Hey guys! Today, we are diving deep into the fascinating world of function operations. Function operations might sound intimidating, but trust me, they're super straightforward once you grasp the basic concepts. We'll be tackling a problem that involves adding, subtracting, and scaling functions. Specifically, we're going to evaluate expressions like (h+k)(2), (h-k)(3), and 3h(2)+2k(3). So, buckle up, grab your thinking caps, and let's get started!
Understanding Function Operations
Before we jump into the problem, let's quickly recap what function operations are all about. In essence, function operations are ways to combine two or more functions to create a new function. The most common operations are addition, subtraction, multiplication, and division. When you perform these operations, you're essentially applying the operation to the outputs of the functions for a given input. This means that to truly master function operations, a solid foundation in understanding function notation and evaluation is essential. When we talk about function notation, we are referring to the way we write functions, typically using letters like f, g, or h, followed by the input variable in parentheses, like f(x). This notation tells us that f is a function that takes x as an input. Evaluating a function, on the other hand, involves substituting a specific value for the input variable and calculating the resulting output. For example, if we have the function f(x) = x² + 1, evaluating f(2) means replacing x with 2, giving us f(2) = 2² + 1 = 5. This fundamental understanding of function notation and evaluation forms the bedrock upon which more complex function operations are built. Without a firm grasp of these basics, tackling problems involving combinations of functions can become unnecessarily challenging. Therefore, before delving into the intricacies of adding, subtracting, multiplying, or dividing functions, it is crucial to ensure that you are completely comfortable with the mechanics of function notation and the process of evaluating functions for specific input values. This foundational knowledge will not only make function operations easier to understand but will also provide a solid platform for more advanced mathematical concepts later on. So, if you ever find yourself struggling with function operations, take a step back and revisit these core concepts – it will make a world of difference!
Addition of Functions (h+k)(x)
When adding functions, denoted as (h+k)(x), we simply add the outputs of the individual functions h(x) and k(x) for the same input x. Mathematically, this is expressed as (h+k)(x) = h(x) + k(x). This operation is very intuitive; you're basically combining the results of two separate functions into one. To truly grasp this concept, let’s break down the mechanics of adding functions. Imagine you have two functions, h(x) and k(x), each representing a different process or relationship. For any given input x, each function will produce an output. The addition of these functions, (h+k)(x), is simply the sum of these individual outputs. In other words, you evaluate h(x) and k(x) separately and then add the results together. This process can be visualized as combining the ‘effects’ of the two functions. For instance, if h(x) represents the cost of materials for a project and k(x) represents the labor cost, then (h+k)(x) would represent the total cost of the project. The beauty of this operation lies in its simplicity and versatility. It allows us to model complex situations by combining simpler, more manageable functions. Furthermore, understanding the addition of functions is a stepping stone to grasping other function operations like subtraction, multiplication, and composition. It’s a foundational concept that underpins much of advanced mathematics and its applications in various fields. Therefore, mastering this basic operation is crucial for anyone looking to delve deeper into the world of functions and their applications. Remember, the key is to think of it as simply adding the outputs – a straightforward yet powerful way to combine mathematical relationships.
Subtraction of Functions (h-k)(x)
Similarly, the subtraction of functions, written as (h-k)(x), involves subtracting the output of k(x) from the output of h(x) for the same input x. This is represented as (h-k)(x) = h(x) - k(x). The order is crucial here, as subtraction is not commutative. When we delve into the subtraction of functions, the order in which we perform the operation becomes paramount. Unlike addition, where the order of the functions doesn't affect the result, subtraction is highly sensitive to the sequence. The expression (h-k)(x) explicitly means that we are subtracting the output of the function k(x) from the output of the function h(x). In mathematical terms, this is represented as (h-k)(x) = h(x) - k(x). Reversing the order, i.e., (k-h)(x), would yield a completely different result because we would then be subtracting h(x) from k(x). This distinction is crucial to understand, as it directly impacts the final outcome. To illustrate this importance, consider two functions, h(x) = x² and k(x) = x. If we calculate (h-k)(2), we get h(2) - k(2) = 2² - 2 = 4 - 2 = 2. However, if we calculate (k-h)(2), we get k(2) - h(2) = 2 - 2² = 2 - 4 = -2. As you can see, the results are different, highlighting the non-commutative nature of function subtraction. This sensitivity to order has practical implications in various fields. For instance, in economics, if h(x) represents revenue and k(x) represents costs, then (h-k)(x) represents profit. Subtracting in the reverse order would give a meaningless or even misleading result. Therefore, when working with function subtraction, always pay close attention to the order of the functions to ensure accurate calculations and interpretations. This meticulous approach is essential for mastering function operations and applying them effectively in real-world scenarios.
