Functions: Language, Inverses, and Composition Notes

Fundamental Language of Functions

Understanding Mappings

A mapping is a process that takes an 'input' from one set of values to an 'output' in another set. There are four types of mappings:

One-one: One 'input' value maps to exactly one 'output' value.
Many-one: Multiple 'input' values map to a single 'output' value.
One-many: One 'input' value maps to multiple 'output' values.
Many-many: Multiple 'input' values map to multiple 'output' values.

Definition of a Function

A function is a specific type of mapping where every 'input' value maps to a single 'output'. Consequently, only one-one and many-one mappings qualify as functions.

Function Notation

Functions are identified using specific notation styles:

Standard Notation: $f(x)$ , $g(x)$ , etc. For example, $f(x) = x^2 - 3x + 2$ . This is pronounced as 'f of x'.
Mapping Notation: $f : x \rightarrow x^2 - 3x + 2$ . This is pronounced as 'the function f maps x to x-squared minus three x plus two'.

Mechanics of Function Operation

A function consists of an input ( $x$ ) and an output ( $f(x)$ or $y$ ). Whatever value or expression replaces $x$ inside the bracket of the function label must replace the $x$ everywhere it appears in the function's definition.

Calculating Outputs from Inputs

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

If the input is $3$ , the output is $f(3) = 2 \times 3 + 1 = 7$ .
If the input is $-4$ , the output is $f(-4) = 2 \times (-4) + 1 = -7$ .
If the input is a variable $a$ , the output is $f(a) = 2a + 1$ .

Calculating Inputs from Outputs

If the output is known, an equation can be solved to find the input. Given $f(x) = 2x + 1$ , if $f(x) = 15$ :

$2x + 1 = 15$
$2x = 14$
$x = 7$

Domain and Range

The Domain

The domain is the set of values allowed to be used as 'inputs'.
A function is not fully defined until its domain is stated.
If a domain is not explicitly provided, it is assumed to be the largest possible set of valid values. For example, the largest set of possible values for $f(x) = \text{x}$ is $x \ge 0$ .
Restrictions applied to the domain can transform a many-one function into a one-one function.

The Range

The range is the set of all possible 'output' values.
The values available in the range depend entirely on the defined domain.

Determining the Range

To find the range from a given domain:

Calculate individual output values for each input in the domain.
Analyze the shape or pattern of the function's graph.
Visualize outputs as $y$ -coordinates. For instance, for $f(x) = x^2$ with a domain of all real numbers, the range is $f(x) \ge 0$ because all $y$ -coordinates on the graph are greater than or equal to zero.

The Modulus Function

The modulus function, also known as the absolute value, ensures any 'input' results in a positive 'output'. It is denoted by vertical lines, such as $|x|$ .

Examples: $|7| = 7$ and $|-7| = 7$ .
If $f(x) \ge 0$ , then $|f(x)| = f(x)$ .
If $f(x) < 0$ , then $|f(x)| = -f(x)$ .

Inverse Functions

Definition and Notation

An inverse function reverses the operations of the original function in the opposite order.

Example: If a function doubles a number and adds $1$ , the inverse subtracts $1$ and halves the result.
Notation: $f^{-1}(x) = \dots$ or $f^{-1} : x \rightarrow \dots$

Finding an Inverse Function (The Step-by-Step Method)

To find the inverse of a function, such as $f(x) = 2x + 1$ :

Write the function in terms of $y$ : $y = 2x + 1$ .
Swap the $x$ and $y$ variables: $x = 2y + 1$ .
Rearrange the equation to make $y$ the subject again:
- $x - 1 = 2y$
- $y = \frac{x - 1}{2}$
Rewrite using inverse notation: $f^{-1}(x) = \frac{x - 1}{2}$ .

Note: Swapping $x$ and $y$ is a specific technique for finding inverses and should not be used in other mathematical contexts.

Requirements and Relationships

Condition for Existence: An inverse function $f^{-1}(x)$ only exists if the original function $f(x)$ is one-to-one. This ensures that every output corresponds back to exactly one unique input.
Undoing a Function: If $f(3) = 10$ , then $f^{-1}(10) = 3$ .
Composition with Inverse: Applying a function and its inverse sequentially returns the original value: $ff^{-1}(x) = f^{-1}f(x) = x$ .
Solving Equations: If $f(x) = 2^x$ and you need to solve $f^{-1}(x) = 5$ , you can take the function of both sides: $ff^{-1}(x) = f(5)$ , leading to $x = f(5) = 2^5 = 32$ .

Domain and Range of Inverses

There is an inverse relationship between the domains and ranges of a function and its inverse:

The range of a function is the domain of its inverse function.
The domain of a function is the range of its inverse function.

Graphs of Inverse Functions

The graph of an inverse function $y = f^{-1}(x)$ is a reflection of the graph of the function $y = f(x)$ in the line $y = x$ .

