Function Composition: Videos & Practice Problems

Function composition involves evaluating one function within another, such as finding f(g(x)). To determine the domain of composed functions, identify restrictions from both the inside and outside functions. For example, if f(x) = 1/(x - 2) and g(x) = √x, ensure g(x) ≥ 0 and g(x) ≠ 2 to find the overall domain. Function decomposition reverses this process, allowing for creative expressions of a function as a composition of two simpler functions, enhancing understanding of their relationships.

Function Composition

Function composition is a fundamental concept in mathematics that involves combining two functions to create a new function. When composing functions, instead of substituting a number into a function, you substitute one function into another. This process can initially seem challenging, but with practice, it becomes clearer.

To evaluate a function at a specific value, you replace the variable with that value. For example, if you have the function \( f(x) = x^2 + 3x - 10 \) and you want to evaluate it at \( f(7) \), you would calculate:

\( f(7) = 7^2 + 3(7) - 10 = 49 + 21 - 10 = 60 \).

In contrast, when composing functions, such as finding \( f(g(x)) \), you replace the variable \( x \) in \( f(x) \) with the entire function \( g(x) \). For instance, if \( g(x) = x - 2 \), then:

\( f(g(x)) = f(x - 2) = (x - 2)^2 + 3(x - 2) - 10 \).

Upon simplifying, this results in:

\( f(g(x)) = x^2 - x - 12 \).

Function composition can also be denoted using the notation \( f \circ g \), which visually resembles the word "fog." In this notation, \( f \) represents the outer function, while \( g \) represents the inner function. Understanding which function is inside and which is outside is crucial for correctly performing the composition.

For example, consider the functions \( f(x) = x + 4 \) and \( g(x) = x^2 - 3 \). To find \( f(g(x)) \), you substitute \( g(x) \) into \( f(x) \):

\( f(g(x)) = f(x^2 - 3) = (x^2 - 3) + 4 = x^2 + 1 \).

Conversely, to find \( g(f(x)) \), you substitute \( f(x) \) into \( g(x) \):

\( g(f(x)) = g(x + 4) = (x + 4)^2 - 3 \).

Expanding this using the FOIL method gives:

\( g(f(x)) = x^2 + 8x + 16 - 3 = x^2 + 8x + 13 \).

In summary, function composition allows you to create new functions by substituting one function into another. Mastering this concept is essential for further studies in mathematics, as it lays the groundwork for understanding more complex operations and relationships between functions.

Given the functions $f\left(x\right)=\sqrt{x+4}$ and $g\left(x\right)=\left(x-2\right)^2-4$ find $\left(f\circ g\right)\left(x\right)$ and $\left(g\circ f\right)\left(x\right)$

• A. $\left(f\circ g\right)\left(x\right)=\sqrt{x-2}$ ; $\left(g\circ f\right)\left(x\right)=\left(x+4\right)-4\sqrt{x+4}$
• B. $\left(f\circ g\right)\left(x\right)=x-2$ ; $\left(g\circ f\right)\left(x\right)=x\left(x+4\right)$
• C. $\left(f\circ g\right)\left(x\right)=x-2$ ; $\left(g\circ f\right)\left(x\right)=4\sqrt{x-4}$
• D. $\left(f\circ g\right)\left(x\right)=x-2$ ; $\left(g\circ f\right)\left(x\right)=\left(x+4\right)-4\sqrt{x+4}$

Answer: (D)

Given the functions $f(x)=\frac{1}{x^2-2}$ and $g(x)=\sqrt{x+2}$ find $(f∘g)(x)$ and $(g\circ f)(x)$.

• A. $(f\circ g)(x)=\frac{1}{x}$ ; $(g\circ f)(x)=\sqrt{\frac{2x^2-3}{x^2-2}}$
• B. $(f\circ g)(x)=\frac{1}{x}$ ; $(g\circ f)(x)=\sqrt{\frac{3}{x^2-2}}$
• C. $(f∘g)(x)=x$ ; $(g\circ f)(x)=\sqrt{x^2-2}$
• D. $(f∘g)(x)=x$ ; $(g\circ f)(x)=\frac{1}{\sqrt{x+2}-2}$

Answer: (A)

Evaluating Composed Functions

In evaluating composed functions, there are multiple methods to approach the problem, each with its own advantages. A composed function is formed by combining two functions, where the output of one function becomes the input of another. For instance, if we have two functions, f(x) = x² and g(x) = x - 1, we can find the composition f(g(x)) by substituting g(x) into f(x).

