Evaluate Functions: Finding G(2a), G(a-1), And G(a+1)-g(a-1)

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In this article, we will delve into the fascinating world of function evaluation, specifically focusing on the function g(x) = 3x² - 4. Function evaluation is a fundamental concept in mathematics, allowing us to determine the output of a function for a given input. We will explore how to evaluate this function for different expressions, including g(2a), g(a-1), and the difference g(a+1) - g(a-1). By understanding these evaluations, we will gain a deeper understanding of the function's behavior and its properties. This exploration is essential for anyone studying algebra, calculus, or any field that utilizes mathematical functions. This article aims to provide a clear and concise explanation of the function evaluation process, making it accessible to students and enthusiasts alike. Whether you are a seasoned mathematician or just beginning your journey, this guide will equip you with the skills and knowledge necessary to confidently evaluate functions. So, let's embark on this exciting journey and unlock the secrets of function evaluation!

a. Finding g(2a)

To find g(2a), we need to substitute the input 'x' in the function g(x) with the expression '2a'. This process involves replacing every instance of 'x' in the function's definition with '2a'. Let's break down the steps:

  1. Start with the original function: g(x) = 3x² - 4
  2. Substitute x with 2a: g(2a) = 3(2a)² - 4
  3. Simplify the expression:
    • (2a)² = (2a) * (2a) = 4a²
    • g(2a) = 3(4a²) - 4
    • g(2a) = 12a² - 4

Therefore, g(2a) = 12a² - 4. This result demonstrates how the function's output changes when the input is a multiple of 'a'. The key takeaway here is the importance of careful substitution and simplification when evaluating functions with algebraic expressions. We've effectively shown how the function g(x) transforms when its input is scaled by a factor of 2. This concept is crucial for understanding the function's behavior and its transformations. Understanding these transformations allows us to predict the function's output for various inputs, which is a fundamental skill in mathematics and its applications. Moreover, this process highlights the power of algebraic manipulation in simplifying complex expressions and arriving at a concise result. By mastering these techniques, we can unlock a deeper understanding of mathematical functions and their properties. This skill is not only valuable in academic settings but also in various real-world applications where mathematical modeling and analysis are essential.

b. Finding g(a-1)

Next, we will determine g(a-1), which involves substituting 'x' with the expression 'a-1' in the function g(x). This substitution is slightly more complex than the previous one, as it involves squaring a binomial expression. Let's go through the steps:

  1. Start with the original function: g(x) = 3x² - 4
  2. Substitute x with (a-1): g(a-1) = 3(a-1)² - 4
  3. Expand the squared binomial (a-1)²:
    • (a-1)² = (a-1) * (a-1) = a² - a - a + 1 = a² - 2a + 1
  4. Substitute the expanded form back into the expression: g(a-1) = 3(a² - 2a + 1) - 4
  5. Distribute the 3: g(a-1) = 3a² - 6a + 3 - 4
  6. Simplify the expression: g(a-1) = 3a² - 6a - 1

Thus, g(a-1) = 3a² - 6a - 1. This result shows how the function's output is affected when the input is a linear expression involving 'a'. The process of expanding the squared binomial is a critical skill in algebra, and this example demonstrates its application in function evaluation. We've successfully navigated the complexities of substituting a binomial expression into the function and simplifying the result. This exercise reinforces the importance of careful algebraic manipulation and attention to detail. Understanding how to handle these types of substitutions is crucial for working with more complex functions and equations. Furthermore, this example highlights the interconnectedness of different algebraic concepts. The ability to expand binomial expressions is not just a standalone skill; it's an essential tool for solving a wide range of mathematical problems. By mastering these fundamental skills, we build a solid foundation for tackling more advanced topics in mathematics.

c. Finding g(a+1) - g(a-1)

Now, we will tackle the most complex part of the problem: finding g(a+1) - g(a-1). This requires us to first evaluate g(a+1) and then subtract g(a-1) from the result. We already know g(a-1) from the previous step, so we just need to find g(a+1).

  1. Start with the original function: g(x) = 3x² - 4
  2. Substitute x with (a+1): g(a+1) = 3(a+1)² - 4
  3. Expand the squared binomial (a+1)²:
    • (a+1)² = (a+1) * (a+1) = a² + a + a + 1 = a² + 2a + 1
  4. Substitute the expanded form back into the expression: g(a+1) = 3(a² + 2a + 1) - 4
  5. Distribute the 3: g(a+1) = 3a² + 6a + 3 - 4
  6. Simplify the expression: g(a+1) = 3a² + 6a - 1

Now that we have g(a+1) = 3a² + 6a - 1 and g(a-1) = 3a² - 6a - 1, we can find the difference:

  1. Subtract g(a-1) from g(a+1):
    • g(a+1) - g(a-1) = (3a² + 6a - 1) - (3a² - 6a - 1)
  2. Distribute the negative sign:
    • g(a+1) - g(a-1) = 3a² + 6a - 1 - 3a² + 6a + 1
  3. Combine like terms:
    • g(a+1) - g(a-1) = (3a² - 3a²) + (6a + 6a) + (-1 + 1)
    • g(a+1) - g(a-1) = 0 + 12a + 0

Therefore, g(a+1) - g(a-1) = 12a. This result is particularly interesting because it shows that the difference between the function's values at (a+1) and (a-1) is a linear function of 'a'. The process of finding this difference highlights the power of algebraic simplification and the cancellation of terms. We've successfully navigated the complexities of evaluating the function at two different inputs and then finding their difference. Understanding these types of operations is crucial for analyzing the function's behavior and its properties. This example reinforces the importance of meticulous algebraic manipulation and attention to detail. Furthermore, the result reveals a hidden property of the function g(x), which is that the difference g(a+1) - g(a-1) is a linear function. This discovery underscores the value of exploring functions and their properties through algebraic manipulation. By mastering these techniques, we can gain deeper insights into the nature of mathematical functions and their applications.

In conclusion, this exercise has provided a comprehensive exploration of function evaluation for the function g(x) = 3x² - 4. We have successfully evaluated the function for different inputs, including g(2a), g(a-1), and g(a+1) - g(a-1). These evaluations have demonstrated the importance of careful substitution, algebraic manipulation, and simplification. By mastering these techniques, we can confidently evaluate functions and gain a deeper understanding of their behavior and properties. This knowledge is essential for anyone studying mathematics or any field that utilizes mathematical functions. Mastering function evaluation is a crucial step in building a strong foundation in mathematics. The skills and concepts learned in this exercise will be invaluable as you progress in your mathematical journey. We encourage you to practice these techniques with other functions and expressions to further solidify your understanding. The ability to evaluate functions is a powerful tool that will serve you well in various mathematical contexts. This exploration has not only enhanced our understanding of function evaluation but has also highlighted the beauty and elegance of algebraic manipulation.