Scalar Multiplication of Functions
Scalar multiplication involves multiplying a function by a constant (a scalar). For example, 3h(x) means multiplying the output of h(x) by 3. Similarly, 2k(x) means multiplying the output of k(x) by 2. Let's break this down further. Scalar multiplication of functions is a fundamental operation in mathematics that involves scaling the output of a function by a constant value. This constant value, often referred to as a scalar, essentially stretches or compresses the function's output, depending on its magnitude. For instance, if we have a function h(x), then 3h(x) means that for every input x, the output of h(x) is multiplied by 3. This multiplication affects the function's graph by vertically stretching it by a factor of 3. Similarly, multiplying by a fraction, like 1/2, would compress the graph vertically. The concept of scalar multiplication extends beyond just single functions. When we encounter expressions like 3h(x) + 2k(x), we are combining scalar multiplication with addition. This means that we first multiply the function h(x) by 3 and the function k(x) by 2, and then we add the resulting functions together. This combination allows us to create more complex and nuanced mathematical models. To illustrate this, consider h(x) representing the number of hours spent studying and k(x) representing the number of hours spent working. If we want to model the total time commitment, but we value studying three times as much as working, we might use the expression 3h(x) + k(x). This shows how scalar multiplication can be used to weigh different functions in a mathematical model. Understanding scalar multiplication is crucial for various mathematical and scientific applications. It allows us to adjust the magnitude of functions, combine them in meaningful ways, and create models that accurately represent real-world phenomena. From physics to economics, scalar multiplication plays a vital role in shaping our understanding of the world through mathematical representations. Therefore, mastering this operation is not just about manipulating equations; it's about developing a powerful tool for analysis and problem-solving.
Problem Breakdown and Solution
Now, let's tackle the problem at hand. We are given two functions:
- h(x) = x² + 1
- k(x) = x - 2
And we need to find:
- (h+k)(2)
- (h-k)(3)
- 3h(2) + 2k(3)
Let's break down each part step-by-step.
1. Evaluating (h+k)(2)
First, we need to find (h+k)(x), which is simply h(x) + k(x). So,
(h+k)(x) = (x² + 1) + (x - 2) = x² + x - 1
Now, we substitute x = 2 into the combined function:
(h+k)(2) = (2)² + 2 - 1 = 4 + 2 - 1 = 5
Therefore, (h+k)(2) = 5. To truly master the evaluation of composite functions like (h+k)(2), it’s beneficial to delve into a step-by-step breakdown of the process. Initially, we establish the combined function (h+k)(x) by adding the individual functions h(x) and k(x). Given that h(x) = x² + 1 and k(x) = x - 2, the addition yields (h+k)(x) = (x² + 1) + (x - 2) = x² + x - 1. This step is crucial as it simplifies the overall evaluation by creating a single function that represents the combined effect of h and k. Once we have the combined function, the next step involves substituting the specific value of x into (h+k)(x). In this case, we are interested in finding (h+k)(2), which means we replace every instance of x in the expression x² + x - 1 with the number 2. This substitution transforms the algebraic expression into an arithmetic one, setting the stage for straightforward calculation. Following the substitution, we perform the arithmetic operations to arrive at the final result. Substituting x = 2 into (h+k)(x) = x² + x - 1 gives us (h+k)(2) = (2)² + 2 - 1. Evaluating this expression involves squaring 2 to get 4, then adding 2, and finally subtracting 1. The sequence of operations, following the order of precedence (exponents before addition and subtraction), leads us to the answer: 4 + 2 - 1 = 5. Thus, the final value of (h+k)(2) is 5. This step-by-step methodology not only provides the correct answer but also reinforces the understanding of how functions operate and interact. By breaking down the problem into manageable parts, we can tackle more complex function evaluations with confidence and accuracy. This approach is particularly valuable when dealing with combinations of functions, as it provides a clear and systematic way to navigate the problem. Therefore, when faced with similar problems, remember to first combine the functions, then substitute the value, and finally, evaluate the expression. This methodical approach is the key to success in function evaluation.
2. Evaluating (h-k)(3)
Next, we find (h-k)(x), which is h(x) - k(x).
