MTH120 College Algebra Chapter 3.4

 3.4 Composition of Functions:

Composition of functions is a fundamental concept in mathematics, particularly in the study of functions and transformations. It involves combining two or more functions to create a new function. The composition of functions is denoted by the symbol "$\circ$".

Here's how you can define and understand the composition of functions:

Composition of Two Functions (f and g):

• Given two functions, $�(�)$ and $�(�)$, the composition of $�$ and $�$ is a new function that applies $�$ to the result of $�$.
• The composition of $�$ and $�$ is denoted as $(�\circ �)(�)$, and it is defined as $(�\circ �)(�)=�(�(�))$.
• In other words, you first apply $�$ to the input $�$, and then you apply $�$ to the result of $�(�)$.

Here's an example to illustrate composition of functions:

Example: Suppose you have two functions:

• $�(�)=2�+3$
• $�(�)={�}^2$

To find $(�\circ �)(�)$, you would first apply $�$ to $�$, and then apply $�$ to the result:

$(�\circ �)(�)=�(�(�))=�({�}^2)=2{�}^2+3$

So, $(�\circ �)(�)=2{�}^2+3$.

Key points to remember about composition of functions:

1. The order in which you compose functions matters. $(�\circ �)(�)$ is not the same as $(�\circ �)(�)$ unless $�$ and $�$ happen to be commutative (i.e., $�(�)�(�)=�(�)�(�)$ for all $�$).

2. Not all pairs of functions can be composed. To be able to compose two functions, the output of the inner function (in this case, $�(�)$) must be within the domain of the outer function (in this case, $�(�)$).

3. Composition of functions is a powerful tool for modeling complex processes and transformations, especially in areas like calculus, linear algebra, and computer science.

4. You can also compose more than two functions. For example, $(�\circ �\circ ℎ)(�)$ represents the composition of three functions: $�(�(ℎ(�)))$.

Overall, composition of functions is a versatile concept used to combine and analyze functions in various mathematical and scientific contexts.

Combining functions using algebraic operations involves performing mathematical operations such as addition, subtraction, multiplication, division, and composition on two or more functions. Here's how you can perform these operations:

1. Addition and Subtraction of Functions:

   • Given two functions $�(�)$ and $�(�)$, you can add or subtract them to create a new function $ℎ(�)$:

     • $ℎ(�)=�(�)+�(�)$ represents the sum of the two functions.
     • $ℎ(�)=�(�)-�(�)$ represents the difference of the two functions.

   • For example:

     • If $�(�)=2�$ and $�(�)=3{�}^2$, then $ℎ(�)=�(�)+�(�)=2�+3{�}^2$ represents their sum.
     • If $�(�)=\sin (�)$ and $�(�)=\cos (�)$, then $ℎ(�)=�(�)-�(�)=\sin (�)-\cos (�)$ represents their difference.

2. Multiplication of Functions:

   • To multiply two functions, $�(�)$ and $�(�)$, you create a new function $ℎ(�)$:

     • $ℎ(�)=�(�)\cdot �(�)$ represents the product of the two functions.

   • For example:

     • If $�(�)=�$ and $�(�)=2{�}^2$, then $ℎ(�)=�(�)\cdot �(�)=�\cdot 2{�}^2=2{�}^3$ represents their product.
     • If $�(�)=\sin (�)$ and $�(�)=\cos (�)$, then $ℎ(�)=�(�)\cdot �(�)=\sin (�)\cdot \cos (�)$ represents their product.

3. Division of Functions:

   • To divide one function $�(�)$ by another function $�(�)$, you create a new function $ℎ(�)$:

     • $ℎ(�)=\frac{�(�)}{�(�)}$ represents the division of $�$ by $�$.

   • Be cautious when dividing functions, as you need to ensure that the denominator $�(�)$ is not zero for the values of $�$ in the domain of interest. Division by zero is undefined.

4. Composition of Functions (already explained in a previous response):

   • Composition combines two functions $�(�)$ and $�(�)$ to create a new function $ℎ(�)$:

     • $ℎ(�)=�(�(�))$ represents the composition of $�$ and $�$, where you apply $�$ to the input and then apply $�$ to the result.

   • For example:

     • If $�(�)=2�$ and $�(�)={�}^2$, then $ℎ(�)=�(�(�))=2{�}^2$ represents their composition.

