Function Operations Explained Addition, Subtraction, Composition And More

In the realm of mathematics, functions serve as fundamental building blocks for modeling relationships and processes. Understanding how to manipulate and combine functions is crucial for solving complex problems and gaining deeper insights into mathematical concepts. This article delves into the world of function operations, providing a comprehensive guide to performing various operations on functions, including addition, subtraction, and composition. We will be working with three specific functions throughout this exploration: f(x) = √(x-1), g(x) = x² + 2, and h(x) = 1/x. These functions will serve as our tools for unraveling the intricacies of function operations.

a. (f+g)(10)

To begin our journey into function operations, let's tackle the first task: evaluating (f+g)(10). This notation represents the sum of the functions f(x) and g(x) evaluated at x = 10. In simpler terms, we need to find the value of f(10) + g(10). The key to solving this lies in understanding the individual functions and their behavior. First, let's evaluate f(10). Substituting x = 10 into the function f(x) = √(x-1), we get f(10) = √(10-1) = √9 = 3. Next, we evaluate g(10). Substituting x = 10 into the function g(x) = x² + 2, we get g(10) = 10² + 2 = 100 + 2 = 102. Now that we have the individual values of f(10) and g(10), we can find their sum. (f+g)(10) = f(10) + g(10) = 3 + 102 = 105. Therefore, the value of (f+g)(10) is 105. This exercise highlights the fundamental principle of function addition: we simply add the corresponding outputs of the individual functions for a given input.

b. (g+h)(1)

Next, let's move on to evaluating (g+h)(1). Similar to the previous example, this notation signifies the sum of the functions g(x) and h(x) evaluated at x = 1. In essence, we need to determine the value of g(1) + h(1). As before, we start by evaluating the individual functions. For g(1), we substitute x = 1 into the function g(x) = x² + 2, which gives us g(1) = 1² + 2 = 1 + 2 = 3. For h(1), we substitute x = 1 into the function h(x) = 1/x, resulting in h(1) = 1/1 = 1. Now that we have g(1) and h(1), we can calculate their sum. (g+h)(1) = g(1) + h(1) = 3 + 1 = 4. Thus, the value of (g+h)(1) is 4. This example further reinforces the concept of function addition as a straightforward process of adding the outputs of the functions for a specific input value. It's crucial to remember to evaluate each function separately before performing the addition.

c. (f+h)(1.25)

Now, let's tackle the evaluation of (f+h)(1.25). This operation involves adding the functions f(x) and h(x) and then evaluating the result at x = 1.25. This means we need to find the value of f(1.25) + h(1.25). To do this, we first evaluate each function individually. For f(1.25), we substitute x = 1.25 into the function f(x) = √(x-1), which gives us f(1.25) = √(1.25-1) = √0.25 = 0.5. Next, we evaluate h(1.25). Substituting x = 1.25 into the function h(x) = 1/x, we get h(1.25) = 1/1.25 = 0.8. Now, we can add the values we obtained. (f+h)(1.25) = f(1.25) + h(1.25) = 0.5 + 0.8 = 1.3. Therefore, the value of (f+h)(1.25) is 1.3. When dealing with decimal inputs, it's important to perform the calculations carefully, especially when dealing with square roots and fractions. This example illustrates the application of function addition with a non-integer input, reinforcing the versatility of this operation.

d. (g-f)(5)

Let's shift our focus to function subtraction and evaluate (g-f)(5). This notation indicates that we need to subtract the function f(x) from the function g(x) and then evaluate the result at x = 5. In other words, we need to find the value of g(5) - f(5). Following our established pattern, we first evaluate the individual functions. For g(5), we substitute x = 5 into the function g(x) = x² + 2, giving us g(5) = 5² + 2 = 25 + 2 = 27. Next, we evaluate f(5). Substituting x = 5 into the function f(x) = √(x-1), we get f(5) = √(5-1) = √4 = 2. Now, we can perform the subtraction. (g-f)(5) = g(5) - f(5) = 27 - 2 = 25. Thus, the value of (g-f)(5) is 25. This example demonstrates the process of function subtraction, which is analogous to addition, but with the crucial difference of subtracting the function outputs. The order of subtraction is essential, as g(x) - f(x) is generally not the same as f(x) - g(x).

e. (h-g)(1/4)

Now, let's evaluate (h-g)(1/4), which involves subtracting the function g(x) from the function h(x) and evaluating the result at x = 1/4. This means we need to calculate h(1/4) - g(1/4). As usual, we begin by evaluating the individual functions. For h(1/4), we substitute x = 1/4 into the function h(x) = 1/x, which yields h(1/4) = 1/(1/4) = 4. Next, we evaluate g(1/4). Substituting x = 1/4 into the function g(x) = x² + 2, we get g(1/4) = (1/4)² + 2 = 1/16 + 2 = 33/16. Now, we perform the subtraction. (h-g)(1/4) = h(1/4) - g(1/4) = 4 - 33/16 = 64/16 - 33/16 = 31/16. Therefore, the value of (h-g)(1/4) is 31/16. This example reinforces the importance of careful fraction arithmetic when dealing with function subtraction, especially when the input is a fraction. It's also a good reminder to always simplify your final answer if possible. Remember, function subtraction involves finding the difference between the outputs of the functions at a given input, maintaining the correct order of operations.