Key features like intercepts, turning points, and asymptotes are all reflected.
Example: The point $(4, 0)$ on $f(x)$ reflects to $(0, 4)$ on $f^{-1}(x)$ .
Example: A horizontal asymptote $y = 4$ on $f(x)$ reflects to a vertical asymptote $x = 4$ on $f^{-1}(x)$ .
Restrictions on the domain of the original function must be maintained in the sketch.
The intersection point of $f(x)$ and $f^{-1}(x)$ always lies on the line $y = x$ . Therefore, you can find the intersection by solving $f(x) = x$ .

Composite Functions

Definition and Notation

A composite function involves applying one function to the result of another function. The 'output' of the first function becomes the 'input' of the second. This is often referred to as a "function-of-a-function".

Notation: $fg(x)$ or $f(g(x))$ . Both mean "f of g of x".

Working with Composition

Order of Operations: The order is critical. In the expression $fg(x)$ , you apply the function closest to the variable first. Thus, apply $g$ to $x$ first, then apply $f$ to the result.
Non-Commutativity: Generally, $fg(x) \neq gf(x)$ .
Self-Composition: The notation $f^2(x)$ represents $ff(x)$ .
Trigonometric Exception: For trigonometric functions, $\sin^2(x)$ refers to $(\sin(x))^2$ , not $\sin(\sin(x))$ .

Domain and Range of Composite Functions

The domain of the composite function $fg(x)$ originates from the domain of the first function applied, which is $g(x)$ .
Compatibility Condition: For $fg(x)$ to exist, the range of $g(x)$ must be contained within the domain of $f(x)$ . If the range of $g(x)$ falls outside the domain of $f(x)$ , the composite function does not exist for those values.
The range of $fg(x)$ is determined by the output of $f(x)$ based on the restricted inputs provided by $g(x)$ .

Worked Examples

Function Evaluation and Substitution

Given $f(x) = 3x^2 - 2x + 1$ :

To find $f(7)$ : $f(7) = 3(7)^2 - 2(7) + 1 = 3(49) - 14 + 1 = 147 - 14 + 1 = 134$
To find $f(x + 3)$ : Substitute $(x + 3)$ for every $x$ : $f(x + 3) = 3(x + 3)^2 - 2(x + 3) + 1$ $f(x + 3) = 3(x^2 + 6x + 9) - 2x - 6 + 1$ $f(x + 3) = 3x^2 + 18x + 27 - 2x - 5$ $f(x + 3) = 3x^2 + 16x + 22$

Solving for Input

Given $g : x \rightarrow 3x - 4$ , find $x$ for which $g(x) = -16$ :

Set the expression equal to $-16$ : $3x - 4 = -16$ $3x = -12$ $x = -4$

Range and Domain Changes

Given $f(x) = (x - 3)^2$ for all values of $x$ :

Range: Since the output is a squared value, it must be zero or positive. Range: $f(x) \ge 0$ .
Modified Domain: If the domain is changed to $x > 5$ , the range changes accordingly. Substitute the boundary: $(5 - 3)^2 = 2^2 = 4$ . Since $x > 5$ , the range is $f(x) > 4$ .

Inverse Calculations

Find the inverse of $f(x) = 5 - 3x$ :

Let $y = 5 - 3x$
Swap variables: $x = 5 - 3y$
Solve for $y$ : $x + 3y = 5$ $3y = 5 - x$ $y = \frac{5 - x}{3}$
Result: $f^{-1}(x) = \frac{5 - x}{3}$

Domain and Range of Inverses

Given $f(x) = \sqrt{3x - 2}, x > \frac{2}{3}$ :

Range of $f(x)$ : The square root results in values $> 0$ . Range: $f > 0$ .
Domain of $f^{-1}(x)$ : This is the range of $f(x)$ . Domain: $x > 0$ .
Range of $f^{-1}(x)$ : This is the domain of $f(x)$ . Range: $f^{-1} > \frac{2}{3}$ .

Composite Function Existence

Given $f(x) = \frac{1}{x}, 0 < x < 1$ and $g(x) = x^2, x > 1$ :

Range of $f(x)$ : For $0 < x < 1$ , $f > 1$ .
Range of $g(x)$ : For $x > 1$ , $g > 1$ .
Does $fg(x)$ exist?: No. The range of $g(x)$ is $g > 1$ , but the domain of $f(x)$ is $0 < x < 1$ . The range of the first function applied ( $g$ ) is not contained within the domain of the second ( $f$ ).
Range of $gf(x)$ : Here, $f$ is applied first. Its range is $f > 1$ , which perfectly matches the domain of $g$ ( $x > 1$ ). The range of $gf(x)$ is thus identical to the range of $g$ , which is $gf > 1$ .