Combining Functions: Videos & Practice Problems

Understanding function operations is crucial in mathematics. Adding and subtracting functions involves combining like terms, while multiplying and dividing requires careful attention to domain restrictions, particularly when denominators are involved. For composed functions, evaluating them can be done by substituting values directly or through a two-step process of finding inner functions first. Key concepts include domain, polynomial behavior, and the importance of identifying restrictions to ensure valid operations. Mastery of these operations enhances problem-solving skills in algebra and calculus.

Adding & Subtracting Functions

Understanding how to perform basic operations with functions, such as addition and subtraction, is essential in algebra. 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 \( f(x) = x^2 + 4 \) and \( g(x) = 5x + 7 \), we can express this as:

\[ f(x) + g(x) = x^2 + 5x + 11 \]

This can also be denoted using function notation as \( (f + g)(x) \). Similarly, subtraction can be represented as \( f(x) - g(x) \) or \( (f - g)(x) \). It is crucial to be aware of different notations that may appear in various contexts.

When determining the domain of the resulting function after addition or subtraction, it is important to identify the common values that both functions can accept. For example, if \( f(x) = \frac{x^2 + 1}{x} \) and \( g(x) = x^2 + x + 2 \), the domain of \( f \) is restricted because the denominator cannot be zero, leading to the condition \( x \neq 0 \). In contrast, \( g \) has a domain of all real numbers, \( (-\infty, \infty) \). Therefore, the domain of \( f + g \) is also restricted to \( x \neq 0 \).

In another example, consider \( g(x) = x^2 + x + 2 \) and \( h(x) = x + \sqrt{x - 8} \). To find \( g(x) - h(x) \), we distribute the negative sign and combine like terms:

\[ g(x) - h(x) = x^2 + x + 2 - x - \sqrt{x - 8} = x^2 + 2 - \sqrt{x - 8} \]

Here, the domain of \( h \) is restricted by the square root, requiring \( x - 8 \geq 0 \) or \( x \geq 8 \). Thus, the domain of \( g - h \) is also \( x \geq 8 \).

In summary, when adding or subtracting functions, it is essential to combine like terms and carefully analyze the domains of the individual functions to determine the domain of the resulting function. This understanding is foundational for further studies in algebra and calculus.

Adding & Subtracting Functions Example 1

In this problem, we explore the addition and subtraction of two functions, defined as follows: \( f(x) = \sqrt{x + 4} + 30 \) and \( g(x) = \sqrt{x + 4} - 2x + 35 \). To find \( f + g \), we combine the two functions by adding their expressions. This results in:

\[ f + g = \sqrt{x + 4} + 30 + \sqrt{x + 4} - 2x + 35 \]

By simplifying, we combine like terms. The square root terms yield \( 2\sqrt{x + 4} \), and the constants combine to give \( 65 \). Thus, the expression simplifies to:

\[ f + g = 2\sqrt{x + 4} - 2x + 65 \]

Next, we determine the domain of this new function. The only restriction arises from the square root, which requires that the expression inside must be non-negative. Therefore, we set up the inequality:

\[ x + 4 \geq 0 \]

Solving this gives \( x \geq -4 \). Hence, the domain of \( f + g \) is \( [-4, \infty) \).

Now, we consider \( f - g \). The expression for this operation is:

\[ f - g = \sqrt{x + 4} + 30 - (\sqrt{x + 4} - 2x + 35) \]

Distributing the negative sign leads to:

\[ f - g = \sqrt{x + 4} + 30 - \sqrt{x + 4} + 2x - 35 \]

Upon simplification, the square root terms cancel out, resulting in:

\[ f - g = 2x - 5 \]

Despite the absence of square roots in the final expression, we must still consider the domain restrictions from the original functions. The requirement that \( x + 4 \geq 0 \) still holds, leading to the same domain of \( [-4, \infty) \) for \( f - g \).

In summary, both operations yield the same domain, demonstrating that the initial conditions of the functions significantly influence the resulting domains, regardless of the simplifications performed later. Understanding these principles is crucial for effectively working with functions and their domains.

Multiplying & Dividing Functions

In the study of function operations, multiplying and dividing functions are essential skills that build on the concepts of addition and subtraction. When multiplying two functions, such as \( f(x) = \sqrt{x} \) and \( g(x) = 3x - 6 \), the product is obtained by distributing one function into the other. This results in \( f(x) \cdot g(x) = \sqrt{x} \cdot (3x - 6) = 3x\sqrt{x} - 6\sqrt{x} \). Understanding the domain of the resulting function is crucial; it is determined by the intersection of the domains of the individual functions. For \( f(x) \), the domain is \( [0, \infty) \) due to the square root, while \( g(x) \) has a domain of all real numbers, \( (-\infty, \infty) \). Therefore, the domain of the product \( f \cdot g \) 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 \( [0, \infty) \), but \( g(x) \) introduces a restriction: the denominator cannot equal zero. Setting \( 3x - 6 = 0 \) gives \( x = 2 \), which must be excluded from the domain. Thus, the domain of the quotient is \( [0, 2) \cup (2, \infty) \), allowing all positive numbers except for 2.