f. (f-h)(17)

Let's move on to evaluating (f-h)(17). This operation requires us to subtract the function h(x) from the function f(x) and then evaluate the result at x = 17. In other words, we need to find the value of f(17) - h(17). We start by evaluating the individual functions. For f(17), we substitute x = 17 into the function f(x) = √(x-1), which gives us f(17) = √(17-1) = √16 = 4. Next, we evaluate h(17). Substituting x = 17 into the function h(x) = 1/x, we get h(17) = 1/17. Now we can perform the subtraction. (f-h)(17) = f(17) - h(17) = 4 - 1/17 = 68/17 - 1/17 = 67/17. Therefore, the value of (f-h)(17) is 67/17. This example further illustrates the process of function subtraction and highlights the need for careful manipulation of fractions. When subtracting fractions, it's crucial to have a common denominator. This problem reinforces the idea that function operations build upon basic arithmetic skills.

g. (g•(f/9))(2)

Now, let's tackle a slightly more complex operation: (g•(f/9))(2). This notation signifies the multiplication of the function g(x) with the function f(x) divided by 9, and then evaluating the result at x=2. In essence, we need to find the value of g(2) * (f(2)/9). As before, we start by evaluating the individual functions. For f(2), we substitute x = 2 into the function f(x) = √(x-1), which gives us f(2) = √(2-1) = √1 = 1. Next, we evaluate g(2). Substituting x = 2 into the function g(x) = x² + 2, we get g(2) = 2² + 2 = 4 + 2 = 6. Now, we can substitute these values into the expression: (g•(f/9))(2) = g(2) * (f(2)/9) = 6 * (1/9) = 6/9 = 2/3. Therefore, the value of (g•(f/9))(2) is 2/3. This example introduces function multiplication combined with division, demonstrating how multiple operations can be combined. It reinforces the importance of following the order of operations and performing calculations step-by-step to avoid errors. The key is to break down the complex operation into simpler components, evaluate each component, and then combine the results.

h. (h/f)(1)

Next, let's evaluate (h/f)(1), which represents the division of function h(x) by function f(x), evaluated at x = 1. This means we need to find the value of h(1)/f(1). Following our established procedure, we first evaluate the individual functions. For f(1), we substitute x = 1 into the function f(x) = √(x-1), which gives us f(1) = √(1-1) = √0 = 0. For h(1), we substitute x = 1 into the function h(x) = 1/x, resulting in h(1) = 1/1 = 1. Now we attempt the division: (h/f)(1) = h(1)/f(1) = 1/0. Here we encounter a critical issue: division by zero is undefined in mathematics. Therefore, the value of (h/f)(1) is undefined. This example highlights a crucial consideration when dealing with function division: we must be mindful of cases where the denominator might be zero. Division by zero leads to an undefined result, and it's important to identify and exclude such cases from the domain of the resulting function. This problem serves as a good reminder to always check for potential division-by-zero errors when performing function division.

i. (g ∘ h)(1)

Let's now explore the concept of function composition and evaluate (g ∘ h)(1). This notation represents the composition of function g(x) with function h(x), evaluated at x = 1. Function composition means applying one function to the result of another. In this case, we first evaluate h(1), and then we use the result as the input for g(x). In other words, (g ∘ h)(1) = g(h(1)). Following this approach, we first find h(1). Substituting x = 1 into the function h(x) = 1/x, we get h(1) = 1/1 = 1. Now, we use this result as the input for g(x). So, we need to find g(1). Substituting x = 1 into the function g(x) = x² + 2, we get g(1) = 1² + 2 = 1 + 2 = 3. Therefore, (g ∘ h)(1) = g(h(1)) = g(1) = 3. Function composition is a powerful tool for building complex functions from simpler ones. The order of operations is critical in composition, as (g ∘ h)(x) is generally not the same as (h ∘ g)(x). This example demonstrates the step-by-step process of evaluating a composite function, starting from the innermost function and working outwards.

Repair Input Keywords

Calculate the following function operations given f(x) = √(x-1), g(x) = x² + 2, and h(x) = 1/x: a. (f+g)(10), b. (g+h)(1), c. (f+h)(1.25), d. (g-f)(5), e. (h-g)(1/4), f. (f-h)(17), g. (g(f/9))(2), h. (h/f)(1), i. (g ∘ h)(1).

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Function Operations Explained: Addition, Subtraction, Composition and More