Function Operations Problem Set 1 Solutions

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In this article, we will delve into the solutions for Problem Set #1, focusing on the fundamental operations involving functions. We will explore addition of functions, evaluating functions at specific points, and provide a step-by-step guide to understanding these concepts. This comprehensive guide aims to clarify any confusion and equip you with the necessary skills to confidently tackle similar problems.

Understanding the Functions

Before we dive into the solutions, let's first define the functions we will be working with. We are given the following functions:

  • f(x) = x + 3
  • p(x) = 2x - 7
  • v(x) = (x² + 5x + 4) / (x² + 2x - 8)
  • h(x) = (x + 1) / (2 - x)
  • t(x) = (x - 2) / (x + 3)

These functions represent different relationships between an input x and an output value. Our task is to perform various operations on these functions, such as addition and evaluation at specific points. Understanding the individual functions is crucial before combining them. Each function has its unique characteristics and domain, which we need to consider when performing operations.

1. (v + g)(x): Adding Rational Functions

The first problem requires us to find the sum of two functions, v(x) and a function g(x) which seems to be missing from the provided function list. Assuming there was a typo and g(x) is meant to be part of v(x)'s definition, which is likely the case given the structure of the problem, we proceed as follows. We treat v(x) as a single function: v(x) = (x² + 5x + 4) / (x² + 2x - 8). This problem highlights the importance of combining functions, particularly rational functions, which involve polynomial expressions in both the numerator and denominator. When adding rational functions, a crucial first step is to factorize the polynomials to simplify the expressions and identify common factors. Factoring the numerator and denominator of v(x) allows us to potentially simplify the expression before adding it to any other function, which is a key strategy in function manipulation.

To determine (v + g)(x), we need to understand the operation of function addition. When we add two functions, say f(x) and g(x), the result is a new function, denoted as (f + g)(x), which is obtained by simply adding the expressions of the individual functions. In other words, (f + g)(x) = f(x) + g(x). However, when dealing with rational functions, the process involves more than just adding the numerators and denominators directly. We need to find a common denominator before we can combine the fractions. This is similar to adding numerical fractions where we first find the least common multiple (LCM) of the denominators. Once we have a common denominator, we can add the numerators and simplify the resulting expression. The simplified expression represents the new function resulting from the addition.

To add v(x) to itself, we write: v(x) + v(x) = (x² + 5x + 4) / (x² + 2x - 8) + (x² + 5x + 4) / (x² + 2x - 8). Since the denominators are already the same, we can directly add the numerators: (2 * (x² + 5x + 4)) / (x² + 2x - 8). Now, we factorize both the numerator and the denominator to simplify the expression further. The numerator can be factored as 2 * (x + 1) * (x + 4), and the denominator can be factored as (x + 4) * (x - 2). Cancelling the common factor of (x + 4), we get the simplified expression: 2 * (x + 1) / (x - 2). This simplified rational function is the result of (v + v)(x), demonstrating the process of adding rational functions and the importance of simplification through factorization.

2. (f + h)(x): Adding a Linear and a Rational Function

Next, we need to determine (f + h)(x). Here, we are adding a linear function f(x) = x + 3 and a rational function h(x) = (x + 1) / (2 - x). This scenario presents a common situation in function operations where functions of different types are combined. The key to adding functions of different types, especially when one is a rational function, is to express all terms with a common denominator. This allows us to combine the expressions into a single, simplified function. The process involves multiplying the non-rational function by a form of 1 that results in the desired common denominator.

To add f(x) and h(x), we first write out the expression: (f + h)(x) = f(x) + h(x) = (x + 3) + (x + 1) / (2 - x). The next step is to find a common denominator. In this case, the denominator of the rational function h(x) is (2 - x), so we need to express f(x) with the same denominator. We achieve this by multiplying f(x) by (2 - x) / (2 - x), which is equivalent to multiplying by 1 and does not change the value of the expression. This gives us: (x + 3) * (2 - x) / (2 - x) + (x + 1) / (2 - x). Now, we expand the numerator of the first term: (2x - x² + 6 - 3x) / (2 - x) + (x + 1) / (2 - x). Simplifying the numerator, we get (-x² - x + 6) / (2 - x) + (x + 1) / (2 - x).

