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Explore function composition properties: non-commutative order, associative grouping, and identity elements. Learn (f∘g) rules with proofs.
---
## Properties of Function Composition
Function composition, which involves combining functions sequentially, has several important properties we need to know. Let's study these properties using the following example functions:
Component: MathContainer
Children:
```math
f(x) = 2x + 1
```
```math
g(x) = x^2 + 4
```
```math
h(x) = \frac{1}{x+1}
```
### Non-Commutative Property
The first and most common property is that the order in which functions are composed **matters**. Changing the order of functions usually results in a different composite function.
In general, $$(f \circ g)(x)$$ is **not equal to** $$(g \circ f)(x)$$.
Visible text: In general, is **not equal to** .
**Example:**
Let's compare $$(g \circ f)(x)$$ and $$(f \circ g)(x)$$.
Visible text: Let's compare and .
1. **Calculating $$(g \circ f)(x)$$:**
```math
(g \circ f)(x) = g(f(x)) = g(2x+1)
```
```math
g(2x+1) = (2x+1)^2 + 4 = (4x^2 + 4x + 1) + 4 = 4x^2 + 4x + 5
```
2. **Calculating $$(f \circ g)(x)$$:**
```math
(f \circ g)(x) = f(g(x)) = f(x^2+4)
```
```math
f(x^2+4) = 2(x^2+4) + 1 = (2x^2 + 8) + 1 = 2x^2 + 9
```
Visible text: 1. **Calculating :**
2. **Calculating :**
Since $$4x^2 + 4x + 5 \neq 2x^2 + 9$$, it is proven that $$(g \circ f)(x) \neq (f \circ g)(x)$$. This property also applies to other compositions, for example $$(f \circ h)(x) \neq (h \circ f)(x)$$ and $$(g \circ h)(x) \neq (h \circ g)(x)$$.
Visible text: Since , it is proven that . This property also applies to other compositions, for example and .
### Associative Property
If we compose three or more functions, the order of **performing** the composition does not affect the final result, as long as the order of the **functions** remains the same.
Mathematically, for functions $$f$$, $$g$$, and $$h$$, the following holds:
Visible text: Mathematically, for functions , , and , the following holds:
```math
((f \circ g) \circ h)(x) = (f \circ (g \circ h))(x)
```
This means we can compose $$f$$ with $$g$$ first, and then compose the result with $$h$$. Alternatively, we can compose $$g$$ with $$h$$ first, and then compose $$f$$ with the result. The outcome will be the same.
Visible text: This means we can compose with first, and then compose the result with . Alternatively, we can compose with first, and then compose with the result. The outcome will be the same.
**Example:**
Let's check if $$((f \circ g) \circ h)(x) = (f \circ (g \circ h))(x)$$.
Visible text: Let's check if .
1. **Calculating $$((f \circ g) \circ h)(x)$$:**
We already know $$(f \circ g)(x) = 2x^2 + 9$$.
```math
((f \circ g) \circ h)(x) = (f \circ g)(h(x)) = (f \circ g)(\frac{1}{x+1})
```
```math
(f \circ g)(\frac{1}{x+1}) = 2\left(\frac{1}{x+1}\right)^2 + 9 = \frac{2}{(x+1)^2} + 9
```
2. **Calculating $$(f \circ (g \circ h))(x)$$:**
First, find $$(g \circ h)(x)$$:
```math
(g \circ h)(x) = g(h(x)) = g(\frac{1}{x+1})
```
```math
g(\frac{1}{x+1}) = \left(\frac{1}{x+1}\right)^2 + 4 = \frac{1}{(x+1)^2} + 4
```
Now, compose $$f$$ with this result:
```math
(f \circ (g \circ h))(x) = f((g \circ h)(x)) = f\left(\frac{1}{(x+1)^2} + 4\right)
```
```math
f\left(\frac{1}{(x+1)^2} + 4\right) = 2\left(\frac{1}{(x+1)^2} + 4\right) + 1 = \frac{2}{(x+1)^2} + 8 + 1 = \frac{2}{(x+1)^2} + 9
```
Visible text: 1. **Calculating :**
We already know .
2. **Calculating :**
First, find :
Now, compose with this result:
Since both results are the same ($$\frac{2}{(x+1)^2} + 9$$), the associative property is proven to hold: $$((f \circ g) \circ h)(x) = (f \circ (g \circ h))(x)$$.
Visible text: Since both results are the same (), the associative property is proven to hold: .
This associative property also applies to other combinations of function order, such as $$((h \circ f) \circ g)(x) = (h \circ (f \circ g))(x)$$ and $$((g \circ f) \circ h)(x) = (g \circ (f \circ h))(x)$$.
Visible text: This associative property also applies to other combinations of function order, such as and .
### Identity Element
There is a special function called the **identity function**, denoted by $$I(x)$$, which is defined as $$I(x) = x$$. This function does not change its input.
Visible text: There is a special function called the **identity function**, denoted by , which is defined as . This function does not change its input.
If a function $$f$$ is composed with the identity function $$I$$ (from either the left or the right), the result is the function $$f$$ itself.
Visible text: If a function is composed with the identity function (from either the left or the right), the result is the function itself.
Component: MathContainer
Children:
```math
(f \circ I)(x) = f(I(x)) = f(x)
```
```math
(I \circ f)(x) = I(f(x)) = f(x)
```
**Example:**
With $$f(x) = 2x + 1$$:
Visible text: With :
- $$(f \circ I)(x) = f(I(x)) = f(x) = 2x + 1$$
- $$(I \circ f)(x) = I(f(x)) = I(2x+1) = 2x + 1$$
Visible text: -
-
Both result in the function $$f(x)$$ again.
Visible text: Both result in the function again.