Evaluating Functions And Finding Domains G(x) And F(x)

by ADMIN 55 views

In the realm of mathematics, functions serve as fundamental building blocks for modeling relationships between variables. Mastering the manipulation and understanding of functions is crucial for success in various mathematical disciplines. This article delves into the intricacies of function operations, focusing on two specific functions, g(x) = (x - 6)(x + 2) and f(x) = 9x + 2. We will explore how to evaluate the quotient of these functions at a given point and determine the values excluded from the domain of the resulting function. Let's embark on this mathematical journey to enhance your understanding of functions.

(a) Evaluating (g/f)(-3): A Step-by-Step Approach

To begin our exploration, let's tackle the first part of the problem: finding (g/f)(-3). This notation represents the quotient of the functions g(x) and f(x) evaluated at x = -3. In simpler terms, we need to divide g(-3) by f(-3). To achieve this, we will follow a step-by-step approach, ensuring clarity and precision in our calculations.

Step 1: Calculate g(-3)

Our first task is to determine the value of g(-3). Recall that g(x) = (x - 6)(x + 2). To find g(-3), we simply substitute -3 for x in the expression for g(x):

g(-3) = (-3 - 6)(-3 + 2)

Now, we perform the arithmetic operations within the parentheses:

g(-3) = (-9)(-1)

Finally, we multiply the two negative numbers, resulting in a positive value:

g(-3) = 9

Therefore, the value of g(-3) is 9. This intermediate result is crucial for the next step in our calculation.

Step 2: Calculate f(-3)

Next, we need to find the value of f(-3). We are given that f(x) = 9x + 2. Similar to the previous step, we substitute -3 for x in the expression for f(x):

f(-3) = 9(-3) + 2

Now, we perform the multiplication:

f(-3) = -27 + 2

Finally, we add the two numbers:

f(-3) = -25

Thus, the value of f(-3) is -25. This result is equally important for calculating the quotient of the functions.

Step 3: Calculate (g/f)(-3)

Now that we have determined g(-3) and f(-3), we can calculate (g/f)(-3). Recall that (g/f)(-3) represents the quotient of g(-3) and f(-3):

(g/f)(-3) = g(-3) / f(-3)

Substituting the values we calculated earlier:

(g/f)(-3) = 9 / (-25)

Therefore, (g/f)(-3) is equal to -9/25. This completes the first part of our problem, demonstrating how to evaluate the quotient of two functions at a specific point.

(b) Identifying Values NOT in the Domain of (g/f)(x)

The second part of our problem requires us to find all values that are NOT in the domain of (g/f)(x). The domain of a function is the set of all possible input values (x-values) for which the function is defined. In the case of a quotient of functions, we need to consider the values that would make the denominator equal to zero, as division by zero is undefined. Therefore, to find the values NOT in the domain of (g/f)(x), we need to determine the values of x for which f(x) = 0.

Step 1: Set f(x) equal to zero

We begin by setting the expression for f(x) equal to zero:

f(x) = 9x + 2 = 0

This equation represents the condition that must be satisfied for x to be excluded from the domain of (g/f)(x).

Step 2: Solve for x

Now, we solve the equation for x. First, we subtract 2 from both sides:

9x = -2

Next, we divide both sides by 9:

x = -2/9

Therefore, the value x = -2/9 makes the denominator, f(x), equal to zero. This value must be excluded from the domain of (g/f)(x).

Step 3: State the values not in the domain

The only value that is not in the domain of (g/f)(x) is x = -2/9. This is because when x = -2/9, f(x) = 0, and division by zero is undefined. Therefore, the function (g/f)(x) is not defined at x = -2/9.

Conclusion: Mastering Function Operations

In this comprehensive guide, we have explored the intricacies of function operations, specifically focusing on the quotient of two functions, g(x) = (x - 6)(x + 2) and f(x) = 9x + 2. We successfully evaluated (g/f)(-3) and determined the value that is not in the domain of (g/f)(x). By understanding these concepts, you can confidently tackle similar problems involving function operations. Remember, the key to success lies in a clear understanding of the definitions and a meticulous approach to calculations. Keep practicing, and you'll master the art of function manipulation in no time!

By understanding the concepts of evaluating functions and identifying values excluded from their domains, you can build a strong foundation in mathematics and excel in your studies. Remember to practice regularly and seek clarification whenever needed. With dedication and perseverance, you can unlock the fascinating world of functions and their applications.

In summary, we have learned how to evaluate the quotient of two functions at a specific point and how to determine the values that are not in the domain of the resulting function. These skills are essential for success in higher-level mathematics courses and various applications in science and engineering. Embrace the challenges, and you'll discover the beauty and power of functions in solving real-world problems.

  • Functions
  • Domain
  • Quotient
  • g(x)
  • f(x)
  • Evaluation
  • Mathematics
  • Algebra
  • Function Operations
  • Excluded Values
  • Function Division
  • Mathematical Functions