f of g of x

• Introduction
• What Is the Meaning of `F` of `G` of `x`?
• Symbol of `f` of `g` of `x`
• How to Solve `f` of `g` of `x`?
• Order of Composite Functions
• How to Find `f` of `g` of `x` From Graph
• How to Find `f` of `g` of `x` From Table
• Self Composition
• Domain and Range of `f` of `g` of `x`
• Properties of Combined Functions
• Solved Examples
• Practice Problems
• Frequently Asked Questions

Introduction

Function composition is like combining two operations into one. Imagine you have a task to outline a picture (operation `f`) followed by filling it with colors (operation `g`), composing them involves outlining the picture initially and then filling it with colors. We'll explore function composition, its mathematical definition, and practical applications across various scenarios.

What Is the Meaning of `F` of `G` of `x`?

Composite functions, also known as `f` of `g` of `x`, are mathematical operations denoted as `f(g(x))` or `(f ∘ g)(x)`. This notation implies that the value of `x` should be substituted into `g(x)` first, and then the result should be plugged into `f(x)`. This concept can be visualized as a series of operations where the output of one function serves as the input for another. 

For example, if we have functions `f(x)` and `g(x)`, and we want to find `f` of `g` of `x`, we would first compute `g(x)` and then use that result as the input for `f(x)`.

Symbol of `f` of `g` of `x`

Composite functions, represented by the symbol `∘`, are combinations of two functions. You might see them written as `(f ∘ g)(x)` or `f(g(x))`. This means you're applying function `g` to the input `x`, and then taking that result and applying function `f` to it. There's another way around too: `(g ∘ f)(x)` or `g(f(x))`. Here, you're first using function `f` on `x`, and then applying function `g` to the result. So, you see, it's like a chain reaction of functions!

For any two functions `f` and `g`, there can be two composite functions:

• `f` of `g` of `x` `= (f ∘ g)(x) = f(g(x))`
• `g` of `f` of `x` `= (g ∘ f)(x) = g(f(x))`

How to Solve `f` of `g` of `x`?

When you see functions within functions, like \( f(g(x)) \), it means you need to perform one function inside another. We need to evaluate the inner function first and then use that value to evaluate the outer function.

Example: If \( f(x) = 2x + 3 \) and \( g(x) = x^2 \) are functions then how do you \( f(g(4)) \)?

Solution: 

Step `1`: Find \( g(4) \). Plug 4 into \( g(x) \), so \( g(4) = 4^2 = 16 \).

Step `2`: Use this result as input for \( f(x) \). So, \( f(g(4)) = f(16) \).

Plug `16` into \( f(x) \), so \( f(16) = 2(16) + 3 = 32 + 3 = 35 \).

NOTE: Need this kind of image with `x=4`, accordingly all values will get changed. 

Order of Composite Functions

When you're dealing with composite functions like \( f(g(x)) \) or \( g(f(x)) \), remember the order matters.

For \( f(g(x)) \), first find \( g(x) \) and then put it into \( f(x) \). For \( g(f(x)) \), it's the reverse.

Example: \( f(x) = x + 2 \) and \( g(x) = (2x)^2 \) are `2` functions. Find \( f(g(3)) \) and \( g(f(3)) \).

Solution:

For \( f(g(3)) \)

Step `1`: First find \( g(3) \). Plug `3` into \( g(x) \), so \( g(3) = (2(3))^2 = 36 \).

Step `2`: Plug `36` into \( f(x) \), so \( f(g(3)) = f(36) = 36 + 2 = 38 \).

For \( g(f(3)) \)

Step `1`: First find \( f(3) \). So \( f(3) = 3 + 2 = 5 \).

Step `2`: Now, put `5` into \( g(x) \), so \( g(5) = (2(5))^2 = 100 \).

Note the the value of \( f(g(3)) \) is different from the value of \( g(f(3)) \).

How to Find `f` of `g` of `x` From Graph

When we're given graphs instead of algebraic equations for functions, we can still find the composite function \( f(g(x)) \) by following a simple process.

`1`. First, we locate the value of \( g(a) \) on the graph of \( g(x) \) to find \( g(a) \). 

