Function Operations And Domain Given F(x) = X^2 - 4x And G(x) = 6 + X

In this article, we delve into the world of function operations, exploring how to combine functions through addition, subtraction, multiplication, and division. We will use the given functions, $f(x) = x^2 - 4x$ and $g(x) = 6 + x$, as examples to illustrate these concepts. Additionally, we will address the crucial aspect of determining the domain of a function, particularly when dealing with rational functions formed by dividing one function by another. Understanding these operations and domain considerations is fundamental in calculus and other advanced mathematical fields.

(a) Summing Functions: $(f+g)(x)$

When we add two functions, $f(x)$ and $g(x)$, we are essentially creating a new function that represents the sum of their individual outputs for any given input $x$. To find $(f+g)(x)$, we simply add the expressions for $f(x)$ and $g(x)$ together.

So, given $f(x) = x^2 - 4x$ and $g(x) = 6 + x$, we can find $(f+g)(x)$ as follows:

$(f+g)(x) = f(x) + g(x)$ $(f+g)(x) = (x^2 - 4x) + (6 + x)$

Now, we combine like terms to simplify the expression:

$(f+g)(x) = x^2 - 4x + x + 6$ $(f+g)(x) = x^2 - 3x + 6$

Therefore, the sum of the functions $f(x)$ and $g(x)$, denoted as $(f+g)(x)$, is the quadratic function $x^2 - 3x + 6$. This new function represents the combined behavior of $f(x)$ and $g(x)$. For any value of x, the output of (f+g)(x) is the sum of the outputs of f(x) and g(x) for the same x-value.

Understanding function addition is crucial because it allows us to model situations where two different factors contribute to an overall result. For example, imagine f(x) represents the cost of materials for a project, and g(x) represents the labor costs. Then, (f+g)(x) would represent the total cost of the project. The result emphasizes the significance of combining functions to get comprehensive results and also illustrates how algebraic manipulation and simplification are essential to obtain the most concise and usable form of the combined function.

(b) Subtracting Functions: $(f-g)(x)$

Similar to addition, subtracting functions involves creating a new function that represents the difference between their outputs. To find $(f-g)(x)$, we subtract the expression for $g(x)$ from the expression for $f(x)$. It's essential to pay close attention to the order of subtraction and to distribute the negative sign correctly.

Using the given functions, $f(x) = x^2 - 4x$ and $g(x) = 6 + x$, we calculate $(f-g)(x)$:

$(f-g)(x) = f(x) - g(x)$ $(f-g)(x) = (x^2 - 4x) - (6 + x)$

Distribute the negative sign to all terms within the parentheses of $g(x)$:

$(f-g)(x) = x^2 - 4x - 6 - x$

Combine like terms:

$(f-g)(x) = x^2 - 5x - 6$

Thus, the difference between the functions $f(x)$ and $g(x)$, written as $(f-g)(x)$, is the quadratic function $x^2 - 5x - 6$. This function shows how the values of $f(x)$ change relative to the values of $g(x)$. It’s crucial to recognize that the order of subtraction matters; $(f-g)(x)$ is generally not the same as $(g-f)(x)$. Understanding function subtraction is valuable because it helps us to see the comparative behavior of two functions.

For instance, if f(x) represents revenue and g(x) represents expenses, then (f-g)(x) would represent the profit. The careful distribution of the negative sign ensures that the subtraction is performed correctly, which is vital for accurate results. This section underscores the importance of order of operations and attention to detail when working with function operations. Moreover, it demonstrates how subtraction can highlight the net effect or difference between two functional relationships, offering insights into rates of change or comparative performance.

(c) Multiplying Functions: $(f imes g)(x)$

Multiplying functions involves creating a new function whose output is the product of the outputs of the original functions. To find $(f imes g)(x)$, we multiply the expressions for $f(x)$ and $g(x)$. This often requires using the distributive property (or the FOIL method) to expand the product.

Given $f(x) = x^2 - 4x$ and $g(x) = 6 + x$, we find $(f imes g)(x)$ as follows:

$(f imes g)(x) = f(x) imes g(x)$ $(f imes g)(x) = (x^2 - 4x)(6 + x)$

Now, we expand the product using the distributive property:

$(f imes g)(x) = x^2(6 + x) - 4x(6 + x)$ $(f imes g)(x) = 6x^2 + x^3 - 24x - 4x^2$

Combine like terms:

$(f imes g)(x) = x^3 + 2x^2 - 24x$

Therefore, the product of the functions $f(x)$ and $g(x)$, denoted as $(f imes g)(x)$, is the cubic function $x^3 + 2x^2 - 24x$. This resulting cubic function can exhibit more complex behavior than the original quadratic and linear functions. It highlights how function multiplication can significantly alter the characteristics of the functions involved. The resulting function's roots and turning points, can be determined through techniques in calculus. For example, f(x) could represent the number of products sold and g(x) could represent the price per product, in which case (f imes g)(x) would represent the total revenue.