These algebraic operations on functions are essential tools for modeling and analyzing complex relationships, transformations, and processes in various mathematical, scientific, and engineering contexts.

Let's create a new function by composing two given functions. Suppose we have the following two functions:

1. $�(�)=2�+1$
2. $�(�)=\sin (�)$

Let's compose these functions to create a new function $ℎ(�)$ defined as $ℎ(�)=�(�(�))$. In other words, we will apply $�(�)$ to the input $�$ and then apply $�(�)$ to the result. Here's how you can do it:

1. Start with $�(�)$:

   • $�(�)=\sin (�)$

2. Apply $�(�)$ to the result of $�(�)$:

   • $ℎ(�)=�(�(�))=�(\sin (�))$

3. Replace $�(�)$ with its expression:

   • $ℎ(�)=2\sin (�)+1$

So, the new function $ℎ(�)$ created by composing $�(�)$ and $�(�)$ is $ℎ(�)=2\sin (�)+1$. This function represents the result of applying $�(�)$ to the output of $�(�)$ for any input $�$.

The commutative property in mathematics refers to the order in which you perform an operation not affecting the result. In the context of functions and composition of functions, the question is whether the composition of functions is commutative, meaning whether the order in which you compose functions affects the result.

In general, the composition of functions is not commutative. This means that the order in which you compose two functions can change the result.

Mathematically, if you have two functions, $�(�)$ and $�(�)$, then:

• $(�\circ �)(�)=�(�(�))$ represents the composition of $�$ and $�$, where you apply $�$ to the input $�$ and then apply $�$ to the result.
• $(�\circ �)(�)=�(�(�))$ represents the composition of $�$ and $�$, where you apply $�$ to the input $�$ and then apply $�$ to the result.

In most cases, $(�\circ �)(�)$ is not equal to $(�\circ �)(�)$. In other words, the order matters, and the composition of functions is not commutative.

However, there are some specific cases where the composition of functions can be commutative. For example:

1. Identity Function: If one of the functions is the identity function $�(�)=�$, then the composition is commutative with any other function. This is because the identity function has no effect on the input.

   Example: $(�\circ �)(�)=�(�)$ and $(�\circ �)(�)=�(�)$ for any function $�(�)$.

2. Functions with Commutative Operations: If the functions involve operations that are commutative, such as addition or multiplication, the composition might be commutative.

   Example: If $�(�)=2�$ and $�(�)=3�$, then $(�\circ �)(�)=2(3�)=6�$ and $(�\circ �)(�)=3(2�)=6�$, so in this case, composition is commutative.

In most mathematical contexts, especially in calculus and algebra, it's safe to assume that composition of functions is not commutative unless you have specific reasons to believe otherwise based on the properties of the functions involved.

Evaluating composite functions involves finding the value of a function that is formed by composing two or more functions. To evaluate a composite function at a specific input, follow these steps:

Suppose you have two functions:

1. $�(�)$
2. $�(�)$

And you want to evaluate the composite function $(�\circ �)(�)$ at a specific value of $�$.

Here are the steps to evaluate $(�\circ �)(�)$:

1. Start with the inner function $�(�)$.

2. Substitute the specific value of $�$ where you want to evaluate $(�\circ �)(�)$ into $�(�)$ to find $�(\text{value of }�)$.

3. Take the result from step 2 and substitute it into the outer function $�(�)$ to find $�(�(\text{value of }�))$.

4. Calculate the value of $�(�(\text{value of }�))$. This will be the result of evaluating $(�\circ �)(�)$ at the specified input value.

Here's a more detailed example:

Suppose you have the following functions:

1. $�(�)=2�+1$
2. $�(�)={�}^2-3$

You want to evaluate the composite function $(�\circ �)(�)$ at $�=2$.

1. Start with the inner function $�(�)$:

   • $�(�)={�}^2-3$

2. Substitute $�=2$ into $�(�)$ to find $�(2)$:

   • $�(2)=2^2-3=4-3=1$

3. Take the result from step 2 and substitute it into the outer function $�(�)$ to find $�(�(2))$:

   • $�(�(2))=�(1)$

4. Calculate $�(�(2))$:

   • $�(1)=2(1)+1=2+1=3$

So, $(�\circ �)(�)$ evaluated at $�=2$ is $3$.