(h-k)(x) = (x² + 1) - (x - 2) = x² + 1 - x + 2 = x² - x + 3
Now, substitute x = 3:
(h-k)(3) = (3)² - 3 + 3 = 9 - 3 + 3 = 9
So, (h-k)(3) = 9. Understanding the intricacies of function subtraction, particularly in the context of evaluating expressions like (h-k)(3), involves a careful and methodical approach. The initial step in this process is to determine the combined function (h-k)(x). This is achieved by subtracting the function k(x) from the function h(x). Given our functions h(x) = x² + 1 and k(x) = x - 2, the subtraction process unfolds as follows: (h-k)(x) = (x² + 1) - (x - 2). The crucial aspect of this step is the correct application of the subtraction operation, especially when dealing with expressions in parentheses. We must distribute the negative sign across the terms inside the parentheses of k(x). This means that we change the sign of each term in (x - 2) before combining it with h(x). Thus, the expression becomes x² + 1 - x + 2. Now, we simplify the expression by combining like terms. In this case, we combine the constant terms 1 and 2, resulting in the simplified combined function: (h-k)(x) = x² - x + 3. This simplified form makes the subsequent evaluation step much more manageable. Once we have the combined function, the next step is to substitute the specified value of x into (h-k)(x). For our problem, we need to find (h-k)(3), so we replace every x in the expression x² - x + 3 with the number 3. This substitution transforms the algebraic expression into an arithmetic one: (h-k)(3) = (3)² - 3 + 3. The final step involves performing the arithmetic operations to arrive at the numerical result. Following the order of operations, we first evaluate the exponent: 3² = 9. Then, we perform the subtraction and addition from left to right: 9 - 3 + 3. This simplifies to 6 + 3, which equals 9. Therefore, the value of (h-k)(3) is 9. This systematic approach to function subtraction ensures accuracy and clarity. By carefully distributing the negative sign, combining like terms, and following the order of operations, we can confidently evaluate complex function expressions. This methodical approach is a cornerstone of mathematical problem-solving and is essential for success in more advanced topics.
3. Evaluating 3h(2) + 2k(3)
Here, we need to find h(2) and k(3) separately first.
h(2) = (2)² + 1 = 4 + 1 = 5
k(3) = 3 - 2 = 1
Now, substitute these values into the expression:
3h(2) + 2k(3) = 3(5) + 2(1) = 15 + 2 = 17
Therefore, 3h(2) + 2k(3) = 17. To successfully evaluate expressions like 3h(2) + 2k(3), it is essential to adopt a methodical approach that breaks down the problem into smaller, more manageable steps. This process involves evaluating each function separately for the given input, then applying scalar multiplication, and finally, adding the results together. The first step in this evaluation is to determine the value of each function at the specified input. For h(2), we substitute x = 2 into the function h(x) = x² + 1. This gives us h(2) = (2)² + 1 = 4 + 1 = 5. Similarly, for k(3), we substitute x = 3 into the function k(x) = x - 2. This yields k(3) = 3 - 2 = 1. Evaluating the functions individually in this manner simplifies the overall expression and reduces the chances of errors. Once we have the values of h(2) and k(3), the next step is to apply scalar multiplication. This involves multiplying the function values by their respective scalar coefficients. In our expression, h(2) is multiplied by 3, and k(3) is multiplied by 2. Thus, we have 3h(2) = 3 * 5 = 15 and 2k(3) = 2 * 1 = 2. Scalar multiplication is a straightforward arithmetic operation, but it is crucial for scaling the function values appropriately before combining them. The final step in evaluating the expression is to add the results of the scalar multiplications. We have 3h(2) = 15 and 2k(3) = 2, so we add these values together: 15 + 2 = 17. This addition combines the scaled function values to give us the final result of the expression. Therefore, 3h(2) + 2k(3) = 17. This step-by-step approach not only provides the correct answer but also reinforces the understanding of how different function operations interact. By breaking down the problem into individual evaluations, scalar multiplications, and a final addition, we can confidently tackle more complex expressions involving multiple functions and operations. This methodical approach is a valuable tool for problem-solving in mathematics and beyond.
Conclusion
And there you have it! We've successfully evaluated (h+k)(2), (h-k)(3), and 3h(2) + 2k(3). Function operations might seem tricky at first, but with a little practice and a step-by-step approach, you'll master them in no time. Remember, the key is to break down the problem into smaller parts and tackle each one individually. Keep practicing, and you'll become a function operation pro in no time! Understanding and mastering function operations is a crucial skill in mathematics. It lays the groundwork for more advanced concepts and provides a powerful tool for solving a wide range of problems. Function operations allow us to combine different functions in various ways, creating new functions that can model complex relationships and phenomena. Whether it's adding, subtracting, multiplying, or dividing functions, each operation provides a unique way to manipulate and combine mathematical relationships. The ability to perform these operations fluently is essential for success in calculus, differential equations, and other advanced mathematical topics. Moreover, function operations have numerous applications in real-world scenarios. In physics, they can be used to describe the combined effect of different forces or the motion of objects under varying conditions. In economics, they can model the interaction of supply and demand or the growth of investments over time. In computer science, they are used in algorithm design and data analysis. The versatility of function operations makes them an indispensable tool in many fields. To truly master function operations, it's important to not only understand the mechanics of each operation but also to develop a conceptual understanding of what they represent. This means thinking about functions as processes that transform inputs into outputs and how these processes can be combined and manipulated. It also means recognizing the connections between different function operations and how they relate to other mathematical concepts. By developing this deeper understanding, you can approach function operations with confidence and creativity. So, keep practicing, keep exploring, and keep pushing your mathematical boundaries. The world of function operations is vast and fascinating, and the more you delve into it, the more you'll discover the power and beauty of mathematics.
