Composition of Functions and Function Operations

Oct 13, 2025 by thesweetdesigns.com 123 views

Section 2.6: Composition of Functions and Function Operations Continued

Composition of Functions and Function Operations

In this section, we will continue to explore the composition of functions and function operations. We will use the given functions $f(x) = \frac{4}{x-7}$ and $g(x) = \frac{1}{x}$ to find and simplify the composition of functions $(f \circ g)(x)$ and $(g \circ f)(x)$ , and determine their domains.

Composition of Functions

The composition of functions is a way of combining two or more functions to create a new function. Given two functions $f(x)$ and $g(x)$ , the composition of functions $(f \circ g)(x)$ is defined as:

(f \circ g)(x) = f(g(x))

This means that we first apply the function $g(x)$ to the input $x$ , and then apply the function $f(x)$ to the result.

Finding and Simplifying the Composition of Functions

To find and simplify the composition of functions $(f \circ g)(x)$ , we need to substitute $g(x)$ into $f(x)$ .

Given $f(x) = \frac{4}{x-7}$ and $g(x) = \frac{1}{x}$ , we can substitute $g(x)$ into $f(x)$ as follows:

(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right)

Now, we can substitute $\frac{1}{x}$ into $f(x)$ :

(f \circ g)(x) = f\left(\frac{1}{x}\right) = \frac{4}{\frac{1}{x}-7}

To simplify this expression, we can multiply the numerator and denominator by $x$ :

(f \circ g)(x) = \frac{4x}{1-7x}

Domain of the Composition of Functions

To determine the domain of the composition of functions $(f \circ g)(x)$ , we need to consider the restrictions on the input $x$ .

The function $g(x) = \frac{1}{x}$ is defined for all $x \neq 0$ . However, the function $f(x) = \frac{4}{x-7}$ is defined for all $x \neq 7$ .

Since the composition of functions $(f \circ g)(x)$ is defined as $f(g(x))$ , we need to ensure that the input $g(x)$ is defined. Therefore, we need to exclude $x = 0$ from the domain of $(f \circ g)(x)$ .

Additionally, we need to ensure that the denominator of $(f \circ g)(x)$ is not equal to zero. Therefore, we need to exclude $x = \frac{1}{7}$ from the domain of $(f \circ g)(x)$ .

Conclusion

In this section, we have found and simplified the composition of functions $(f \circ g)(x)$ and determined its domain. We have also considered the restrictions on the input $x$ and excluded $x = 0$ and $x = \frac{1}{7}$ from the domain of $(f \circ g)(x)$ .

Finding and Simplifying the Composition of Functions $(g \circ f)(x)$

To find and simplify the composition of functions $(g \circ f)(x)$ , we need to substitute $f(x)$ into $g(x)$ .

Given $f(x) = \frac{4}{x-7}$ and $g(x) = \frac{1}{x}$ , we can substitute $f(x)$ into $g(x)$ as follows:

(g \circ f)(x) = g(f(x)) = g\left(\frac{4}{x-7}\right)

Now, we can substitute $\frac{4}{x-7}$ into $g(x)$ :

(g \circ f)(x) = g\left(\frac{4}{x-7}\right) = \frac{1}{\frac{4}{x-7}}

To simplify this expression, we can multiply the numerator and denominator by $x-7$ :

(g \circ f)(x) = \frac{x-7}{4}

Domain of the Composition of Functions $(g \circ f)(x)$

To determine the domain of the composition of functions $(g \circ f)(x)$ , we need to consider the restrictions on the input $x$ .

The function $f(x) = \frac{4}{x-7}$ is defined for all $x \neq 7$ . However, the function $g(x) = \frac{1}{x}$ is defined for all $x \neq 0$ .

Since the composition of functions $(g \circ f)(x)$ is defined as $g(f(x))$ , we need to ensure that the input $f(x)$ is defined. Therefore, we need to exclude $x = 7$ from the domain of $(g \circ f)(x)$ .

Additionally, we need to ensure that the denominator of $(g \circ f)(x)$ is not equal to zero. Therefore, we need to exclude $x = 0$ from the domain of $(g \circ f)(x)$ .

Conclusion

In this section, we have found and simplified the composition of functions $(g \circ f)(x)$ and determined its domain. We have also considered the restrictions on the input $x$ and excluded $x = 7$ and $x = 0$ from the domain of $(g \circ f)(x)$ .

Key Takeaways

The composition of functions $(f \circ g)(x)$ is defined as $f(g(x))$ .
The composition of functions $(g \circ f)(x)$ is defined as $g(f(x))$ .
The domain of the composition of functions $(f \circ g)(x)$ is all $x \neq 0$ and $x \neq \frac{1}{7}$ .
The domain of the composition of functions $(g \circ f)(x)$ is all $x \neq 7$ and $x \neq 0$ .