`2`. Then, we find the corresponding value of \( f(x) \) at \( x = g(a) \) on the graph of \( f(x) \) to determine \( f(g(a)) \).

Example: Find \( f(g(-2)) \) from the graph given below. 

Solution: 

`1`. As per the graph of `g(x)`, when `x = -2`, `y = 2`. Hence locate the point \( (-2, 2) \) on the graph of \( g(x) \) as \( g(-2) = 2 \), i.e. `2` being the `y`-coordinate on the graph of `g(x)` at `x = -2`.

`2`. Then find the corresponding value on the graph of \( f(x) \) at \( x = 2 \) to find \( f(g(-2)) \),  i.e., the `y`-coordinate on the graph of `f(x)` at `x = 2` which is `4`. Hence, \( f(g(-2)) = 4\).

This approach simplifies finding composite functions even when only graphical information is available.

How to find `f` of `g` of `x` From Table

When you're given tables representing functions \( f(x) \) and \( g(x) \), you can easily find the value of \( f(g(x)) \) or \( g(f(x)) \) at a specific point. 

Example: Calculate \( g(f(3)) \).

Solution:

To find \( g(f(3)) \):

`1`. \( f(3) = 5 \) (from the first table)

`2`. \( g(5) = 17 \) (from the second table)

`3`. Therefore, \( g(f(3)) = g(5) = 17 \).

Self Composition

When a function is combined with itself, it's called a self-composite function. For any function \( f(x) \), its self-composition is written as \( f(f(x)) \) or \( (f∘f)(x) \). 

Example: Let's consider the function \( f(x) = 2x + 1 \). We'll find \( (f∘f)(x) \), which means we're composing the function \( f \) with itself.

Solution:

Given: \( f(x) = 2x + 1 \)

To find \( (f∘f)(x) \), we substitute \( f(x) \) into \( f(f(x)) \):

\( (f∘f)(x) = f(f(x)) \)

\( = f(2x + 1) \)

Now, applying the function \( f \) again to \( 2x + 1 \):

\( (f∘f)(x) = 2(2x + 1) + 1 \)

\( = 4x + 2 + 1 \)

\( = 4x + 3 \)

So, \( (f∘f)(x) = 4x + 3 \).

Domain and Range of `f` of `g` of `x`

The domain of a function \( y = f(x) \) is all the \( x \) values where it's defined, meaning all the different inputs you can use. The range is all the \( y \)-values that the function can give you, which are all the different outputs.

If we have two functions, let's call them \( g \) and \( f \), where \( g \) goes from set \( A \) to set \( B \), and \( f \) goes from set \( B \) to set \( C \), then when we combine them \( f(g(x)) \), we create a new function. For most of the cases, this new function's domain will be the same as \( g \)'s domain, and its range will be the same as \( f \)'s range. However, there might be cases where this doesn't hold true. Let's learn how to figure out the domain and range of \( f(g(x)) \).

Domain of Composite Functions

If we think of \( g(x) \) as the inner function and \( f(x) \) as the outer function, then the domain of the combined function \( f(g(x)) \) is just the same as the domain of \( g(x) \). This is the set of numbers we can put into \( g(x) \) to get some result. 

To find this domain of \( f(g(x)) \):

`1`. First, we figure out the domain of the inner function, \( g(x) \) and represent it as a set of numbers `P`. 

`2`. Then, we see what numbers we get when we plug those numbers into \( g(x) \). These numbers become the domain of the composed function \( f(g(x)) \) and represent this set of numbers as `Q`. 

`3`. The intersection of the set of numbers `(P ∩ Q)` gives us the domain of the whole combined function `f(g(x))`.

Range of Composite Functions

The range of the combined function doesn't change much because of the inner function \( g(x) \). It mostly depends on the outer function \( f(x) \), so we can find it using the usual methods.

We can understand this better with an example.

Example: Find domain and range of \( f(x) = 2x + 3 \) and \( g(x) = x^3 \).

Solution:

For \( f(g(x)) \), first, we check the inner function, \( g(x) \), and find its domain, which includes all real numbers.

Then, \( f(g(x)) = f(x^3) = 2(x^3) + 3 \).

