Problem 16 - Calculus with Concepts in Calculus

Question

Find the domains and rules of \(f+g, f g\), and \(f / g .\) \(f(x)=\frac{x+2}{x-3}\) and \(g(x)=\frac{x+3}{x^{2}-4}\)

Step-by-Step Solution

Answer

Domains: \( f+g \), \( fg \): \( x \neq 3, 2, -2 \); \( \frac{f}{g} \): \( x \neq 3, 2, -2, -3 \).

Step 1: Determine the Domain of Function f

The function is given as \( f(x) = \frac{x+2}{x-3} \). The domain of \( f(x) \) is all the real numbers except where the denominator is zero. Thus, set \( x-3 = 0 \) which gives \( x = 3 \). Therefore, the domain of \( f(x) \) is all real numbers except \( x = 3 \).

Step 2: Determine the Domain of Function g

The function is given as \( g(x) = \frac{x+3}{x^2-4} \). The domain of \( g(x) \) is all the real numbers except where the denominator is zero, which is found by setting \( x^2-4 = 0 \). Solving \( x^2-4 = 0 \) gives \( x = 2 \) and \( x = -2 \). Thus, the domain of \( g(x) \) is all real numbers except \( x = 2 \) and \( x = -2 \).

Step 3: Find the Domain of \( f+g \)

The domain of \( f+g \) is the intersection of the domains of \( f \) and \( g \). From Steps 1 and 2, \( x eq 3, 2, -2 \). Thus, the domain of \( f+g \) is all real numbers except \( x = 3, 2, -2 \).

Step 4: Find the Rule of \( f+g \)

To find \( f+g \), add the functions: \[ (f+g)(x) = \frac{x+2}{x-3} + \frac{x+3}{x^2-4} \]. To combine these fractions, find the common denominator, which is \( (x-3)(x^2-4) \). Combine and simplify to obtain the rule for \( f+g \).

Step 5: Find the Domain of \( fg \)

The domain of \( fg \) is also the intersection of the domains of \( f \) and \( g \). Thus, the domain of \( fg \) is all real numbers except \( x = 3, 2, -2 \).

Step 6: Find the Rule of \( fg \)

To find \( fg \), multiply the functions: \[ (fg)(x) = \left( \frac{x+2}{x-3} \right) \times \left( \frac{x+3}{x^2-4} \right) \]. Multiply the numerators and the denominators to find \( (fg)(x) = \frac{(x+2)(x+3)}{(x-3)(x^2-4)} \). Simplify if possible.

Step 7: Find the Domain of \( \frac{f}{g} \)

The domain for \( \frac{f}{g} \) is all real numbers except where \( g(x) = 0 \), and wherever \( f(x) \) and \( g(x) \) are undefined. From earlier steps, \( x eq 3, 2, -2 \). Additionally, \( g(x) \) is zero when its numerator is zero, thus at \( x = -3 \) (also exclude this point). Hence, the domain is all real numbers except \( x = 3, 2, -2, -3 \).

Step 8: Find the Rule of \( \frac{f}{g} \)

To find \( \frac{f}{g}\), divide the functions: \[ \frac{f}{g}(x) = \frac{\frac{x+2}{x-3}}{\frac{x+3}{x^2-4}} = \frac{x+2}{x-3} \times \frac{x^2-4}{x+3} \]. This simplifies to \( \frac{(x+2)(x^2-4)}{(x-3)(x+3)} \). Simplify further if possible.

Key Concepts

Rational Functions

Rational functions are an intriguing part of mathematics. They are essentially the quotient of two polynomials. Think of them as the division of one polynomial by another. For instance, in the function given by \( f(x) = \frac{x+2}{x-3} \), the numerator is \( x+2 \) and the denominator is \( x-3 \). However, be mindful of the denominators. They must not be zero, as division by zero isn't defined.
The complexity often lies in finding the domain, which we'll discuss further. Simply put, the domain of a rational function includes all real numbers except those that make the denominator zero.

Function Operations

Function operations provide a way to form new functions from existing ones. Operations like addition, multiplication, and division of functions broaden your understanding of mathematical concepts.

• Addition: When tasked with \( f+g \), like our exercise, you are adding two functions. Here, you find a common denominator before adding.
• Multiplication: For \( fg \), you multiply the numerators and the denominators, simplifying where possible.
• Division: With \( \frac{f}{g} \), you replace division with multiplication involving the reciprocal of \( g(x) \). Always remember to check the denominator to ensure it's not zero.

Understanding these operations helps in solving problems related to function interaction. They may seem complex, but becoming comfortable with basic arithmetic operations on functions simplifies them.

Domain Restriction

Domain restriction is pivotal in the world of rational functions. It's all about figuring out the set of valid inputs. Put simply, it's setting the boundaries for your function. While determining the domain, focus on:

• Where the denominator is not zero. These are restrictions where the function becomes undefined.
• Finding common limitations across combined functions. For example, the domain of \( f+g \) is restricted by both \( f \) and \( g \).

In this exercise, for \( f(x) = \frac{x+2}{x-3} \), \( x \) can't be 3. Similarly, \( g(x) = \frac{x+3}{x^{2}-4} \) restricts \( x \) from being 2 or -2. Therefore, you'll exclude these from your domain to avoid undefined arithmetic.

Mathematics Education

In mathematics education, grasping the concept of domains in rational functions enriches the learning experience. Often, learners find understanding domains challenging, but with practice, the concepts become clearer.

It's important to integrate the theory with practical exercises, like finding the domain for operations between functions. This builds a deeper understanding of not just what to exclude from a domain, but why those exclusions matter.

• Practice consistently with various examples to strengthen your confidence.
• Exploring mistakes can enhance the learning process. Identifying why a solution is incorrect helps cement the concept in a student's mind.

Hence, combining mathematical theories and exercises ensures a comprehensive educational approach, paving the way for deeper understanding and application of rational functions in real-world problems.