Now that we have a common denominator, we can add the numerators: (-x² - x + 6 + x + 1) / (2 - x). Simplifying the numerator further, we obtain (-x² + 7) / (2 - x). This resulting rational function represents the sum of f(x) and h(x). This process demonstrates the importance of finding a common denominator when adding functions, especially when dealing with rational functions. The final expression (-x² + 7) / (2 - x) is the simplified form of (f + h)(x), showcasing the result of combining a linear function and a rational function.

3. (t + h)(x): Adding Two Rational Functions

Here, we are tasked with finding the sum of two rational functions, t(x) = (x - 2) / (x + 3) and h(x) = (x + 1) / (2 - x). This problem further reinforces the concept of adding rational functions, which involves finding a common denominator and combining the numerators. The challenge here is that the denominators, (x + 3) and (2 - x), are different and do not share any obvious common factors, making the process of finding the common denominator a bit more involved. This type of problem is essential for building a strong foundation in algebraic manipulation and function operations.

To determine (t + h)(x), we first write the expression for the sum: (t + h)(x) = t(x) + h(x) = (x - 2) / (x + 3) + (x + 1) / (2 - x). Since the denominators are different, we need to find a common denominator before we can add the fractions. The common denominator is the product of the two denominators, which is (x + 3) * (2 - x). We then rewrite each fraction with this common denominator. For the first fraction, we multiply both the numerator and the denominator by (2 - x), and for the second fraction, we multiply both the numerator and the denominator by (x + 3). This gives us: ((x - 2) * (2 - x)) / ((x + 3) * (2 - x)) + ((x + 1) * (x + 3)) / ((x + 3) * (2 - x)).

Next, we expand the numerators: (2x - x² - 4 + 2x) / ((x + 3) * (2 - x)) + (x² + 3x + x + 3) / ((x + 3) * (2 - x)). Simplifying the numerators, we get (-x² + 4x - 4) / ((x + 3) * (2 - x)) + (x² + 4x + 3) / ((x + 3) * (2 - x)). Now that we have a common denominator, we can add the numerators: (-x² + 4x - 4 + x² + 4x + 3) / ((x + 3) * (2 - x)). Combining like terms in the numerator, we obtain (8x - 1) / ((x + 3) * (2 - x)). This resulting rational function represents the sum of t(x) and h(x). This process demonstrates the importance of carefully expanding and simplifying expressions when adding rational functions. The final expression (8x - 1) / ((x + 3) * (2 - x)) is the simplified form of (t + h)(x), showcasing the result of combining two rational functions.

4. (p + v)(2): Evaluating the Sum of Functions

This problem involves evaluating the sum of two functions, p(x) = 2x - 7 and v(x) = (x² + 5x + 4) / (x² + 2x - 8), at a specific point, x = 2. This is a crucial aspect of function operations as it combines the concept of function addition with the evaluation of a function at a given input value. Evaluating functions at specific points allows us to understand the behavior of the function and its output for particular inputs. This skill is fundamental in various applications of functions, including modeling real-world phenomena and solving equations.

To determine (p + v)(2), we first need to find the expression for (p + v)(x), which is the sum of the two functions: (p + v)(x) = p(x) + v(x) = (2x - 7) + (x² + 5x + 4) / (x² + 2x - 8). Now, we need to add these two expressions. To do this, we find a common denominator, which in this case is x² + 2x - 8. We rewrite the first term with this denominator: (2x - 7) * (x² + 2x - 8) / (x² + 2x - 8) + (x² + 5x + 4) / (x² + 2x - 8). Expanding the numerator of the first term, we get (2x³ + 4x² - 16x - 7x² - 14x + 56) / (x² + 2x - 8) + (x² + 5x + 4) / (x² + 2x - 8). Simplifying the numerator, we have (2x³ - 3x² - 30x + 56) / (x² + 2x - 8) + (x² + 5x + 4) / (x² + 2x - 8).