In practice, the multiplication of functions can help model situations where one quantity influences the scale or magnitude of another quantity. This section highlights the application of the distributive property in algebraic manipulation and shows how multiplying functions can produce a new function with different properties and characteristics than its factors. Understanding these transformations is key to advanced calculus concepts and real-world applications. The resulting cubic function illustrates that the multiplication of a quadratic and a linear function yields a cubic function, demonstrating a foundational concept in polynomial algebra.

(d) Dividing Functions: $\left(\frac{f}{g}\right)(x)$

Dividing functions, represented as $\left(\frac{f}{g}\right)(x)$, involves creating a new function where the output is the quotient of the outputs of the original functions. This is written as $\frac{f(x)}{g(x)}$. However, there's a crucial restriction: the denominator, $g(x)$, cannot be equal to zero, as division by zero is undefined.

Using $f(x) = x^2 - 4x$ and $g(x) = 6 + x$, we find $\left(\frac{f}{g}\right)(x)$:

$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$ $\left(\frac{f}{g}\right)(x) = \frac{x^2 - 4x}{6 + x}$

We can factor the numerator to see if any simplification is possible:

$\left(\frac{f}{g}\right)(x) = \frac{x(x - 4)}{6 + x}$

In this case, there are no common factors between the numerator and denominator that can be canceled. Therefore, the quotient function is:

$\left(\frac{f}{g}\right)(x) = \frac{x(x - 4)}{6 + x}$

This rational function represents the division of $f(x)$ by $g(x)$. Understanding function division is important because it introduces the concept of rational functions, which have unique behaviors and asymptotes. The simplified form of the divided function clearly shows the relationship between the original functions. For instance, this operation could model the concentration of a substance over time, where f(x) represents the amount of substance and g(x) represents the volume of the solution.

The crucial aspect here is the consideration of the denominator. The function is undefined when $g(x) = 0$, which leads us to the next section on determining the domain of this rational function. This section showcases the formation of a rational function through division and the importance of considering the possibility of simplification through factoring. Furthermore, it emphasizes the necessity of identifying values of x that would make the denominator zero, thus highlighting the concept of excluded values in the domain.

(e) The Domain of $\frac{f}{g}$

The domain of a function is the set of all possible input values (x-values) for which the function produces a real number output. When dealing with rational functions, like $\frac{f}{g}$, we must exclude any values of $x$ that make the denominator, $g(x)$, equal to zero. These values would result in division by zero, which is undefined.

In our case, we have $\left(\frac{f}{g}\right)(x) = \frac{x(x - 4)}{6 + x}$. To find the domain, we need to determine the values of $x$ for which the denominator, $6 + x$, is not equal to zero.

Set the denominator equal to zero and solve for $x$:

$6 + x = 0$ $x = -6$

This means that $x = -6$ is the value that makes the denominator zero, and thus, it must be excluded from the domain. The domain of $\frac{f}{g}$ is all real numbers except $x = -6$. We can express this in interval notation as:

$(-\infty, -6) \cup (-6, \infty)$

The domain represents the set of all x-values for which the function is defined. This restriction is a fundamental aspect of rational functions and highlights the importance of considering potential discontinuities. Understanding the domain is essential in various applications. For instance, if we are modeling a physical process, the domain might represent the range of feasible input values. This section underscores the significance of identifying and excluding values that lead to undefined operations, thus ensuring the function provides meaningful outputs. Understanding the domain of the function, particularly for rational functions, helps to interpret the function's behavior and its applicability within real-world contexts.

Conclusion

In this comprehensive guide, we explored various operations on functions, including addition, subtraction, multiplication, and division, using the functions $f(x) = x^2 - 4x$ and $g(x) = 6 + x$ as examples. We also emphasized the critical concept of determining the domain of a function, particularly for rational functions, to avoid undefined operations. Mastering these function operations and domain considerations is fundamental for success in calculus and other advanced mathematical fields. Understanding these concepts provides a strong foundation for exploring more complex mathematical models and applications in real-world scenarios. The ability to perform these operations and to correctly identify the domain of the resulting functions are foundational skills in algebra and calculus, with wide-ranging implications across various scientific and engineering disciplines.