In summary, to evaluate a composite function, start with the inner function, substitute the specific value of $�$ into it, then take the result and substitute it into the outer function. Finally, calculate the value to find the result of the composite function at that particular input.

Evaluating composite functions using tables involves using tables of values for the individual functions and then combining these values to find the values of the composite function. Here's a step-by-step guide on how to do it:

Suppose you have two functions:

1. $�(�)$
2. $�(�)$

And you want to evaluate the composite function $(�\circ �)(�)$ or $(�\circ �)(�)$ using tables of values for $�$ and $�$.

Here are the steps to evaluate the composite function using tables:

1. Create tables of values for both $�$ and $�$ for the range of $�$ values you are interested in.

2. For each function, you should have a table with two columns: one for $�$ values and another for $�(�)$ or $�(�)$ values.

3. Calculate the values of $�(�(�))$ or $�(�(�))$ by substituting the values of $�$ from the tables into the corresponding functions. This will give you intermediate values for the composite function.

4. Create a new table with three columns: one for $�$ values, one for $�(�(�))$ values (or $�(�(�))$ values), and one for the final result.

5. Fill in the $�$ values from the original tables into the new table.

6. Fill in the intermediate values you calculated in step 3 into the appropriate column in the new table.

7. Calculate the final values of $(�\circ �)(�)$ or $(�\circ �)(�)$ by performing the operations on the intermediate values (e.g., addition, multiplication) if necessary. This will give you the values of the composite function.

8. The final column in the new table will contain the values of the composite function for the corresponding $�$ values.

Here's a simplified example:

Suppose you have the following functions and their respective tables:

1. $�(�)=2�$

   $�$	$�(�)$
   1	2
   2	4
   3	6

2. $�(�)={�}^2$

   $�$	$�(�)$
   1	1
   2	4
   3	9

You want to evaluate $(�\circ �)(�)$. Here are the steps:

3. Calculate intermediate values by substituting the $�$ values from the $�(�)$ table into $�(�)$:

   $�$	$�(�(�))$
   1	2
   4	8
   9	18

4. Create a new table:

   $�$	$�(�(�))$	$(�\circ �)(�)$
   1	2
   4	8
   9	18

5. Fill in the $�$ values from the original tables:

   $�$	$�(�(�))$	$(�\circ �)(�)$
   1	2
   4	8
   9	18

6. Fill in the intermediate values:

   $�$	$�(�(�))$	$(�\circ �)(�)$
   1	2	2
   4	8	8
   9	18	18

7. The final column contains the values of $(�\circ �)(�)$:

   $�$	$�(�(�))$	$(�\circ �)(�)$
   1	2	2
   4	8	8
   9	18	18

So, you've evaluated $(�\circ �)(�)$ using tables of values for $�$ and $�$.

To evaluate a composite function using the information provided by the graphs of its individual functions, follow these steps:

Suppose you have two functions:

1. $�(�)$
2. $�(�)$

And you want to evaluate the composite function $(�\circ �)(�)$ using the information provided by the graphs.

Here are the steps:

1. Examine the Graphs: First, study the graphs of the individual functions $�(�)$ and $�(�)$. Understand the behavior of each function and identify specific points or values of interest.

2. Identify the Input Value: Determine the value of $�$ at which you want to evaluate $(�\circ �)(�)$. This is the input value for your composite function.

3. Locate the Input Value on the Graph of $�(�)$: Find the point on the graph of $�(�)$ where the input value $�$ corresponds. This will give you the corresponding $�(�)$ value.

4. Use the $�(�)$ Value to Evaluate $�(�(�))$: Take the $�(�)$ value obtained from step 3 and find it on the graph of $�(�)$. This will give you the corresponding $�(�(�))$ value.

5. The Result is $(�\circ �)(�)$: The $�(�(�))$ value you found in step 4 is the result of evaluating $(�\circ �)(�)$ at the specified input value $�$.

Let's illustrate this process with an example:

Suppose you have the following functions and their respective graphs:

1. $�(�)={�}^2$

   • The graph of $�(�)$ is a parabola opening upward.

2. $�(�)=\sin (�)$

   • The graph of $�(�)$ is a sinusoidal curve.

You want to evaluate $(�\circ �)(�)$ at $�=�\mathrm{/}4$.