The domain of \( 2(x^3) + 3 \) is still all real numbers.

So, the domain of \( f(g(x)) \) is also all real numbers.

And the range of \( f(g(x)) \) is the range of \( 2(x^3) + 3 \).

In this case, the range is also all real numbers.the range i.e. `(-∞,∞)`.

Properties of Composite Functions

Associative Property:

This property says that when you combine three functions \( f \), \( g \), and \( h \), it doesn't matter how you group them. Whether you first combine \( f \) with \( g \), and then with \( h \), or you first combine \( g \) with \( h \), and then with \( f \), you'll end up with the same result.

\((f \circ g) \circ h = f \circ (g \circ h)\)

Commutative Property:

This property states that if you can switch the order of two functions and still get the same result, those functions are commutative.If \( f \) combined with \( g \) is the same as \( g \) combined with \( f \), then \( f \) and \( g \) are commutative.

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

However, please note that for most of the functions \((f \circ g)\) is not equal to \((f \circ g)\).

Identity Property:

Every function has an identity function \( I(x) = x \). When you combine any function with the identity function, you get back the original function.It's like multiplying a number by `1`.

\( f \circ I = I \circ f = f \)

Inverse Property:

If a function \( f \) has an inverse function \( f^{-1} \), combining them will give us identity function.

\((f \circ f^{-1}) = I = (f^{-1} \circ f)\)

These properties help us understand how functions combine and behave under composition.

Solved Examples

Example `1`: Let \( f(x) = 3x + 1 \) and \( g(x) = x^2 \). Find \( f(g(2)) \).

Solution:

\( g(2) = 2^2 = 4 \).

Substitute \( g(2) \) into \( f(x) \): \( f(g(2)) = 3(4) + 1 = 12 + 1 = 13 \).

Example `2`: Let \( f(x) = \sqrt{x + 2} \).Find \( f(f(4)) \).

Solution: 

\( f(4) \): \( f(4) = \sqrt{4 + 2} = \sqrt{6} \).

Substitute \( f(4) \) into \( f(x) \): \( f(f(4)) = \sqrt{\sqrt{6} + 2} \).

Example `3`: Let \( f(x) = 2x \) and \( g(x) = x^2 + 1 \).Find \( g(f(3)) \).

Solution: 

\( f(3) \): \( f(3) = 2(3) = 6 \).

Substitute \( f(3) \) into \( g(x) \): \( g(f(3)) = (6)^2 + 1 = 36 + 1 = 37 \).

Example `4`: Let \( f(x) = \sqrt{x} \) and \( g(x) = x^2 - 4 \).Find \( f(g(x)) \) and determine its domain and range.

Solution:

`1`. First, find \( g(x) \): \( g(x) = x^2 - 4 \).

`2`. Now, substitute \( g(x) \) into \( f(x) \): \( f(g(x)) = \sqrt{x^2 - 4} \).

`3`. Domain of \( f(g(x)) \): Since \( f(g(x)) \) involves a square root, the expression under the square root must be non-negative. So, \( x^2 - 4 \geq 0 \). Solving this inequality gives \( x \geq -2 \) and \( x \leq 2 \). Therefore, the domain of \( f(g(x)) \) is \( -2 \leq x \leq 2 \).

`4`. Range of \( f(g(x)) \): For any real number \( x \), \( x^2 - 4 \) will always be non-negative. Therefore, the square root of \( x^2 - 4 \) will always be real. Hence, the range of \( f(g(x)) \) is all real numbers \( \geq 0 \).

Example `5`: Let \( f(x) = x^3 \) and \( g(x) = \sqrt[3]{x} \).Determine whether \( f(g(x)) \) and \( g(f(x)) \) are equal.

Solution:

Composite Function \( f(g(x)) \):

Substitute \( g(x) \) into \( f(x) \): \( f(g(x)) = (g(x))^3 = (\sqrt[3]{x})^3 = x \).

Composite Function \( g(f(x)) \):

Substitute \( f(x) \) into \( g(x) \): \( g(f(x)) = \sqrt[3]{f(x)} = \sqrt[3]{x^3} = x \).