In another example, consider \( f(x) = x^2 - 4 \) and \( g(x) = x + 2 \). The product \( f \cdot g \) can be calculated using the FOIL method, yielding \( f(x) \cdot g(x) = (x^2 - 4)(x + 2) = x^3 + 2x^2 - 4x - 8 \). Since both functions are polynomials, their domains are all real numbers, resulting in a combined domain of \( (-\infty, \infty) \) for the product.

For the division \( \frac{f(x)}{g(x)} = \frac{x^2 - 4}{x + 2} \), we first identify the domain restrictions. The denominator \( x + 2 \) cannot be zero, leading to the restriction \( x \neq -2 \). Simplifying the expression reveals that \( x^2 - 4 \) can be factored as \( (x + 2)(x - 2) \), allowing the \( x + 2 \) terms to cancel, resulting in \( x - 2 \). However, the original restriction must still be acknowledged, so the domain of the simplified function is \( (-\infty, -2) \cup (-2, \infty) \).

In summary, when performing operations on functions, it is vital to consider both the algebraic manipulation and the implications for the domain. This ensures a comprehensive understanding of the resulting function and its valid input values.

Given the functions $h\left(x\right)=2x^3-4$ and $k\left(x\right)=x^2+2$, find and fully simplify $h\cdot k\left(x\right)$

• A. $h\cdot k\left(x\right)=2\left(x^5+2x^3-2x^2-4\right)$
• B. $h\cdot k\left(x\right)=2x^5-8$
• C. $h\cdot k\left(x\right)=2x^5+4x^3-8$
• D. $h\cdot k\left(x\right)=x^2+4x+4$

Answer: (A)

Given the functions $L\left(x\right)=x-2$ and $M\left(x\right)=x^2$, calculate $\frac{L}{M}\left(5\right)$

• A. $\frac{L}{M}\left(5\right)=\frac{25}{3}$
• B. $\frac{L}{M}\left(5\right)=\frac53$
• C. $\frac{L}{M}\left(5\right)=\frac{3}{25}$
• D. $\frac{L}{M}\left(5\right)=\frac35$

Answer: (C)

Evaluating Composed Functions

Evaluating composed functions involves understanding how to combine two functions and then determine their output based on 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 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 importance of function composition, while the shortcut method offers a quicker alternative when appropriate. Mastery of these techniques enhances problem-solving skills in function evaluation.

Given the functions $f(x)=x+3$ and $g(x)= x^2$ find $(f∘g)(2)$ and $(g∘f)(2)$.

• A. $(f∘g)(2)=5$ ; $(g∘f)(2)=25$
• B. $(f\circ g)(2)=7;(g\circ f)(2)=4$
• C. $(f∘g)(2)=7$ ; $(g∘f)(2)=25$
• D. $(f∘g)(2)=1$ ; $(g∘f)(2)=1$

Answer: (C)

Here’s what students ask on this topic:

What is the process for adding and subtracting functions?

Adding and subtracting functions involves combining like terms, similar to adding and subtracting polynomials. For example, if you have two functions, $f\left(x\right)=x^2+4$ and $g\left(x\right)=5x+7$, you can add them to get $x^2+5x+11$. Subtraction follows a similar process, but you distribute the negative sign across the terms of the second function. It's crucial to find the common domain of the individual functions to ensure the resulting function is defined for all values in the domain.

How do you multiply and divide functions?

Multiplying functions involves distributing terms, similar to polynomial multiplication. For example, if $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=3x-6$, their product is $3x\sqrt{x}-6\sqrt{x}$. The domain is the intersection of the domains of $f$ and $g$. Dividing functions requires ensuring the denominator is not zero. For $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=3x-6$, the quotient is $\frac{\sqrt{x}}{3x-6}$, with domain restrictions from both functions and the division.

What are composed functions and how are they evaluated?

Composed functions involve applying one function to the result of another. For example, if $f\left(x\right)=x^2$ and $g\left(x\right)=x-1$, then $f\left(g\right(x\left)\right)$ is ${x-1}^2$. To evaluate at a specific value, substitute the value into the inner function first, then into the outer function. For $f\left(g\right(3\left)\right)$, calculate $g\left(3\right)$ first, then $f$ of that result. This method can be faster but may not always be applicable.

How do you find the domain of combined functions?

Finding the domain of combined functions requires identifying restrictions from each function and the operation performed. For addition and subtraction, find the common domain of the individual functions. For multiplication, the domain is the intersection of the domains of the two functions. For division, ensure the denominator is not zero, adding restrictions from the denominator. For example, if $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=3x-6$, the domain for $\frac{\sqrt{x}}{3x-6}$ is $(0,2)\cup (2,\infty )$, excluding $2$ due to division.