Evaluating Composite Functions With Polynomials Step-by-Step

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In the realm of mathematics, composite functions play a crucial role in understanding how different functions interact with each other. This article delves into the evaluation of composite functions, specifically focusing on polynomial functions. We will explore how to add, subtract, and perform scalar multiplication on these functions, and then evaluate the results at specific points. By understanding these operations, we can gain a deeper insight into the behavior and properties of polynomial functions. In this article, we'll walk through a step-by-step evaluation of composite functions involving polynomials. We'll tackle expressions like (h + k)(2), (h - k)(3), and 3h(2) + 2k(3), where h(x) = x² + 1 and k(x) = x - 2. This exploration will not only solidify your understanding of function operations but also enhance your problem-solving skills in algebra and beyond. Let's begin by defining the basic functions and then move on to evaluating the composite functions.

Let's start by clearly defining the polynomial functions we will be working with. We have two functions:

  • h(x) = x² + 1: This is a quadratic function, characterized by its parabolic shape when graphed. The '+ 1' shifts the parabola upwards by one unit.
  • k(x) = x - 2: This is a linear function, which will appear as a straight line when graphed. The '- 2' indicates the y-intercept of the line.

Understanding the nature of these individual functions is crucial before we combine them. The quadratic function h(x) will always produce a non-negative value due to the square term, while the linear function k(x) will produce values that vary linearly with x. When we combine these functions, we'll see how their individual properties influence the overall result. Now, let's move on to evaluating the composite functions, starting with the sum of the functions.

To evaluate (h + k)(2), we first need to understand what this notation means. The expression (h + k)(x) represents the sum of the two functions, h(x) and k(x). Therefore, (h + k)(x) = h(x) + k(x). Once we find the combined function, we can substitute x = 2 into the resulting expression. This process involves a few key steps:

  1. Find the sum of the functions:

    • Substitute the expressions for h(x) and k(x):
      (h + k)(x) = (x² + 1) + (x - 2)
    • Combine like terms:
      (h + k)(x) = x² + x - 1

  2. Substitute x = 2:

    • Replace x with 2 in the combined function:
      (h + k)(2) = (2)² + (2) - 1
    • Simplify the expression:
      (h + k)(2) = 4 + 2 - 1 (h + k)(2) = 5

Therefore, (h + k)(2) = 5. This result tells us the value of the combined function h + k at the point x = 2. By following this step-by-step approach, we can accurately evaluate composite functions formed by addition. The key is to first find the general form of the combined function and then substitute the specific value of x. Now, let's move on to evaluating the difference of the functions.

Next, let's evaluate (h - k)(3). Similar to the previous example, this notation represents the difference between the two functions, h(x) and k(x). So, (h - k)(x) = h(x) - k(x). Again, we'll first find the combined function and then substitute x = 3 into the resulting expression. The steps are as follows:

  1. Find the difference of the functions:

    • Substitute the expressions for h(x) and k(x):
      (h - k)(x) = (x² + 1) - (x - 2)
    • Distribute the negative sign and combine like terms:
      (h - k)(x) = x² + 1 - x + 2 (h - k)(x) = x² - x + 3

  2. Substitute x = 3:

    • Replace x with 3 in the combined function:
      (h - k)(3) = (3)² - (3) + 3
    • Simplify the expression:
      (h - k)(3) = 9 - 3 + 3 (h - k)(3) = 9

Therefore, (h - k)(3) = 9. This shows us the value of the function h - k at x = 3. Notice the importance of distributing the negative sign correctly when subtracting functions; it's a common area for errors. Now, we move on to evaluating a more complex expression involving scalar multiplication and addition of functions.

Now, let's tackle the expression 3h(2) + 2k(3). This expression involves scalar multiplication of the functions h(x) and k(x), followed by addition. This means we need to evaluate h(2) and k(3) separately, multiply them by the respective scalars, and then add the results. Here’s the breakdown:

  1. Evaluate h(2):

    • Substitute x = 2 into h(x) = x² + 1:
      h(2) = (2)² + 1
    • Simplify:
      h(2) = 4 + 1 h(2) = 5

  2. Evaluate k(3):

    • Substitute x = 3 into k(x) = x - 2:
      k(3) = (3) - 2
    • Simplify:
      k(3) = 1

  3. Multiply by the scalars and add:

    • Substitute the values of h(2) and k(3) into the expression:
      3h(2) + 2k(3) = 3(5) + 2(1)
    • Simplify:
      3h(2) + 2k(3) = 15 + 2 3h(2) + 2k(3) = 17

Therefore, 3h(2) + 2k(3) = 17. This example demonstrates how to combine different operations on functions, including scalar multiplication and addition. It's crucial to follow the order of operations and evaluate each component separately before combining them. By mastering these steps, you can handle more complex composite function evaluations. Now, let's summarize the key takeaways from this discussion.

In this article, we've explored how to evaluate composite functions involving polynomials. We've covered the following key operations:

  • Addition of Functions (h + k)(x): We learned to add two functions by combining their expressions and then substituting a specific value for x.
  • Subtraction of Functions (h - k)(x): We saw the importance of distributing the negative sign correctly when subtracting one function from another.
  • Scalar Multiplication and Addition: We evaluated expressions like 3h(2) + 2k(3) by first finding the function values at specific points, multiplying them by scalars, and then adding the results.

These operations form the foundation for more advanced topics in function analysis and calculus. By practicing these techniques, you can develop a strong understanding of how functions interact and how to manipulate them effectively. Remember, the key to success in mathematics is consistent practice and a clear understanding of the underlying concepts. Now, let's conclude with a brief summary of the entire process.

Evaluating composite functions is a fundamental skill in mathematics. By understanding how to add, subtract, and perform scalar multiplication on functions, we can solve a wide range of problems. In this article, we've walked through step-by-step examples of evaluating (h + k)(2), (h - k)(3), and 3h(2) + 2k(3). These examples illustrate the importance of following the order of operations and paying attention to details such as distributing negative signs. With practice, you can become proficient in evaluating composite functions and confidently tackle more complex mathematical challenges. This understanding not only strengthens your algebraic skills but also prepares you for more advanced concepts in calculus and beyond. Keep practicing, and you'll find that composite functions become less daunting and more manageable.