To illustrate, let's evaluate f(g(3)). First, we find the composition f(g(x)) by substituting g(x) into f(x):

f(g(x)) = f(x - 1) = (x - 1)²

Next, we simplify this expression:

(x - 1)² = (x - 1)(x - 1) = x² - 2x + 1

Now, to evaluate f(g(3)), we substitute x = 3 into the simplified expression:

f(g(3)) = 3² - 2(3) + 1 = 9 - 6 + 1 = 4

Thus, f(g(3)) = 4.

Alternatively, a shortcut method can be employed. Instead of finding the entire composition first, we can evaluate the inside function directly. For g(3), we compute:

g(3) = 3 - 1 = 2

Then, we substitute this result into f(x):

f(g(3)) = f(2) = 2² = 4

This method is often quicker, but it is essential to note that it may not always be applicable. In some cases, you will be required to find the composition f(g(x)) before evaluating it at a specific number. Therefore, while the shortcut can save time, understanding the full composition is crucial for certain problems.

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)

Domain Restrictions of Composed Functions

Understanding the domain of composed functions is essential for evaluating them correctly. When dealing with composed functions, such as \( f(g(x)) \), it is crucial to identify the restrictions imposed by both the inner function \( g(x) \) and the outer function \( f(x) \).

Consider the functions \( f(x) = \frac{1}{x - 2} \) and \( g(x) = \sqrt{x} \). To find the domain of the composition \( f(g(x)) \), we follow a systematic approach. First, we determine the restrictions for the inner function \( g(x) \). Since \( g(x) = \sqrt{x} \), it is clear that \( x \) must be greater than or equal to 0, as the square root of a negative number is undefined. Thus, the restriction for \( g(x) \) is:

\( x \geq 0 \)

Next, we substitute \( g(x) \) into \( f(x) \) to form the composed function:

\( f(g(x)) = f(\sqrt{x}) = \frac{1}{\sqrt{x} - 2} \)

Now, we need to identify any additional restrictions that arise from this composition. Specifically, the denominator of the fraction cannot equal zero. Therefore, we set up the equation:

\( \sqrt{x} - 2 \neq 0 \)

Solving this gives:

\( \sqrt{x} \neq 2 \)

Squaring both sides results in:

\( x \neq 4 \)

At this point, we have two restrictions: \( x \geq 0 \) from \( g(x) \) and \( x \neq 4 \) from the composition. To express the domain of the composed function \( f(g(x)) \), we combine these restrictions. The domain can be described as all values from 0 to 4, excluding 4, and all values greater than 4:

Domain: \( [0, 4) \cup (4, \infty) \)

This comprehensive approach ensures that we account for all restrictions when determining the domain of composed functions, allowing for accurate evaluations and further mathematical operations.

Given the functions $f(x) = x^2$ and $g(x)=\sqrt{x-8}$ find $(f∘g)(x)$ and determine its domain.

• A. $(f∘g)(x)=x-8$ ; $Dom:(-\infty,\infty)$
• B. $(f\circ g)(x)=\sqrt{x^2-8}$ ; $Dom:(-∞,∞)$
• C. $(f∘g)(x)=x-8$ ; $Dom:[8,∞)$
• D. $(f\circ g)(x)=\sqrt{x^2-8}$ ; $Dom:[8,∞)$

Answer: (C)

Decomposition of Functions

Function decomposition is the process of breaking down a composite function into its individual components. This is essentially the reverse of function composition, where two functions are combined to form a new function. Understanding how to decompose functions can be challenging, but with practice, it can become an enjoyable and creative exercise.

To illustrate function decomposition, consider the function \( h(x) = \sqrt{x - 2} \). The goal is to express this function as a composition of two functions, \( f \) and \( g \), such that \( h(x) = f(g(x)) \). A common strategy is to identify the inner function first. In this case, since \( h(x) \) involves a square root, we can set \( g(x) \) equal to the expression inside the square root, which is \( x - 2 \). Consequently, we define \( f(x) \) as the square root function applied to \( x \), specifically \( f(x) = \sqrt{x} \). Thus, we have:

\( g(x) = x - 2 \)

\( f(x) = \sqrt{x} \)

When we compose these functions, \( f(g(x)) = f(x - 2) = \sqrt{x - 2} \), which matches our original function \( h(x) \).

However, there are multiple valid ways to decompose a function. For instance, we could also define \( g(x) \) as the entire function \( \sqrt{x - 2} \) and set \( f(x) = x \). This gives us:

\( g(x) = \sqrt{x - 2} \)

\( f(x) = x \)

When composed, \( f(g(x)) = f(\sqrt{x - 2}) = \sqrt{x - 2} \), which again results in \( h(x) \).