Steps:

1. Examine the graphs of $�(�)$ and $�(�$) to understand their behavior.

2. Identify $�=�\mathrm{/}4$ as the input value.

3. Locate $�\mathrm{/}4$ on the graph of $�(�)$ and find the corresponding $�(�)$ value, which is approximately $�(�\mathrm{/}4)\approx 0.71$.

4. Use the $�(�)$ value to find $�(�(�))$ on the graph of $�(�)$. It corresponds to $�(0.71)\approx 0.51$.

5. The result is $(�\circ �)(�\mathrm{/}4)=0.51$.

So, $(�\circ �)(�\mathrm{/}4)=0.51$ based on the information provided by the graphs of $�(�)$ and $�(�)$.

Evaluating composite functions using formulas involves applying the composition of functions algebraically. Here's how you can do it:

Suppose you have two functions:

1. $�(�)$
2. $�(�)$

And you want to evaluate the composite function $(�\circ �)(�)$ using their formulas.

Here are the steps:

1. Start with the inner function $�(�)$.

2. Substitute the formula of $�$ for $�(�)$.

3. Take the result from step 2 and substitute it into the formula of $�(�)$.

4. Calculate the final value of $(�\circ �)(�)$ based on the formula.

Here's an example:

Suppose you have the following functions:

1. $�(�)=3{�}^2$
2. $�(�)=\sqrt{�+1}$

You want to evaluate the composite function $(�\circ �)(�)$.

Steps:

1. Start with the inner function $�(�)$:

   • $�(�)=\sqrt{�+1}$

2. Substitute the formula of $�$ for $�(�)$:

   • $�(�)=\sqrt{�+1}$

3. Take the result from step 2 and substitute it into the formula of $�(�)$:

   • $(�\circ �)(�)=�(�(�))=3(\sqrt{�+1}{)}^2$

4. Calculate the final value:

   • $(�\circ �)(�)=3(�+1)$

So, $(�\circ �)(�)=3(�+1)$.

Now, you have evaluated the composite function $(�\circ �)(�)$ using their formulas, and it is expressed as $3(�+1)$. You can further simplify or substitute values of $�$ into this expression to find specific results.

To find the domain of a composite function $(�\circ �)(�)$, you need to consider the domains of the individual functions $�(�)$ and $�(�)$ and determine where their compositions are valid. The domain of $(�\circ �)(�)$ is the set of all $�$ values for which both $�(�(�))$ and $�(�)$ are defined.

Here's how you can find the domain of a composite function:

1. Start with the Domains of the Individual Functions:

   • Determine the domain of $�(�)$, denoted as ${�}_{�}$, which consists of all $�$ values for which $�(�)$ is defined.
   • Determine the domain of $�(�)$, denoted as ${�}_{�}$, which consists of all $�$ values for which $�(�)$ is defined.

2. Consider the Composition:

   • The domain of the composite function $(�\circ �)(�)$ consists of all $�$ values for which both $�(�(�))$ and $�(�)$ are defined.

3. Intersection of Domains:

   • Find the intersection of the domains ${�}_{�}$ and ${�}_{�}$. This means you're looking for the values that are in both sets ${�}_{�}$ and ${�}_{�}$.
   • The intersection of ${�}_{�}$ and ${�}_{�}$ is the domain of the composite function.

4. Write the Domain of the Composite Function:

   • Express the domain of the composite function $(�\circ �)(�)$ in set notation, if necessary.

Let's illustrate this process with an example:

Suppose you have the following functions:

1. $�(�)=\sqrt{�}$

   • The domain of $�(�)$ is ${�}_{�}=[0,\mathrm{\infty })$ because the square root is defined only for non-negative real numbers.

2. $�(�)=\frac{1}{�}$

   • The domain of $�(�)$ is ${�}_{�}=(-\mathrm{\infty },0)\cup (0,\mathrm{\infty })$ because $�(�)$ is undefined at $�=0$.

Now, you want to find the domain of $(�\circ �)(�)$, which is $(�\circ �)(�)=�(�(�))=\sqrt{\frac{1}{�}}$.

Steps:

1. Start with the domains of the individual functions:

   • ${�}_{�}=[0,\mathrm{\infty })$
   • ${�}_{�}=(-\mathrm{\infty },0)\cup (0,\mathrm{\infty })$

2. Consider the composition:

   • The domain of $(�\circ �)(�)$ consists of values for which both $�(�(�))$ and $�(�)$ are defined.

