sqrt[3]{\frac{1}{2}} \), a point where the function is undefined. Hence, the domain excludes this solution, allowing all other real numbers.

This process is crucial for understanding restrictions on domains, particularly in division scenarios, ensuring that the function remains valid across the defined set of input values.

Cubic Root Functions

Cubic root functions are based on taking the cube root of a number or expression. The function \( f(x) = \sqrt[3]{6-3x} \), for instance, involves the cube root, which is more forgiving than its even root counterparts, like the square root. Since cube roots are defined for all real numbers, \( f(x) \) has no restrictions on its domain.

Cube root functions allow negative inputs because each negative value has a real cube root. This is why the function \( f(x) \) participates easily in addition, subtraction, multiplication, and composition with little to no restriction concerns.

Understanding cubic root functions is essential for tackling algebraic problems that involve transforming data or inputs, as these functions provide a broad spectrum of applicable real numbers, simplifying many operations.