Solving For Quotient Of Functions If R(x) = 3x - 1 And S(x) = 2x + 1

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In the realm of mathematics, functions are fundamental building blocks. They describe relationships between inputs and outputs, and mathematical operations can be performed on them. One such operation is finding the quotient of two functions. In this article, we'll explore how to determine the expression equivalent to the quotient of two functions, specifically when given two linear functions, r(x) = 3x - 1 and s(x) = 2x + 1, and asked to evaluate (r/s)(6). This comprehensive guide will walk you through the process step-by-step, ensuring a clear understanding of the concepts involved. We'll break down the problem, discuss the relevant mathematical principles, and provide a detailed solution. Understanding the quotient of functions is a crucial skill in algebra and calculus, laying the groundwork for more advanced mathematical concepts. So, let's dive in and unravel this problem together!

Defining the Quotient of Functions

The quotient of two functions, denoted as (r/s)(x), is essentially dividing one function by another. Mathematically, it's expressed as (r/s)(x) = r(x) / s(x), where r(x) and s(x) are the two functions, and crucially, s(x) cannot be equal to zero. This is because division by zero is undefined in mathematics. The domain of the quotient function (r/s)(x) is the set of all x values that are in the domains of both r(x) and s(x), excluding any x values that make s(x) = 0. To truly grasp the concept of the quotient of functions, it's important to understand the individual functions themselves. Functions can be linear, quadratic, exponential, trigonometric, and more. Each type of function has its own unique properties and characteristics. When dealing with the quotient of functions, we're essentially combining the behaviors of these individual functions through division. This can lead to interesting and sometimes complex results, especially when considering the domain and potential discontinuities caused by the denominator function (s(x)) becoming zero. The quotient of functions is a fundamental concept that appears in various areas of mathematics, including calculus, where it plays a crucial role in finding derivatives and limits. By mastering this concept, you'll gain a valuable tool for analyzing and understanding mathematical relationships.

Applying the Concept to the Given Functions

In our specific problem, we are given two functions: r(x) = 3x - 1 and s(x) = 2x + 1. To find (r/s)(6), we need to first determine the general expression for (r/s)(x) and then substitute x = 6. The expression for (r/s)(x) is simply r(x) divided by s(x), which gives us (r/s)(x) = (3x - 1) / (2x + 1). This is a rational function, which is a function expressed as the ratio of two polynomials. Now, to evaluate (r/s)(6), we substitute x = 6 into the expression we just found. This gives us (r/s)(6) = (3(6) - 1) / (2(6) + 1). It's crucial to follow the order of operations (PEMDAS/BODMAS) to correctly evaluate this expression. First, we perform the multiplications in the numerator and denominator: 3(6) = 18 and 2(6) = 12. Then, we substitute these values back into the expression: (r/s)(6) = (18 - 1) / (12 + 1). Finally, we perform the subtractions and additions in the numerator and denominator to obtain the numerical value of (r/s)(6). This process of substituting a specific value into a function and evaluating the result is a fundamental skill in mathematics. It allows us to determine the output of a function for a given input and is essential for understanding the behavior of functions and their applications in real-world scenarios.

Step-by-Step Solution

Let's break down the calculation of (r/s)(6) step-by-step:

  1. Write the expression for (r/s)(x): As we established earlier, (r/s)(x) = r(x) / s(x). Substituting the given functions, we get (r/s)(x) = (3x - 1) / (2x + 1).
  2. Substitute x = 6 into the expression: Replace every instance of x in the expression with 6: (r/s)(6) = (3(6) - 1) / (2(6) + 1).
  3. Perform the multiplications: Multiply 3 by 6 in the numerator and 2 by 6 in the denominator: (r/s)(6) = (18 - 1) / (12 + 1).
  4. Perform the subtractions and additions: Subtract 1 from 18 in the numerator and add 1 to 12 in the denominator: (r/s)(6) = 17 / 13.
  5. Simplify the fraction (if possible): In this case, 17/13 is already in its simplest form as 17 and 13 are both prime numbers and have no common factors other than 1. Therefore, (r/s)(6) = 17 / 13. This step-by-step approach ensures that we arrive at the correct answer by systematically applying the definition of the quotient of functions and following the order of operations. By breaking down the problem into smaller, manageable steps, we can minimize the chances of making errors and gain a deeper understanding of the process involved. This methodical approach is crucial for solving mathematical problems effectively and efficiently.

