Relations, Functions, Evaluation, and Function Notation

Relations and Functions

• Definition of an Ordered Pair: Points on a graph are expressed as ordered pairs $(x, y)$, where the first value represents the horizontal coordinate ($x$) and the second value represents the vertical coordinate ($y$).
• Tables of Values: A table listing $x$ and $y$ coordinates is a tool used to select specific points to plot and connect when graphing equations such as $y = 2x - 4$.
• Definition of a Relation: A relation is any set of ordered pairs that establishes a relationship between two variables.
• Definition of a Function: A function is a specialized type of relation in which every input value ($x$) corresponds to exactly one well-defined output value ($y$).
  • The Single-Output Rule: A single $x$ value cannot map to two or more different $y$ values.
  • Permissible Mappings: Multiple distinct $x$ values are permitted to map to the exact same $y$ value. This does not violate the definition of a function because knowing the $x$ value still determines a unique $y$ value without uncertainty.

Identifying Functions from Sets and Point Lists

• Mapping Diagram Method:

  • Elements of the domain ($x$ values) and range ($y$ values) are placed into distinct sets, with directed arrows showing the mappings.
  • Example 1a: Consider the relation with points $(0, 5)$, $(1, -3)$, $(4, 2)$, $(3, 0)$, and $(1, -5)$.
  • Domain set: $\{0, 1, 4, 3\}$
  • Range set: $\{5, -3, 2, 0, -5\}$
  • Mappings:
    • $0 \rightarrow 5$
    • $1 \rightarrow -3$
    • $4 \rightarrow 2$
    • $3 \rightarrow 0$
    • $1 \rightarrow -5$
  • Analysis: The single input $x = 1$ maps to two distinct outputs ($y = -3$ and $y = -5$).
  • Conclusion: This relation is not a function.
  • Example 1b: Consider the relation with points $(1, 2)$, $(2, 2)$, $(3, 2)$, $(4, 2)$, and $(5, 2)$.
  • Domain set: $\{1, 2, 3, 4, 5\}$
  • Range set: $\{2\}$
  • Mappings:
    • $1 \rightarrow 2$
    • $2 \rightarrow 2$
    • $3 \rightarrow 2$
    • $4 \rightarrow 2$
    • $5 \rightarrow 2$
  • Analysis: Each individual $x$ value points strictly to a single $y$ value ($2$).
  • Conclusion: This relation is a function.

• Direct Inspection of Duplicate Inputs:

  • If an input $x = 2$ yields two outputs $y = 5$ and $y = 0$, the relation fails to be a function due to output uncertainty.
  • If all $x$ values in a list of points are unique, the relation is guaranteed to be a function.

Graphical Analysis and the Vertical Line Test

• The Vertical Line Test (VLT): A graph represents a function if and only if no vertical line intersects the graph more than once.

  • If a vertical line intersects a graph at two or more points, those points share the exact same $x$ coordinate but have different $y$ coordinates, violating the definition of a function.
  • Threshold Criteria: The vertical line test only needs to be failed at a single location for the entire graph to be classified as non-functional.
  • Zero Intersections: A vertical line passing through a region where the graph does not exist intersects $0$ times. This does not violate the function definition because there is no output uncertainty.

• Graphical Examples:

  • Graph A: Possesses points $(0, 5)$ and $(0, -5)$. A vertical line drawn along $x = 0$ intersects both points (2 intersections). Not a function.
  • Graph B: Every vertical line intersects the graph at most once. Is a function.
  • Graph C: Passes the vertical line test at every point. Is a function.

Piecewise Functions and Point Continuity

• Graphical Symbols for Endpoints:

  • Solid Circle (Filled Circle): Indicates that the endpoint is included as an active point on the graph.
  • Open Circle: Indicates a boundary that the graph approaches infinitely closely, but the specific point itself is excluded from the graph.

• Analysis of Discontinuous/Piecewise Example:

  • Graph Features:
  • A upper-right segment starting with a solid circle at $(3, 5)$ and extending to the right.
  • A middle line segment stretching from $(0, 0)$ up to an open circle at $(3, 3)$.
  • VLT Application at $x = 3$:
  • The point $(3, 5)$ is included on the graph due to the solid circle.
  • The point $(3, 3)$ is excluded from the graph due to the open circle.
  • Therefore, a vertical line drawn at $x = 3$ intersects the graph at exactly one point, $(3, 5)$.
  • Conclusion: The relation passes the vertical line test across its entire domain and is a function.

• Real-World Applications of Piecewise Functions:

  • Income Tax Brackets: Tax policy dictates varying percentage rates depending on income tiers.
  • Payroll / Social Security Taxes: Social Security tax applies only up to a fixed income threshold (e.g., $\$160{,}000$), above which only Medicare tax applies, introducing structural breaks in financial formulas.
  • Financial Support Cliffs: Systems where earning an additional dollar results in a abrupt loss of financial assistance benefits.

