Advanced Function Composition and Inverse Functions Notes

The Concept of Function Composition

Function composition is the process of "smashing" two operations together. It is a mathematical process where the output of one function becomes the input of another.
We are primarily concerned with the transition from the initial input value to the final output value of the chain. This relationship defines the composition function.
The order of operations is critical. Changing the order in which functions are applied often results in entirely different outputs, meaning function composition is generally not commutative.

Evaluating Composition Functions Numerically

Example 1: $f(x) = 2x - 1$ and $g(x) = x + 4$
- Evaluation of $g(f(-2))$ :
  - Step 1: Input $-2$ into $f(x)$ . Calculation: $2 \times (-2) - 1 = -5$ .
  - Step 2: Take the output $-5$ and input it into $g(x)$ . Calculation: $-5 + 4 = -1$ .
  - Result: $g(f(-2)) = -1$ .
- Evaluation of $f(g(3))$ :
  - Step 1: Input $3$ into $g(x)$ . Calculation: $3 + 4 = 7$ .
  - Step 2: Input $7$ into $f(x)$ . Calculation: $2 \times 7 - 1 = 13$ .
  - Result: $f(g(3)) = 13$ .
- Evaluation of $g(f(3))$ :
  - Step 1: Input $3$ into $f(x)$ . Calculation: $2 \times 3 - 1 = 5$ .
  - Step 2: Input $5$ into $g(x)$ . Calculation: $5 + 4 = 9$ .
  - Result: $g(f(3)) = 9$ .
- Comparative Observations:
  - $f(g(3)) = 13$ while $g(f(3)) = 9$ . This demonstrates that switching the order of functions ( $f$ of $g$ vs $g$ of $f$ ) produces different answers almost all the time.

Constructing the Algebraic Composition Function

Instead of performing a two-stage numerical calculation, we can build a single function that represents the entire composition.
Algorithm for Construction:
1. Start with the outer function.
2. Replace the variable in the outer function with sets of parentheses.
3. Insert the expression for the inner function inside those parentheses.
4. Simplify the resulting algebraic expression.
Continuing Example 1 ( $f(x) = 2x - 1$ , $g(x) = x + 4$ ):
- Building $g(f(x))$ :
  - Outer function $g(x)$ is $( \text{something} ) + 4$ .
  - Insert $f(x) = 2x - 1$ : $(2x - 1) + 4$ .
  - Simplify: $2x + 3$ .
  - Verification: $-2 \rightarrow 2(-2) + 3 = -1$ (Matches previous calculation).
- Building $f(g(x))$ :
  - Outer function $f(x)$ is $2 \times ( \text{something} ) - 1$ .
  - Insert $g(x) = x + 4$ : $2(x + 4) - 1$ .
  - Simplify: Distribute multiplication: $2x + 8 - 1 = 2x + 7$ .
  - Note the difference: $2x + 3$ is not the same as $2x + 7$ .

Terminology and Notation of Composition

Functions are identified by specific notation to determine the correct order:
- $f \text{ composed with } g$ is written $(f \text{ o } g)(x)$ or $f(g(x))$ . This is often nicknamed the "fog" function.
- $g \text{ composed with } f$ is written $(g \text{ o } f)(x)$ or $g(f(x))$ . This is nicknamed the "gof" function.
Self-Composition: A function can be composed with itself, such as $g(g(x))$ . This is an iterative process, not a squaring process. For example, if $g(x) = 2x - 3$ , then $g(g(x)) = 2(2x - 3) - 3 = 4x - 6 - 3 = 4x - 9$ .

Interpreting Composition in Context

Composition allows us to bridge two related rates or data sets.
Example: Exercise and Calories
- $C(s)$ provides the number of calories burned when doing $s$ sit-ups.
- $s(t)$ provides the number of sit-ups doable in $t$ minutes.
- $C(s(3))$ means calculating the number of sit-ups done in $3$ minutes, then calculating the calories burned for that quantity of sit-ups. Effectively, the composition relates calories burned ( $C$ ) directly to time ( $t$ ).
Example: Driving and Fuel
- $f(x)$ is miles driven in $x$ hours.
- $g(y)$ is gallons of gas used in driving $y$ miles.
- The composition $g(f(x))$ is meaningful because the output of $f$ (miles) matches the required input for $g$ (miles). This calculates gallons of gas per hours of driving.
- The composition $f(g(y))$ is questionable because the input for $f$ expects time, but it would receive the output of $g$ (gallons of gas).

