8.1 Function Notation, Transformations, Compositions & Inverses

Function Notation, Transformations, Compositions & Inverses

Core Advanced Algebra: Functions form the foundational language of advanced mathematics on the ACCUPLACER Advanced Algebra and Functions (AAF) assessment. Mastery requires analyzing input-output mappings, establishing analytical domain constraints, executing algebraic combinations and compositions, applying multi-step parent function transformations, and deriving inverse functions algebraically and graphically.

---

Function Fundamentals, Relations & The Vertical Line Test

A relation is any set of ordered pairs $(x, y)$ establishing a relationship between an input set and an output set. A function is a specialized relation with a strict rule: each input element $x$ in the domain maps to exactly one output value $y$ in the range.

  FUNCTION (Valid Mapping)               NOT A FUNCTION (Invalid Mapping)
  Domain (Inputs)    Range (Outputs)    Domain (Inputs)    Range (Outputs)
     [ 1 ] ─────────> [ 4 ]                [ 1 ] ─────────> [ 4 ]
     [ 2 ] ─────────> [ 5 ]                [ 2 ] ──┬──────> [ 5 ]
     [ 3 ] ─────────> [ 6 ]                        └──────> [ 6 ] (One input maps
  (Each input has exactly one output)                              to two outputs!)

The Vertical Line Test (VLT)

Geometrically, a curve plotted in the Cartesian coordinate plane represents $y$ as a function of $x$ if and only if no vertical line intersects the curve at more than one point.

• If a vertical line $x = c$ intersects a curve at two or more distinct points $(c, y_1)$ and $(c, y_2)$ where $y_1 \neq y_2$, the single input $c$ produces multiple outputs, violating the definition of a function.
• Example: The circle $x^2 + y^2 = 25$ fails the vertical line test (at $x = 0$, $y = 5$ and $y = -5$). However, the upper semicircle $y = \sqrt{25 - x^2}$ passes the vertical line test and is a valid function.

---

Determining Domain & Range Restrictions

The domain of a real-valued function $f(x)$ is the complete set of all real input values $x$ for which the function produces a real output. The range is the resulting set of all real output values $y = f(x)$.

The Two Universal Algebraic Domain Restrictions

When working in the real number system ($\mathbb{R}$), domain restrictions arise primarily from two algebraic conditions:

1. Rational Denominator Constraint (Division by Zero):
   • Division by zero is mathematically undefined.
   • For any rational algebraic expression $f(x) = \frac{p(x)}{q(x)}$, set the denominator equal to zero ($q(x) = 0$) and exclude those roots from the domain.
2. Even Radical Constraint (Negative Radicands):
   • Even-indexed roots ($\sqrt{\phantom{x}}, \sqrt[4]{\phantom{x}}, \sqrt[6]{\phantom{x}}$) of negative numbers produce imaginary numbers and are excluded from real-valued domains.
   • For $f(x) = \sqrt[2n]{g(x)}$, enforce the inequality $g(x) \ge 0$.
   • Note: Odd-indexed roots ($\sqrt[3]{\phantom{x}}, \sqrt[5]{\phantom{x}}$) are defined for all real numbers (e.g., $\sqrt[3]{-8} = -2$), so odd radicals introduce no domain restrictions.
3. Combined Denominator and Radical Constraint:
   • If an even radical appears in a denominator, $f(x) = \frac{1}{\sqrt{g(x)}}$, the radicand cannot be negative AND cannot equal zero, establishing the strict inequality $g(x) > 0$.

Worked Example: Determining Analytical Domain

Find the domain of the function $f(x) = \frac{\sqrt{2x + 10}}{x^2 - 2x - 15}$ in interval notation.

• Step 1 (Radicand constraint): The numerator contains a square root, requiring the radicand to be non-negative: $$2x + 10 \ge 0 \implies 2x \ge -10 \implies x \ge -5$$
• Step 2 (Denominator constraint): The denominator cannot equal zero: $$x^2 - 2x - 15 \neq 0 \implies (x - 5)(x + 3) \neq 0 \implies x \neq 5 \quad \text{and} \quad x \neq -3$$
• Step 3 (Intersection of conditions): Combine $x \ge -5$ with the exclusions $x \neq -3$ and $x \neq 5$: $$\text{Domain} = [-5, -3) \cup (-3, 5) \cup (5, \infty)$$

---

Function Evaluation & The Algebra of Functions

Function evaluation involves substituting a specified numerical value or algebraic expression into every occurrence of the independent variable $x$.

Evaluating Algebraic Inputs & The Difference Quotient

Given $f(x) = 3x^2 - 4x + 1$:

• $f(-2) = 3(-2)^2 - 4(-2) + 1 = 3(4) + 8 + 1 = 21$
• $f(a + 2) = 3(a + 2)^2 - 4(a + 2) + 1 = 3(a^2 + 4a + 4) - 4a - 8 + 1 = 3a^2 + 12a + 12 - 4a - 7 = 3a^2 + 8a + 5$

The Difference Quotient measures the average rate of change across an interval of width $h$:

$$\frac{f(x + h) - f(x)}{h} \quad (h \neq 0)$$ $$\frac{[3(x+h)^2 - 4(x+h) + 1] - [3x^2 - 4x + 1]}{h} = \frac{3x^2 + 6xh + 3h^2 - 4x - 4h + 1 - 3x^2 + 4x - 1}{h} = \frac{6xh + 3h^2 - 4h}{h} = 6x + 3h - 4$$

Operations on Functions

For functions $f(x)$ and $g(x)$, standard arithmetic operations produce new functions:

---

Composite Functions: Evaluation & Domain Restrictions

A composite function $(f \circ g)(x)$ applies the output of an inner function $g(x)$ as the direct input to an outer function $f(x)$:

$$(f \circ g)(x) = f(g(x)) \quad \text{("f of g of x")}$$

                        Composite Function Pipeline
  Input x  ───> [ Inner Function g(x) ] ───> Output g(x) ───> [ Outer Function f(u) ] ───> f(g(x))

Non-Commutative Property of Composition

In general, function composition is not commutative: $(f \circ g)(x) \neq (g \circ f)(x)$.

