Finding (g ⋅ H)(1) A Step-by-Step Guide With Examples

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 across various disciplines. This article delves into the concept of function composition, specifically focusing on the product of two functions, g(n) and h(n). We will explore how to evaluate the product of these functions at a given point, providing a comprehensive analysis of the expression (g ⋅ h)(1). This exploration will not only enhance your understanding of function operations but also equip you with the tools to tackle similar problems with confidence.

Before we dive into the computation of (g ⋅ h)(1), let's first clearly define the functions g(n) and h(n). We are given:

• g(n) = n² + 4 + 2n
• h(n) = -3n + 2

These functions represent different types of relationships. g(n) is a quadratic function, characterized by the n² term, which implies a parabolic shape when graphed. The presence of the 2n and constant terms (4) shifts and scales the parabola. On the other hand, h(n) is a linear function, indicated by the -3n term, resulting in a straight line when graphed. The +2 term represents the y-intercept of this line. Understanding the nature of these functions is essential for predicting their behavior and how they interact when combined.

The product of two functions, denoted as (g ⋅ h)(n), is a new function formed by multiplying the outputs of the individual functions for the same input value. Mathematically, this is expressed as:

(g ⋅ h)(n) = g(n) ⋅ h(n)

This operation is a fundamental way of combining functions, and it's crucial to understand its implications. When we multiply functions, we are essentially creating a new function that reflects the combined behavior of the original functions. For instance, if g(n) represents the area of a rectangle and h(n) represents its height, then (g ⋅ h)(n) might represent the volume of a prism with that rectangular base. The product of functions can exhibit complex behaviors depending on the nature of the original functions. The resulting function's graph and properties can be quite different from those of the individual functions, showcasing the intricate interplay between mathematical expressions.

Now that we understand the concept of the product of functions, let's evaluate (g ⋅ h)(1). This means we need to find the value of the combined function when n is equal to 1. To do this, we'll follow these steps:

Step 1: Evaluate g(1)

Substitute n = 1 into the expression for g(n):

g(1) = (1)² + 4 + 2(1) = 1 + 4 + 2 = 7

So, the value of g(1) is 7. This is a straightforward application of function evaluation, where we replace the variable with the given input and perform the arithmetic operations. This step highlights the fundamental process of determining a function's output for a specific input, a cornerstone of function analysis.

Step 2: Evaluate h(1)

Similarly, substitute n = 1 into the expression for h(n):

h(1) = -3(1) + 2 = -3 + 2 = -1

Thus, the value of h(1) is -1. This step mirrors the previous one, reinforcing the process of function evaluation. Here, we see a linear function producing a negative output for the input of 1, which provides insight into the function's behavior and slope.

Step 3: Multiply g(1) and h(1)

Now, we multiply the values we found in the previous steps:

(g ⋅ h)(1) = g(1) ⋅ h(1) = 7 ⋅ (-1) = -7

Therefore, (g ⋅ h)(1) = -7. This final step combines the individual function evaluations to produce the value of the product function at the specified input. The result, -7, represents the combined output of the two functions when n = 1. This calculation demonstrates how the product of functions integrates the behavior of its components, leading to a potentially different outcome than either function would produce on its own.

While we evaluated (g ⋅ h)(1) by first finding g(1) and h(1) separately, there's another approach: we can first find the general expression for (g ⋅ h)(n) and then substitute n = 1. Let's explore this method.

Step 1: Find the Expression for (g ⋅ h)(n)

Multiply the expressions for g(n) and h(n):

(g ⋅ h)(n) = g(n) ⋅ h(n) = (n² + 4 + 2n) ⋅ (-3n + 2)

Now, we need to expand this product. This involves multiplying each term in the first expression by each term in the second expression:

(g ⋅ h)(n) = n²(-3n + 2) + 4(-3n + 2) + 2n(-3n + 2)

Expanding each term gives:

(g ⋅ h)(n) = -3n³ + 2n² - 12n + 8 - 6n² + 4n

Step 2: Simplify the Expression

Combine like terms to simplify the expression:

(g ⋅ h)(n) = -3n³ + (2n² - 6n²) + (4n - 12n) + 8

(g ⋅ h)(n) = -3n³ - 4n² - 8n + 8

Now we have the general expression for (g ⋅ h)(n), which is a cubic function.

Step 3: Evaluate (g ⋅ h)(1)

Substitute n = 1 into the simplified expression:

(g ⋅ h)(1) = -3(1)³ - 4(1)² - 8(1) + 8

(g ⋅ h)(1) = -3 - 4 - 8 + 8 = -7

As we found earlier, (g ⋅ h)(1) = -7. This alternative approach demonstrates the power of algebraic manipulation in simplifying complex expressions. By first finding the general expression for the product function, we can efficiently evaluate it at any point. This method is particularly useful when we need to evaluate the product function at multiple points, as it avoids repeated multiplication of the original function expressions.

We've explored two methods for evaluating (g ⋅ h)(1), and each offers its own advantages:

1. Direct Substitution: This method involves evaluating g(1) and h(1) separately and then multiplying the results. It's straightforward and easy to understand, making it ideal for simple evaluations or when only a single point needs to be computed.
2. Finding the Product Function First: This method involves finding the general expression for (g ⋅ h)(n) and then substituting n = 1. It requires more algebraic manipulation but is more efficient when evaluating the product function at multiple points. It also provides a deeper understanding of the resulting function's behavior.

The choice between these methods depends on the specific problem and the desired level of understanding. For a single evaluation, direct substitution is often quicker. However, if multiple evaluations are needed or if a thorough analysis of the product function is desired, finding the general expression is the preferred approach. Both methods provide valuable insights into function operations and enhance problem-solving skills.

The concept of the product of functions has wide-ranging applications in various fields, including:

• Physics: Describing the combined effect of multiple forces or factors.
• Engineering: Modeling systems where multiple components interact.
• Economics: Analyzing the interplay of supply and demand curves.
• Computer Graphics: Creating complex transformations and effects.
• Data Analysis: Combining different data sets to reveal insights.

Understanding function operations, such as the product of functions, is essential for building mathematical models that accurately represent real-world phenomena. These operations allow us to combine simpler functions into more complex ones, capturing intricate relationships and behaviors. The significance of these concepts extends beyond theoretical mathematics, providing practical tools for solving problems across diverse domains. By mastering function operations, individuals can develop a deeper understanding of the world around them and contribute to advancements in various fields.

In this article, we've provided a detailed exploration of the product of functions, focusing on the evaluation of (g ⋅ h)(1). We defined the functions g(n) and h(n), explained the concept of the product of functions, and demonstrated two methods for evaluating (g ⋅ h)(1). We also discussed the applications and significance of this concept in various fields. Through this comprehensive analysis, we hope to have enhanced your understanding of function operations and equipped you with the skills to tackle similar problems with confidence. The product of functions is a powerful tool in the mathematical arsenal, enabling us to model and analyze complex relationships in a clear and concise manner. Its applications are vast and varied, making it a fundamental concept for anyone seeking to understand the mathematical foundations of our world.