Evaluating Compositions from Graphs

To evaluate $f(g(x))$ from a graph:
1. Locate the specified $x$ value on the horizontal axis of the $g(x)$ graph.
2. Move vertically to the curve and then horizontally to the $y$ axis to find the output value.
3. Take that output value and use it as the $x$ value on the $f(x)$ graph.
4. Repeat the process on the second graph to find the final answer.
Specific Case: If $g(1) = 3$ , then to find $f(g(1))$ , you look for the point on the $f$ graph where the input is $3$ . If the point on that curve is $(3, 3)$ , the final result is $3$ .

Domain Restrictions for Composite Functions

Finding the domain of a composition $g(f(x))$ involves addressing two distinct problems:

Direct Restrictions of the Inner Function: Any value of $x$ that causes the inner function $f(x)$ to be undefined (e.g., division by zero, square root of a negative) is automatically restricted from the composition.
Output Conflicts: Any value of $x$ that results in an output from $f(x)$ which is restricted in the outer function $g(x)$ . Essentially, the output of the first stage cannot be the "problem child" for the second stage.

Algebraic Process for Domain Restrictions:
- Identify restrictions for $f(x)$ (e.g., $x - 2 = 0 \rightarrow x = 2$ ).
- Identify restrictions for $g(x)$ (e.g., $x + 2 = 0 \rightarrow x = -2$ ).
- Set the inner function equal to the restriction of the outer function ( $f(x) = \text{restriction of } g$ ) and solve for $x$ . These $x$ values are also excluded from the domain.

Algebraic Methods for Squaring Binomials

When simplifying compositions, students often encounter squared binomials (e.g., $(3x + 2)^2$ ). Three methods for solving these are:

Identity/Pattern Recognition: $(a + b)^2 = a^2 + 2ab + b^2$ .
- Example: $(3x + 2)^2 \rightarrow (3x)^2 + 2(3x)(2) + (2)^2 = 9x^2 + 12x + 4$ .
FOIL (First, Outside, Inside, Last):
- $(3x + 2)(3x + 2) \rightarrow 3x \times 3x + 3x \times 2 + 2 \times 3x + 2 \times 2 = 9x^2 + 6x + 6x + 4 = 9x^2 + 12x + 4$ .
Area Model (Box Method):
- Create a $2 \times 2$ grid with components of the binomial on the sides. Multiply each intersecting square ( $3x \times 3x, 3x \times 2, 2 \times 3x, 2 \times 2$ ) and add them together.

Introduction to Inverse Functions

Definition: An inverse function reverses the operation of the original function. If a function takes $x$ to $y$ , the inverse takes $y$ back to $x$ .
One-to-One (1:1) Property: Only functions that are one-to-one have inverses. A function is one-to-one if for every output there is a unique input.
Horizontal Line Test: To check if a function is one-to-one, draw a horizontal line through the graph. If it crosses the graph more than once, the function is not one-to-one and does not have a unique inverse.
- Example: $f(x) = x^2$ is not one-to-one because a horizontal line crosses it twice (e.g., at $x = -2$ and $x = 2$ ).

Properties and Verification of Inverses

Ordered Pairs: If $(a, b)$ is on the graph of $f(x)$ , then $(b, a)$ is on the graph of $f^{-1}(x)$ .
Graphical Reflection: The graph of an inverse function is a reflection of the original function's graph across the diagonal line $y = x$ .
Domain and Range: The domain of $f$ is the range of $f^{-1}$ , and the range of $f$ is the domain of $f^{-1}$ .
Algebraic Verification: Two functions are inverses of each other if and only if their composition results in $x$ : $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ .

Procedures for Finding Inverse Functions

To find the inverse of a function (e.g., $f(x) = \frac{3}{2}(x - 2) + 1$ ) algebraically:

Notation Swap: Replace $f(x)$ with $y$ .
Variable Swap: Literally swap the roles of $x$ and $y$ ( $x$ becomes $y$ , and $y$ becomes $x$ ).
Isolation: Solve the new equation for $y$ using algebraic operations (PEMDAS in reverse).
Final Form: Replace the final isolated $y$ with the notation $f^{-1}(x)$ .

Application - Temperature Conversion:
- Celsius from Fahrenheit: $C = \frac{5}{9}(F - 32)$ .
- To find Fahrenheit from Celsius, find the inverse function: $F = \frac{9}{5}C + 32$ .
- Example: $30^\text{o}C \rightarrow \frac{9}{5}(30) + 32 = 54 + 32 = 86^\text{o}F$ .