Moreover, creativity in decomposition allows for even more unconventional approaches. For example, we could define \( g(x) = \sqrt{x - 2} - 1000 \) and \( f(x) = x + 1000 \). This results in:

\( g(x) = \sqrt{x - 2} - 1000 \)

\( f(x) = x + 1000 \)

When composed, \( f(g(x)) = f(\sqrt{x - 2} - 1000) = (\sqrt{x - 2} - 1000) + 1000 = \sqrt{x - 2} \), which is still valid.

In summary, function decomposition allows for various interpretations and methods to express a composite function. The key is to identify the inner function and creatively define the outer function, ensuring that the composition returns to the original function. With practice, this process can enhance your understanding of functions and their relationships.

Decomposition of Functions Example 1

To solve the problem of decomposing the function \( h(x) = \frac{1}{x^2 + 3x - 10} \) into a composition of two functions \( f \) and \( g \), we need to identify these functions such that \( h(x) = f(g(x)) \). The goal is to express \( h(x) \) in terms of \( f \) and \( g \).

First, we recognize that \( h(x) \) is a fraction, where the denominator can be treated as the inner function \( g(x) \). Thus, we can define:

\( g(x) = x^2 + 3x - 10 \)

Next, the outer function \( f(x) \) will be the operation applied to \( g(x) \). Since \( h(x) \) is the reciprocal of \( g(x) \), we can express \( f(x) \) as:

\( f(x) = \frac{1}{x} \)

Putting it all together, we have:

\( h(x) = f(g(x)) = f(x^2 + 3x - 10) = \frac{1}{x^2 + 3x - 10} \)

This decomposition shows that \( g(x) \) captures the polynomial in the denominator, while \( f(x) \) represents the reciprocal function. This method of decomposition is intuitive and effectively illustrates the relationship between the functions.

In summary, the functions are:

\( g(x) = x^2 + 3x - 10 \)

\( f(x) = \frac{1}{x} \)

By composing these functions, we can reconstruct the original function \( h(x) \). This approach highlights the importance of understanding function composition in algebra.

Here’s what students ask on this topic:

What is function composition in mathematics?

Function composition in mathematics involves applying one function to the results of another function. If you have two functions, f(x) and g(x), the composition of these functions is written as (f ∘ g)(x) or f(g(x)). This means you first apply g to x, and then apply f to the result of g(x). For example, if f(x) = x² and g(x) = x + 1, then f(g(x)) = f(x + 1) = (x + 1)². Function composition is a fundamental concept in mathematics, particularly in calculus and algebra, as it allows for the combination of multiple functions into a single operation.

How do you find the domain of a composed function?

To find the domain of a composed function, such as f(g(x)), follow these steps: First, determine the domain of the inside function g(x). This includes all x-values for which g(x) is defined. Next, find the domain of the outside function f(x), but apply it to the range of g(x). This means identifying any restrictions on f(x) based on the output values of g(x). Finally, combine these restrictions to determine the overall domain of the composed function. For example, if f(x) = 1/(x - 2) and g(x) = √x, the domain of g(x) is x ≥ 0, and the domain of f(g(x)) requires g(x) ≠ 2, so the overall domain is x ≥ 0 and x ≠ 4.

What are some common methods for evaluating composed functions?

There are two common methods for evaluating composed functions: the direct method and the shortcut method. The direct method involves first finding the composition of the functions, such as f(g(x)), and then substituting the given value into the resulting function. For example, if f(x) = x² and g(x) = x - 1, to find f(g(3)), first compute f(g(x)) = (x - 1)², then substitute x = 3 to get (3 - 1)² = 4. The shortcut method involves evaluating the inside function first, then using its result in the outside function. For the same example, find g(3) = 2, then f(2) = 2² = 4. Both methods yield the same result.

How do you decompose a function into two simpler functions?

To decompose a function into two simpler functions, identify an inner function g(x) and an outer function f(x) such that the original function h(x) can be written as f(g(x)). For example, if h(x) = √(x - 2), you can set g(x) = x - 2 and f(x) = √x. This way, h(x) = f(g(x)) = √(x - 2). Another approach is to choose g(x) as a more complex expression and f(x) as a simpler one. For instance, if h(x) = √(x - 2) - 1000, you could set g(x) = √(x - 2) - 1000 and f(x) = x + 1000. Decomposition allows for creative and multiple correct solutions.

What are some examples of function composition problems?

Here are a few examples of function composition problems: 1) Given f(x) = x² + 3 and g(x) = 2x - 1, find f(g(x)). Solution: f(g(x)) = (2x - 1)² + 3. 2) Given f(x) = √x and g(x) = x² - 4, find g(f(x)). Solution: g(f(x)) = (√x)² - 4 = x - 4. 3) Given f(x) = 1/x and g(x) = x + 2, find f(g(x)). Solution: f(g(x)) = 1/(x + 2). These problems illustrate how to substitute one function into another and simplify the resulting expression.