Now, we can add the numerators: (2x³ - 3x² - 30x + 56 + x² + 5x + 4) / (x² + 2x - 8). Combining like terms, we get (2x³ - 2x² - 25x + 60) / (x² + 2x - 8). This is the expression for (p + v)(x). To evaluate (p + v)(2), we substitute x = 2 into this expression: (2(2)³ - 2(2)² - 25(2) + 60) / ((2)² + 2*(2) - 8)*. Simplifying, we have (28 - 24 - 50 + 60) / (4 + 4 - 8), which is (16 - 8 - 50 + 60) / (0). We get 18 / 0, which is undefined. This indicates that the function (p + v)(x) is not defined at x = 2, possibly due to a discontinuity or a zero in the denominator. This highlights the importance of checking the domain of the function when evaluating it at a specific point. The result shows that (p + v)(2) is undefined because the denominator becomes zero at x = 2.

5. (f + h): Discussion

The final part of the problem set asks for a discussion on (f + h). This prompts us to reflect on the process of adding functions and the characteristics of the resulting function. A discussion can cover various aspects, such as the domain of the new function, its behavior, and any notable features. In this case, we are discussing the sum of a linear function f(x) = x + 3 and a rational function h(x) = (x + 1) / (2 - x). This discussion is essential for a deeper understanding of function operations beyond just the mechanical steps of addition. It encourages critical thinking about the properties of functions and how they interact when combined.

From our earlier calculation in problem 2, we found that (f + h)(x) = (-x² + 7) / (2 - x). This is a rational function, which means it is a ratio of two polynomials. To discuss this function, we can consider several key aspects. First, the domain of a rational function is all real numbers except for the values that make the denominator zero. In this case, the denominator is (2 - x), which is zero when x = 2. Therefore, the domain of (f + h)(x) is all real numbers except x = 2. This exclusion is crucial because the function is undefined at x = 2 due to division by zero. Understanding the domain is fundamental in analyzing the behavior of the function.

Next, we can discuss the behavior of the function. As x approaches 2, the denominator (2 - x) approaches zero, which means the function will approach infinity or negative infinity, depending on the direction from which x approaches 2. This indicates a vertical asymptote at x = 2. The numerator, (-x² + 7), does not become zero at x = 2, so the function does not have a hole at this point. We can also analyze the end behavior of the function. As x approaches positive or negative infinity, the function behaves like the ratio of the leading terms of the numerator and denominator, which is (-x²) / (-x) = x. This suggests that the function will behave like a linear function as x becomes very large or very small. However, since the degree of the numerator is one greater than the degree of the denominator, there is no horizontal asymptote. Instead, there is an oblique or slant asymptote.

Furthermore, we can discuss the intercepts of the function. The y-intercept is the value of the function when x = 0. Substituting x = 0 into (f + h)(x), we get (-0² + 7) / (2 - 0) = 7 / 2. So, the y-intercept is at the point (0, 3.5). To find the x-intercepts, we set the numerator equal to zero and solve for x: -x² + 7 = 0. This gives us x² = 7, so x = ±√7. Therefore, the x-intercepts are at the points (-√7, 0) and (√7, 0). These intercepts provide valuable information about where the function crosses the x- and y-axes. In summary, the discussion of (f + h)(x) involves analyzing its domain, behavior near discontinuities, end behavior, and intercepts. This comprehensive approach provides a deeper understanding of the function and its characteristics.

This detailed exploration of Problem Set #1 provides a solid understanding of function operations, including addition and evaluation. By carefully working through each problem and understanding the underlying concepts, you can build a strong foundation in function manipulation. Remember to always consider the domain of the functions and simplify expressions whenever possible. These skills are essential for success in more advanced mathematical topics.