Determining the Domain of a Composite Function $(f \circ g)(x)$

The domain of $(f \circ g)(x)$ consists of all real numbers $x$ satisfying two strict criteria:

1. $x$ must be in the domain of the inner function $g(x)$.
2. The output $g(x)$ must be in the domain of the outer function $f(x)$.

Worked Example: Composite Function and Composite Domain

Let $f(x) = \frac{4}{x - 3}$ and $g(x) = \sqrt{x + 2}$. Find $(f \circ g)(x)$ and state its domain.

• Step 1 (Form composite equation): $$(f \circ g)(x) = f(g(x)) = f(\sqrt{x + 2}) = \frac{4}{\sqrt{x + 2} - 3}$$
• Step 2 (Inner function domain): For $g(x) = \sqrt{x + 2}$, the radicand must be non-negative: $x + 2 \ge 0 \implies x \ge -2$.
• Step 3 (Outer function restriction): For $f(u) = \frac{4}{u - 3}$, the denominator cannot equal zero: $u \neq 3 \implies g(x) \neq 3$. $$\sqrt{x + 2} \neq 3 \implies x + 2 \neq 9 \implies x \neq 7$$
• Step 4 (Combine restrictions): Combine $x \ge -2$ with $x \neq 7$: $$\text{Domain}(f \circ g) = [-2, 7) \cup (7, \infty)$$

---

Master Reference: Function Transformations

Transformations modify parent functions geometrically through shifts, scalings, and reflections. The generalized transformation format for any parent function $f(x)$ is:

$$g(x) = a \cdot f(b(x - h)) + k$$

Transformation Rules Reference Table

The Standard Order of Transformations

When applying multiple transformations to a parent function, execute operations in the following order:

1. Horizontal Shifts ($x - h$)
2. Horizontal Stretch / Compression / Reflection ($b \cdot x$ and negative sign inside)
3. Vertical Stretch / Compression / Reflection ($a \cdot f(u)$ and negative sign outside)
4. Vertical Shifts ($+ k$)

---

One-to-One Functions, Horizontal Line Test & Inverse Functions

A function $f$ is one-to-one (injective) if no two distinct inputs produce the same output: $f(x_1) = f(x_2) \implies x_1 = x_2$.

The Horizontal Line Test (HLT)

A function $f(x)$ possesses an inverse function $f^{-1}(x)$ if and only if no horizontal line intersects its graph at more than one point.

• Quadratic parent $f(x) = x^2$ fails the HLT (horizontal line $y = 4$ intersects at $x = -2$ and $x = 2$). It has no inverse over $(-\infty, \infty)$ unless its domain is restricted (e.g., $x \ge 0$).
• Cubic parent $f(x) = x^3$ passes the HLT and is one-to-one over its entire real domain.

Fundamental Properties of Inverse Functions

1. Cancellation Identity: $f(f^{-1}(x)) = x$ for all $x$ in the domain of $f^{-1}$, and $f^{-1}(f(x)) = x$ for all $x$ in the domain of $f$.
2. Domain-Range Reversal: $$\text{Domain}(f^{-1}) = \text{Range}(f) \quad \text{and} \quad \text{Range}(f^{-1}) = \text{Domain}(f)$$
3. Geometric Symmetry: The graph of $y = f^{-1}(x)$ is the reflection of the graph of $y = f(x)$ across the identity line $y = x$. Every coordinate pair $(a, b)$ on $f$ becomes $(b, a)$ on $f^{-1}$.

The 4-Step Algebraic Procedure for Finding $f^{-1}(x)$

1. Replace $f(x)$ with $y$.
2. Swap variables $x$ and $y$ (interchange every $x$ with $y$ and every $y$ with $x$).
3. Solve the resulting equation algebraically for $y$.
4. Replace $y$ with the formal inverse notation $f^{-1}(x)$.

Worked Example: Finding the Inverse of a Rational Function

Find the inverse function $f^{-1}(x)$ for $f(x) = \frac{3x + 2}{5x - 4}$ where $x \neq \frac{4}{5}$.

• Step 1 (Set $y = f(x)$): $y = \frac{3x + 2}{5x - 4}$
• Step 2 (Swap $x$ and $y$): $x = \frac{3y + 2}{5y - 4}$
• Step 3 (Clear denominator): Multiply both sides by $(5y - 4)$: $$x(5y - 4) = 3y + 2 \implies 5xy - 4x = 3y + 2$$
• Step 4 (Collect $y$-terms): Move all terms containing $y$ to the left and non-$y$ terms to the right: $$5xy - 3y = 4x + 2$$
• Step 5 (Factor and isolate $y$): $$y(5x - 3) = 4x + 2 \implies y = \frac{4x + 2}{5x - 3}$$
• Step 6 (Formal notation): $f^{-1}(x) = \frac{4x + 2}{5x - 3}$ with domain $x \neq \frac{3}{5}$.