Combining Functions: Videos & Practice Problems

In mathematics, understanding how to perform operations on functions, such as addition and subtraction, is essential. When adding or subtracting functions, the process mirrors that of combining polynomials. For instance, if we have two polynomials, we can combine like terms to simplify the expression. For example, if we add the polynomials \(x^2 + 4\) and \(5x + 7\), we combine the \(x^2\) terms, the \(x\) terms, and the constant terms, resulting in \(x^2 + 5x + 11\).

Functions can be represented in different notations. For example, \(f(x) + g(x)\) can also be expressed as \((f + g)(x)\), and similarly, \(f(x) - g(x)\) can be written as \((f - g)(x)\). It is crucial to be aware of these notations as they may appear in various contexts.

When determining the domain of the sum or difference of functions, one must identify the common values for both functions. The domain of a function is the set of all possible input values (x-values) that do not cause any undefined behavior, such as division by zero. For instance, if \(f(x) = \frac{x^2 + 1}{x}\), the domain excludes \(x = 0\) because division by zero is undefined. Conversely, a polynomial function like \(g(x) = x^2 + x + 2\) has a domain of all real numbers, as there are no restrictions.

To find the domain of the combined function \(f + g\), we consider the restrictions from both functions. In this case, since \(f\) has a restriction at \(x = 0\), the domain of \(f + g\) is also \(x \neq 0\).

In another example, when subtracting functions, such as \(g(x) - h(x)\), we first distribute the negative sign across the second function. For instance, if \(g(x) = x^2 + x + 2\) and \(h(x) = x + \sqrt{x - 8}\), the subtraction leads to \(g(x) - h(x) = x^2 + 2 - \sqrt{x - 8}\). Here, the domain of \(h(x)\) is restricted by the square root, which requires that \(x - 8 \geq 0\), or \(x \geq 8\). Thus, the domain of \(g - h\) is also \(x \geq 8\).

In summary, adding and subtracting functions involves combining like terms and carefully considering the domains of the individual functions to determine the domain of the resulting function. This understanding is fundamental in algebra and lays the groundwork for more advanced mathematical concepts.

In this exploration of function operations, we analyze two functions: \( f(x) = \sqrt{x + 4} + 30 \) and \( g(x) = \sqrt{x + 4} - 2x + 35 \). The goal is to perform addition and subtraction on these functions while determining the domain of the resulting expressions.

To find \( f + g \), we combine the two functions: \[f + g = f(x) + g(x) = \left(\sqrt{x + 4} + 30\right) + \left(\sqrt{x + 4} - 2x + 35\right).\]This simplifies to:\[f + g = 2\sqrt{x + 4} - 2x + 65.\]Next, we determine the domain. The only restriction arises from the square root, which requires that the expression inside must be non-negative:\[x + 4 \geq 0 \implies x \geq -4.\]Thus, the domain for \( f + g \) is \( [-4, \infty) \).

Next, we consider \( f - g \):\[f - g = f(x) - g(x) = \left(\sqrt{x + 4} + 30\right) - \left(\sqrt{x + 4} - 2x + 35\right).\]Distributing the negative sign gives:\[f - g = \sqrt{x + 4} + 30 - \sqrt{x + 4} + 2x - 35.\]The square root terms cancel, leading to:\[f - g = 2x - 5.\]Despite the absence of square roots in the final expression, we must still consider the original functions' domains. The requirement from the square root remains:\[x + 4 \geq 0 \implies x \geq -4.\]Therefore, the domain for \( f - g \) is also \( [-4, \infty) \).

In summary, both operations yield the same domain of \( [-4, \infty) \), highlighting the importance of considering the original functions when determining domain restrictions, even after simplification. This understanding is crucial for working with functions and their operations effectively.

In the study of functions, understanding how to multiply and divide them is essential. When multiplying two functions, such as \( f(x) = \sqrt{x} \) and \( g(x) = 3x - 6 \), the product is expressed as \( f(x) \cdot g(x) = \sqrt{x} \cdot (3x - 6) \). This can be simplified to \( 3x\sqrt{x} - 6\sqrt{x} \). The domain of the resulting function is determined by the intersection of the domains of the individual functions. For \( f(x) \), the domain is \( [0, \infty) \) since the square root function is defined for non-negative values. For \( g(x) \), which is a polynomial, the domain is all real numbers, \( (-\infty, \infty) \). Therefore, the domain of the product \( f(x) \cdot g(x) \) is \( [0, \infty) \), as it is the more restrictive of the two domains.

When dividing functions, the process is similar but requires additional caution regarding the denominator. For the same functions, the division is expressed as \( \frac{f(x)}{g(x)} = \frac{\sqrt{x}}{3x - 6} \). Here, the domain of \( f(x) \) remains \( [0, \infty) \), while \( g(x) \) must not equal zero. Setting \( 3x - 6 = 0 \) gives \( x = 2 \), which is a restriction. Thus, the domain of the quotient is \( [0, 2) \cup (2, \infty) \), excluding the point where the denominator is zero.

In another example, consider \( f(x) = x^2 - 4 \) and \( g(x) = x + 2 \). The product \( f(x) \cdot g(x) \) can be calculated using the FOIL method, resulting in \( x^3 + 2x^2 - 4x - 8 \). Since both functions are polynomials, their domain is all real numbers, \( (-\infty, \infty) \). When dividing these functions, the expression becomes \( \frac{x^2 - 4}{x + 2} \). The domain restriction arises from the denominator, which cannot be zero, leading to \( x \neq -2 \). Thus, the domain for the division is \( (-\infty, -2) \cup (-2, \infty) \).

In summary, when performing operations on functions, it is crucial to consider the domains carefully. The resulting domain after multiplication is the intersection of the individual domains, while for division, it is the intersection adjusted for any restrictions imposed by the denominator. This understanding is fundamental for successfully navigating function operations in mathematics.

Evaluating composed functions involves understanding how to combine two functions and then determine their output for a specific input. When given two functions, such as f(x) = x² and g(x) = x - 1, the composition f(g(x)) can be found by substituting g(x) into f(x). This process begins by calculating f(g(x)) = f(x - 1), which simplifies to (x - 1)². To further simplify, we can apply the FOIL method, resulting in x² - 2x + 1.

Once the composed function is established, evaluating it at a specific value, such as 3, involves substituting 3 into the simplified expression. Thus, f(g(3)) = 3² - 2(3) + 1 simplifies to 9 - 6 + 1 = 4. This method illustrates a systematic approach to evaluating composed functions.

Alternatively, a shortcut method can be employed for efficiency. Instead of first finding the composition, you can evaluate the inside function directly. For instance, calculating g(3) gives 3 - 1 = 2. Then, substituting this result into f yields f(g(3)) = f(2) = 2² = 4. This method can save time but should be used cautiously, as some problems may require the full composition to be found before evaluation.

In summary, while both methods yield the same result, understanding when to apply each is crucial. The first method emphasizes the composition process, while the second offers a quicker evaluation route, provided the problem allows for it.