Function Notation and Evaluation

• Function Notation: Expressed as $f(x)$, replacing the traditional dependent variable $y$.

  • $f(x)$ is read as "$f$ of $x$".
  • Important Clarification: The expression $f(x)$ represents function evaluation rule/formula, not multiplication of $f$ by $x$.
  • Function notation defines an input-output machine where inserting an input $x$ executes a specific algebraic formula to produce a single output $y$.

• Evaluating Quadratic Functions:

  • Given the function:     $f(x) = x^2 - 4x + 2$

  • Evaluating $f(3)$:     $f(3) = (3)^2 - 4(3) + 2$$f(3) = 9 - 12 + 2$$f(3) = -1$

  • The corresponding point on the graph is $(3, -1)$.

  • Evaluating $f(-2)$:

  • Enclosing negative values in parentheses is essential to preserve correct signs under order of operations.     $f(-2) = (-2)^2 - 4(-2) + 2$$f(-2) = 4 + 8 + 2$$f(-2) = 14$

  • The corresponding point on the graph is $(-2, 14)$.

  • Evaluating $f(0)$:     $f(0) = (0)^2 - 4(0) + 2$$f(0) = 0 - 0 + 2$$f(0) = 2$

  • Substitution of $0$ eliminates all terms containing $x$, leaving only the constant term.

Complex Substitutions and Algebraic Expansions

• Evaluating $f\left(\frac{6}{x}\right)$ for $f(x) = x^2 - 4x + 2$:

  • Substitute the entire quotient $\left(\frac{6}{x}\right)$ into every instance of $x$:     $f\left(\frac{6}{x}\right) = \left(\frac{6}{x}\right)^2 - 4\left(\frac{6}{x}\right) + 2$

  • Exponent Rule Application:     $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$$\left(\frac{6}{x}\right)^2 = \frac{6^2}{x^2} = \frac{36}{x^2}$

  • Fraction Multiplication Application:     $-4\left(\frac{6}{x}\right) = -\frac{4}{1} \times \frac{6}{x} = -\frac{24}{x}$

  • Final Simplified Form:     $f\left(\frac{6}{x}\right) = \frac{36}{x^2} - \frac{24}{x} + 2$

• Evaluating $f(x + h)$ for $f(x) = x^2 - 4x + 2$:

  • Substitute the binomial quantity $(x + h)$ into every instance of $x$:     $f(x + h) = (x + h)^2 - 4(x + h) + 2$

  • Expanding $(x + h)^2$:

  • Binomial expansion requires distribution across all terms: $(x + h)(x + h)$.

  • FOIL Method:       $\text{First}: x \times x = x^2$$\text{Outer}: x \times h = xh$$\text{Inner}: h \times x = hx$$\text{Last}: h \times h = h^2$$\text{Sum} = x^2 + 2xh + h^2$

  • Box Multiplication Method:

    • Construct a $2 \times 2$ grid with headers $x$ and $h$:
    • Top-left cell: $x \cdot x = x^2$
    • Top-right cell: $x \cdot h = xh$
    • Bottom-left cell: $h \cdot x = hx$
    • Bottom-right cell: $h \cdot h = h^2$
    • Summing all terms yields $x^2 + 2xh + h^2$

  • Distributing $-4$ across $(x + h)$:     $-4(x + h) = -4x - 4h$

  • Full Expansion:     $f(x + h) = x^2 + 2xh + h^2 - 4x - 4h + 2$

  • Calculus Application: This specific expansion form represents a key component of the difference quotient used to determine derivatives by definition.

Rational Functions Evaluation and Simplification

• Given Rational Function:   $g(t) = \frac{1 - 2t}{t + 4}$

• Evaluating $g(17)$:   $g(17) = \frac{1 - 2(17)}{17 + 4}$$g(17) = \frac{1 - 34}{21}$$g(17) = \frac{-33}{21}$

  • Reducing by Common Factor $3$:     $\frac{-33}{21} = \frac{3 \times (-11)}{3 \times 7} = -\frac{11}{7}$
  • Order of Operations Warning: A rational expression implies hidden parentheses around the entire numerator and the entire denominator: $\frac{(1 - 2t)}{(t + 4)}$.