Practice Problems

Find and simplify the composition of functions $(f \circ g)(x)$ and $(g \circ f)(x)$ , where $f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{x-1}$ .
Determine the domain of the composition of functions $(f \circ g)(x)$ and $(g \circ f)(x)$ , where $f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{x-1}$ .

Answer Key

$(f \circ g)(x) = \frac{1}{1-x}$ and $(g \circ f)(x) = \frac{1}{1-x}$
The domain of $(f \circ g)(x)$ is all $x \neq 1$ and $x \neq 0$ . The domain of $(g \circ f)(x)$ is all $x \neq 1$ and $x \neq 0$ .

Conclusion

In this section, we have explored the composition of functions and function operations. We have found and simplified the composition of functions $(f \circ g)(x)$ and $(g \circ f)(x)$ , and determined their domains. We have also considered the restrictions on the input $x$ and excluded $x = 0$ and $x = \frac{1}{7}$ from the domain of $(f \circ g)(x)$ , and excluded $x = 7$ and $x = 0$ from the domain of $(g \circ f)(x)$ .
Q&A: Composition of Functions and Function Operations

Q: What is the composition of functions?

A: The composition of functions is a way of combining two or more functions to create a new function. Given two functions $f(x)$ and $g(x)$ , the composition of functions $(f \circ g)(x)$ is defined as:

(f \circ g)(x) = f(g(x))

Q: How do I find and simplify the composition of functions?

A: To find and simplify the composition of functions, you need to substitute $g(x)$ into $f(x)$ . For example, given $f(x) = \frac{4}{x-7}$ and $g(x) = \frac{1}{x}$ , you can substitute $g(x)$ into $f(x)$ as follows:

(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right)

Now, you can substitute $\frac{1}{x}$ into $f(x)$ :

(f \circ g)(x) = f\left(\frac{1}{x}\right) = \frac{4}{\frac{1}{x}-7}

To simplify this expression, you can multiply the numerator and denominator by $x$ :

(f \circ g)(x) = \frac{4x}{1-7x}

Q: What is the domain of the composition of functions?

A: The domain of the composition of functions is all $x$ for which the input $g(x)$ is defined. For example, given $f(x) = \frac{4}{x-7}$ and $g(x) = \frac{1}{x}$ , the domain of $(f \circ g)(x)$ is all $x \neq 0$ and $x \neq \frac{1}{7}$ .

Q: How do I determine the domain of the composition of functions?

A: To determine the domain of the composition of functions, you need to consider the restrictions on the input $x$ . You need to ensure that the input $g(x)$ is defined, and that the denominator of the composition of functions is not equal to zero.

Q: What is the difference between $(f \circ g)(x)$ and $(g \circ f)(x)$ ?

A: The composition of functions $(f \circ g)(x)$ is defined as $f(g(x))$ , while the composition of functions $(g \circ f)(x)$ is defined as $g(f(x))$ . This means that the order of the functions is reversed.

Q: How do I find and simplify the composition of functions $(g \circ f)(x)$ ?

A: To find and simplify the composition of functions $(g \circ f)(x)$ , you need to substitute $f(x)$ into $g(x)$ . For example, given $f(x) = \frac{4}{x-7}$ and $g(x) = \frac{1}{x}$ , you can substitute $f(x)$ into $g(x)$ as follows:

(g \circ f)(x) = g(f(x)) = g\left(\frac{4}{x-7}\right)

Now, you can substitute $\frac{4}{x-7}$ into $g(x)$ :

(g \circ f)(x) = g\left(\frac{4}{x-7}\right) = \frac{1}{\frac{4}{x-7}}

To simplify this expression, you can multiply the numerator and denominator by $x-7$ :

(g \circ f)(x) = \frac{x-7}{4}

Q: What is the domain of the composition of functions $(g \circ f)(x)$ ?

A: The domain of the composition of functions $(g \circ f)(x)$ is all $x$ for which the input $f(x)$ is defined. For example, given $f(x) = \frac{4}{x-7}$ and $g(x) = \frac{1}{x}$ , the domain of $(g \circ f)(x)$ is all $x \neq 7$ and $x \neq 0$ .

Q: How do I determine the domain of the composition of functions $(g \circ f)(x)$ ?

A: To determine the domain of the composition of functions $(g \circ f)(x)$ , you need to consider the restrictions on the input $x$ . You need to ensure that the input $f(x)$ is defined, and that the denominator of the composition of functions is not equal to zero.

Conclusion

In this Q&A article, we have explored the composition of functions and function operations. We have answered questions about finding and simplifying the composition of functions, determining the domain of the composition of functions, and the difference between $(f \circ g)(x)$ and $(g \circ f)(x)$ . We hope that this article has been helpful in understanding the composition of functions and function operations.