Analyzing the Answer Choices

Now, let's relate our solution to the given answer choices. The expression we derived for (r/s)(6) is 17/13. We need to identify which of the provided options is equivalent to this value. Looking at the answer choices, we can see that:

  • The option (3(6) - 1) / (2(6) + 1) directly represents the substitution of x = 6 into the expression (r/s)(x) = (3x - 1) / (2x + 1). This is the exact step we performed in our solution, making it the correct answer.
  • The option (8) / (2(6) + 1) is incorrect because it seems to have an error in calculating the numerator. 3(6) - 1 is equal to 17, not 8.
  • The option (36 - 1) / (26 + 1) is also incorrect. It appears there's a misunderstanding of function notation; 3(6) means 3 multiplied by 6, not 36, and similarly for 2(6).
  • The option (6) - 1 / (6) + 1 is incorrect as well. This option seems to have incorrectly applied the value of 6 to the functions and also has an incorrect placement of parentheses, leading to a different calculation altogether.

Therefore, the only option that accurately represents the value of (r/s)(6) is (3(6) - 1) / (2(6) + 1). This exercise highlights the importance of carefully following the steps in the solution and accurately interpreting mathematical expressions and notations. By comparing our derived solution with the answer choices, we can confidently identify the correct answer and reinforce our understanding of the concepts involved.

Key Takeaways and Generalization

This problem demonstrates a fundamental concept in mathematics: the quotient of functions. We learned that (r/s)(x) is obtained by dividing the function r(x) by the function s(x), expressed as (r/s)(x) = r(x) / s(x). When evaluating (r/s)(a) for a specific value a, we substitute x = a into the expression for (r/s)(x) and simplify. In this case, we found that (r/s)(6) = (3(6) - 1) / (2(6) + 1) = 17/13. This process can be generalized to any two functions, r(x) and s(x), and any value a within their domains (excluding values that make s(a) = 0). The key steps remain the same:

  1. Determine the expression for (r/s)(x): Divide r(x) by s(x).
  2. Substitute x = a: Replace every x in the expression with a.
  3. Simplify: Follow the order of operations to evaluate the expression.

Understanding the quotient of functions is crucial for various mathematical applications, including calculus, where it's used in finding derivatives and limits. Mastering this concept provides a solid foundation for tackling more complex mathematical problems. Furthermore, this problem highlights the importance of careful substitution, following the order of operations, and accurately interpreting mathematical notation. These skills are essential for success in mathematics and other STEM fields. By practicing and applying these concepts, you can build your mathematical fluency and problem-solving abilities.

Conclusion

In conclusion, we have successfully determined the expression equivalent to (r/s)(6), given the functions r(x) = 3x - 1 and s(x) = 2x + 1. By understanding the concept of the quotient of functions, applying the correct substitution, and following the order of operations, we arrived at the solution: (r/s)(6) = (3(6) - 1) / (2(6) + 1). This exercise demonstrates the importance of mastering fundamental mathematical concepts and applying them systematically to solve problems. The quotient of functions is a key concept that appears in various areas of mathematics, and a solid understanding of it is crucial for further studies in algebra, calculus, and beyond. By practicing similar problems and applying the step-by-step approach outlined in this article, you can strengthen your mathematical skills and build confidence in your ability to tackle complex problems. Remember, mathematics is a building process, and each concept learned lays the foundation for future learning. So, continue to explore, practice, and challenge yourself, and you'll be well on your way to mathematical success.