• Evaluating $g\left(\frac{7}{5}\right)$:   $g\left(\frac{7}{5}\right) = \frac{1 - 2\left(\frac{7}{5}\right)}{\frac{7}{5} + 4} = \frac{1 - \frac{14}{5}}{\frac{7}{5} + 4}$

  • Numerator Common Denominator:     $1 - \frac{14}{5} = \frac{5}{5} - \frac{14}{5} = -\frac{9}{5}$

  • Denominator Common Denominator:     $\frac{7}{5} + 4 = \frac{7}{5} + \frac{20}{5} = \frac{27}{5}$

  • Division of Fractions (Reciprocal Multiplication):     $g\left(\frac{7}{5}\right) = \frac{-\frac{9}{5}}{\frac{27}{5}} = -\frac{9}{5} \times \frac{5}{27}$

  • Simplification:

  • The factor $5$ in the numerator and denominator cancels out.

  • Reducing $\frac{-9}{27}$ by factor $9$ yields $-\frac{1}{3}$.     $g\left(\frac{7}{5}\right) = -\frac{1}{3}$

Undefined Values and Algebraic Expressions in Rational Functions

• Evaluating $g(-4)$ for $g(t) = \frac{1 - 2t}{t + 4}$:   $g(-4) = \frac{1 - 2(-4)}{-4 + 4} = \frac{1 + 8}{0} = \frac{9}{0}$

  • Conclusion: Division by zero is mathematically impossible within real and complex numbers. Thus, $g(-4)$ is undefined.

• Evaluating $g(4 - x)$ for $g(t) = \frac{1 - 2t}{t + 4}$:   $g(4 - x) = \frac{1 - 2(4 - x)}{(4 - x) + 4}$

  • Numerator Expansion:     $1 - 2(4 - x) = 1 - 8 + 2x = 2x - 7$

  • Denominator Expansion:     $(4 - x) + 4 = 8 - x$

  • Final Expression:     $g(4 - x) = \frac{2x - 7}{8 - x}$

Reading Graphical Functions and Solving Equations

• Extracting Values from Graphs:

  • Given graph features:
  • When $x = -3$, the graph touches the horizontal axis ($y = 0$), so $f(-3) = 0$
  • When $x = 3$, the graph is at height $y = 4$, so $f(3) = 4$

• Solving $f(x) = 3$ Graphically:

  • This statement assigns $y = 3$ and requires finding the input $x$ that corresponds to that height.
  • Locating $y = 3$ on the graph shows an $x$ value of $0$.
  • Solution: $x = 0$.

• Structural Difference Between Evaluation and Solving:

  • Evaluating $f(\text{number})$: Produces at most one single output value due to the definition of a function.
  • Solving $f(x) = \text{number}$: Can yield multiple distinct $x$ values, a single $x$ value, or no $x$ values at all.

• Expressing Implicit Equations in Function Notation:

  • Given the implicit relation:     $3x^2 + 2y^3 = 432$

  • Step 1: Isolate $y^3$ by subtracting $3x^2$ from both sides:     $2y^3 = 432 - 3x^2$

  • Step 2: Divide both sides by $2$:     $y^3 = \frac{432 - 3x^2}{2}$

  • Step 3: Take the cube root of both sides:     $y = \sqrt[3]{\frac{432 - 3x^2}{2}}$

  • Note: Odd roots (like cube roots) do not produce a $\pm$ sign, whereas even roots require a $\pm$ sign.

  • Step 4: Rewrite using function notation:     $f(x) = \sqrt[3]{\frac{432 - 3x^2}{2}}$

Course Context and Overview of Function Types

• Course Purpose: College Algebra provides foundational preparation for Calculus, where derivatives and integrals are computed on functions.
• Primary Function Types Covered:
  • Quadratic Functions and Polynomial Functions: Functions containing non-negative integer powers of $x$ ($x^n$).
  • Rational Functions: Functions formed by dividing one polynomial by another polynomial.
  • Exponential Functions: Functions modeling growth or decay rates proportional to their current value, characterized by constant doubling times (e.g., compound interest, unconstrained bacterial growth).
  • Logarithmic Functions: The inverse functions of exponential expressions.

Questions and Discussion

• Question: On a piecewise graph with a segment extending horizontally to the right from $(3, 5)$, why does it extend horizontally rather than diagonally?

  • Response: The path of the graph is dictated entirely by its underlying piecewise formula $g(x)$. In real-world contexts, flat or jumping piecewise segments represent tier structures such as tax brackets or benefits cliffs in public support programs.

• Question: Are factor orders flexible in expanded terms like $2xh$?

  • Response: Yes. Because multiplication is commutative ($a \cdot b = b \cdot a$), $2xh$ is completely equivalent to $2hx$. Standard convention lists variables alphabetically ($x$ before $h$).

• Question: What are the policies regarding online homework assignments on MyOpenMath?

  • Response: Rolling due dates are utilized. Assignment A1 is due on the evening of the lecture, and Assignment A2 is due Wednesday. Extensions can be requested via email. Online problem sets consist of multi-part practice questions worth 2 points per sub-question.