Simplification of Compositions

Oct 13, 2025 by talynt.org 136 views

Introduction

In mathematics, composition of functions is a fundamental concept that allows us to combine two or more functions to create a new function. This concept is crucial in various branches of mathematics, including algebra, calculus, and analysis. In this article, we will explore the composition of functions, specifically the composition of the functions $h$ and $f$ defined as follows:

$h(x) = x^2 - 6$

$f(x) = \frac{x}{6}$

We will find the compositions $h \circ h$ and $f \circ f$ , and simplify our answers.

Composition of Functions

The composition of two functions $f$ and $g$ is denoted by $f \circ g$ and is defined as:

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

In other words, we first apply the function $g$ to the input $x$ , and then apply the function $f$ to the result.

Composition of $h$ and $h$

To find the composition $h \circ h$ , we need to substitute $h(x)$ into the function $h$ .

$h \circ h = h(h(x))$

$= h(x^2 - 6)$

$= (x^2 - 6)^2 - 6$

$= x^4 - 12x^2 + 36 - 6$

$= x^4 - 12x^2 + 30$

Therefore, the composition $h \circ h$ is given by:

$h \circ h = x^4 - 12x^2 + 30$

Composition of $f$ and $f$

To find the composition $f \circ f$ , we need to substitute $f(x)$ into the function $f$ .

$f \circ f = f(f(x))$

$= f\left(\frac{x}{6}\right)$

$= \frac{\frac{x}{6}}{6}$

$= \frac{x}{36}$

Therefore, the composition $f \circ f$ is given by:

$f \circ f = \frac{x}{36}$

Simplification of Compositions

In the previous sections, we found the compositions $h \circ h$ and $f \circ f$ . However, we can simplify these compositions further.

For the composition $h \circ h$ , we can simplify it as follows:

$h \circ h = x^4 - 12x^2 + 30$

$= (x^2 - 6)^2 - 6$

This shows that the composition $h \circ h$ can be expressed in terms of the original function $h$ .

For the composition $f \circ f$ , we can simplify it as follows:

$f \circ f = \frac{x}{36}$

$= \frac{1}{36}x$

This shows that the composition $f \circ f$ can be expressed in terms of the original function $f$ .

Conclusion

In this article, we explored the composition of functions, specifically the composition of the functions $h$ and $f$ defined as follows:

$h(x) = x^2 - 6$

$f(x) = \frac{x}{6}$

We found the compositions $h \circ h$ and $f \circ f$ , and simplified our answers. We also showed that the compositions can be expressed in terms of the original functions $h$ and $f$ .

Final Answer

The final answer is:

$h \circ h = x^4 - 12x^2 + 30$

Introduction

In our previous article, we explored the composition of functions, specifically the composition of the functions $h$ and $f$ defined as follows:

$h(x) = x^2 - 6$

$f(x) = \frac{x}{6}$

We found the compositions $h \circ h$ and $f \circ f$ , and simplified our answers. In this article, we will answer some frequently asked questions about composition of functions.

Q&A

Q: What is the composition of two functions?

A: The composition of two functions $f$ and $g$ is denoted by $f \circ g$ and is defined as:

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

In other words, we first apply the function $g$ to the input $x$ , and then apply the function $f$ to the result.

Q: How do I find the composition of two functions?

A: To find the composition of two functions $f$ and $g$ , you need to substitute $g(x)$ into the function $f$ . This can be done by replacing $x$ in the function $f$ with $g(x)$ .

Q: What is the difference between the composition of two functions and the product of two functions?

A: The composition of two functions $f$ and $g$ is denoted by $f \circ g$ , while the product of two functions is denoted by $f \cdot g$ . The composition of two functions is a new function that is obtained by applying one function to the output of the other function, while the product of two functions is a new function that is obtained by multiplying the two functions.

Q: Can the composition of two functions be commutative?

A: No, the composition of two functions is not commutative. This means that the order of the functions in the composition matters. For example, $(f \circ g)(x) \neq (g \circ f)(x)$ in general.

Q: Can the composition of two functions be associative?

A: Yes, the composition of two functions is associative. This means that the order in which we apply the functions does not matter. For example, $((f \circ g) \circ h)(x) = (f \circ (g \circ h))(x)$ .

Q: How do I simplify the composition of two functions?

A: To simplify the composition of two functions, you can try to factor out common terms or use algebraic manipulations to simplify the expression.

Q: Can the composition of two functions be used to solve equations?

A: Yes, the composition of two functions can be used to solve equations. For example, if we have an equation of the form $f(g(x)) = c$ , we can use the composition of functions to solve for $x$ .

Examples

Example 1: Composition of two linear functions

Suppose we have two linear functions $f(x) = 2x + 1$ and $g(x) = 3x - 2$ . Find the composition $f \circ g$ .

Solution:

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

$= f(3x - 2)$

$= 2(3x - 2) + 1$

$= 6x - 4 + 1$

$= 6x - 3$

Therefore, the composition $f \circ g$ is given by:

$f \circ g = 6x - 3$

Example 2: Composition of two quadratic functions

Suppose we have two quadratic functions $f(x) = x^2 + 2x + 1$ and $g(x) = x^2 - 3x + 2$ . Find the composition $f \circ g$ .

Solution:

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

$= f(x^2 - 3x + 2)$

$= (x^2 - 3x + 2)^2 + 2(x^2 - 3x + 2) + 1$

$= x^4 - 6x^3 + 13x^2 - 12x + 4 + 2x^2 - 6x + 4 + 1$

$= x^4 - 6x^3 + 15x^2 - 18x + 9$

Therefore, the composition $f \circ g$ is given by:

$f \circ g = x^4 - 6x^3 + 15x^2 - 18x + 9$

Conclusion

In this article, we answered some frequently asked questions about composition of functions. We also provided examples of how to find the composition of two functions and how to simplify the composition of two functions. We hope that this article has been helpful in understanding the concept of composition of functions.