3. Find the intersection of domains:

   • The intersection of ${�}_{�}$ and ${�}_{�}$ is $(0,\mathrm{\infty })$.

4. Write the domain of the composite function:

   • The domain of $(�\circ �)(�)$ is $(0,\mathrm{\infty })$.

So, the domain of $(�\circ �)(�)$ is $(0,\mathrm{\infty })$. This means that the composite function $(�\circ �)(�)$ is defined for all positive real numbers but not for zero or negative numbers.

Decomposing a composite function into its component functions involves reversing the process of function composition. When you have a composite function $(�\circ �)(�)$, you can find the component functions $�(�)$ and $�(�)$ by solving for them.

Here's how you can decompose a composite function into its component functions:

1. Start with the Composite Function:

   • You have a composite function $(�\circ �)(�)$, which is a combination of two functions.

2. Use a Variable for the Intermediate Result:

   • To decompose the function, introduce a new variable, say $�$, and write the composition as $�(�)=�(�(�))$.

3. Solve for the Inner Function $�(�)$:

   • Isolate the inner function $�(�)$ by expressing it in terms of $�$:
     • $�(�)=�(�)$

4. Solve for the Outer Function $�(�)$:

   • Isolate the outer function $�(�)$ by expressing it in terms of $�$:
     • $�(�)=�(�)$

Now, you have decomposed the composite function $(�\circ �)(�)$ into its component functions $�(�)$ and $�(�)$ in terms of the variable $�$.

Here's an example:

Suppose you have the composite function $(�\circ �)(�)=(2{�}^2+1{)}^3$. You want to decompose it into its component functions $�(�)$ and $�(�)$.

Steps:

1. Start with the composite function:

   • $(�\circ �)(�)=(2{�}^2+1{)}^3$

2. Introduce a new variable $�$ and write the composition as $�(�)=(2{�}^2+1{)}^3$.

3. Solve for the inner function $�(�)$ in terms of $�$:

   • $�(�)={�}^{1\mathrm{/}3}-1$

4. Solve for the outer function $�(�)$ in terms of $�$:

   • $�(�)={�}^3$

Now you have decomposed the composite function into its component functions:

• $�(�)={�}^{1\mathrm{/}3}-1$
• $�(�)={�}^3$

You can see that $�(�)$ is the inner function that operates on $�$ first, and $�(�)$ is the outer function that operates on the result of $�(�)$.

Let's explore a couple of examples where we decompose composite functions into their component functions.

Example 1:

Suppose you have the composite function $(�\circ �)(�)=\sqrt{3�+2}$. You want to decompose it into its component functions $�(�)$ and $�(�)$.

1. Start with the composite function:

   • $(�\circ �)(�)=\sqrt{3�+2}$

2. Introduce a new variable $�$ and write the composition as $�(�)=\sqrt{3�+2}$.

3. Solve for the inner function $�(�)$ in terms of $�$:

   • $�(�)=\frac{{�}^2-2}{3}$

4. Solve for the outer function $�(�)$ in terms of $�$:

   • $�(�)=\sqrt{�}$

Now you have decomposed the composite function into its component functions:

• $�(�)=\frac{{�}^2-2}{3}$
• $�(�)=\sqrt{�}$

In this case, $�(�)$ represents the inner function, and $�(�)$ represents the outer function.

Example 2:

Suppose you have the composite function $(ℎ\circ �)(�)=(4{�}^3-2�{)}^2$. You want to decompose it into its component functions $ℎ(�)$ and $�(�)$.

1. Start with the composite function:

   • $(ℎ\circ �)(�)=(4{�}^3-2�{)}^2$

2. Introduce a new variable $�$ and write the composition as $�(�)=(4{�}^3-2�{)}^2$.

3. Solve for the inner function $�(�)$ in terms of $�$:

   • $�(�)=\sqrt[2]{�}$

4. Solve for the outer function $ℎ(�)$ in terms of $�$:

   • $ℎ(�)=�$

Now you have decomposed the composite function into its component functions:

• $�(�)=\sqrt[2]{�}$
• $ℎ(�)=�$

In this case, $�(�)$ represents the inner function, and $ℎ(�)$ represents the outer function.

These examples demonstrate how to decompose composite functions into their component functions using a new variable and solving for the inner and outer functions in terms of that variable.