Since both \( f(g(x)) \) and \( g(f(x)) \) simplify to \( x \), they are equal.

This indicates that the given functions \( f(x) \) and \( g(x) \) are inverses of each other.

Practice Problems

Q`1`. Let \( f(x) = 2x + 1 \) and \( g(x) = x^2 - 3 \). Find \( f(g(x)) \).

1. \( 2x^2 - 5 \)
2. \( 4x^2 + 2x - 1 \)
3. \( x^4 - 6x^2 + 4x - 2 \)
4. \( 2x^4 - 5x^2 - 6 \)

Answer: a

Q`2`. Consider \( f(x) = \sqrt{x + 2} \) and \( g(x) = 3x - 4 \). Determine \( f(g(x)) \).

1. \( \sqrt{3x - 2} \)
2. \( \sqrt{9x^2 - 24x + 16} \)
3. \( 3\sqrt{x - 4} \)
4. \( \sqrt{9x - 12} \)

Answer: a

Q`3`. Suppose \( f(x) = x^2 \) and \( g(x) = 2x + 3 \). Find \( f(g(x)) \).

1. \( (2x + 3)^2 \)
2. \( 4x^2 - 12x + 9 \)
3. \( 4x^2 + 6x + 9 \)
4. \( 2x^2 + 3 \)

Answer: a

Q`4`. Let \( f(x) = \frac{1}{x} \) and \( g(x) = x^2 + 5 \). Determine \( f(g(x)) \).

1. \( \frac{1}{x^2 + 5} \)
2. \( \frac{x^2 + 5}{x} \)
3. \( x^2 + 5x \)
4. \( \frac{1}{x^2} + 5 \)

Answer: a

Q`5`. Assume \( f(x) = e^x \) and \( g(x) = 2x - 1 \). Find \( f(g(x)) \).

1. \( e^{2x - 1} \)
2. \( e^{2x} - 1 \)
3. \( e^{x^2 - 1} \)
4. \( e^{4x - 2} \)

Answer: a

Frequently Asked Questions

Q`1`. What is the composition of functions?

Answer: "`F` of `G` of `X`," denoted as \( f(g(x)) \), represents a composite function. This function arises by substituting \( g(x) \) into \( f(x) \), effectively replacing the variable \( x \) in \( f(x) \) with \( g(x) \).

Q`2`. What distinguishes \( f(g(x)) \) from \( g(f(x)) \)?

Answer: \( f(g(x)) \), written as \( f(g(x)) \), represents a composite function. It is derived by substituting \( g(x) \) into \( f(x) \), effectively replacing the variable \( x \) in \( f(x) \) with \( g(x) \).

On the other hand, \( g(f(x)) \), denoted as \( g(f(x)) \), also represents a composite function. It is formed by substituting \( f(x) \) into \( g(x) \), replacing the variable \( x \) in \( g(x) \) with \( f(x) \).

For instance, consider \( f(x) = x^2 \) and \( g(x) = \sin(x) \):

\( f(g(x)) \) yields \( f(\sin(x)) = (\sin(x))^2 = \sin^2(x) \).

\( g(f(x)) \) results in \( g(x^2) = \sin(x^2) \).

Q`3`. What is the significance of the order of composition in functions?

Answer: Yes, the sequence matters a lot in composite functions because \( f(g(x)) \) might or might not be the same as \( g(f(x)) \). This relies on the functions \( f(x) \) and \( g(x) \).

Q`4`. How do you figure out the domain of a composite function?

Answer: To find the domain of a composite function, we look at the values we can put into the composite function. It's like finding where two sets of numbers overlap: the domain of the inner function and the domain of the function after they're combined.

Q`5`. How can we determine the range of a composite function?

Answer: The range of a composite function is simply the set of all possible output values generated by that composite function. Importantly, it doesn't rely on the specifics of the inner function used within the composite function.

Q`6`. How does the composition of functions relate to real-world scenarios?

Answer: The composition of functions reflects how different actions or processes are interconnected in real-world scenarios, such as sequential operations in manufacturing or compound interest calculations in finance. Understanding this concept aids in modeling complex systems, analyzing processes, or solving engineering problems.