4.1 Concept of a Function, Evaluation, Domain & Range

4.1 Concept of a Function, Evaluation, Domain & Range

Functions represent the foundational language of college algebra, precalculus, and calculus. In real-world application, functions model cause-and-effect relationships where one variable uniquely determines another—such as time determining position, temperature determining pressure, or price determining consumer demand.

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1. Formal Definition of a Function and the Vertical Line Test

Mathematically, a relation is any set of ordered pairs $(x, y)$ that associates elements from an input set $X$ with elements from an output set $Y$. A function $f$ is a restricted type of relation that enforces strict single-valued determinism.

$\text{Formal Mapping: } f: X \to Y \quad \text{where } \forall x \in X, \exists! y \in Y \text{ such that } y = f(x)$

• Domain ($X$): The set of all permissible independent input values $x$.
• Codomain ($Y$): The target set containing all possible output values.
• Range ($f(X)$): The actual subset of $Y$ containing all realized output values $y = f(x)$ generated by evaluating inputs from the domain.

┌─────────────────────────────────────────────────────────┐
│               Relation vs. Function Mapping             │
├────────────────────────────┬────────────────────────────┤
│   General Relation (Set)   │    Mathematical Function   │
│  - Inputs can map to       - Each input x maps to       │
│    MULTIPLE outputs        │    EXACTLY ONE output y    │
│  - Example: (2, 5), (2, 8) │  - Example: (2, 5), (3, 8) │
└────────────────────────────┴────────────────────────────┘

The Vertical Line Test (VLT)

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

• Passing VLT: If every vertical line $x = c$ intersects the graph at most once, each input $c$ yields a unique output $f(c)$. Parabolas opening upward ($y = x^2 - 3$), lines ($y = 2x + 1$), and exponential curves ($y = e^x$) pass the VLT and are valid functions.
• Failing VLT: If a single vertical line $x = c$ intersects the graph at two or more points, the input $c$ is assigned multiple distinct outputs, violating the definition of a function. Circles ($x^2 + y^2 = 25$, where $x = 0$ intersects at $(0, 5)$ and $(0, -5)$) and sideways parabolas ($x = y^2$) fail the VLT and are not functions of $x$.

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2. Function Notation, Evaluation, and the Difference Quotient

The standard notation $y = f(x)$ is read as "$y$ equals $f$ of $x$." Here, $f$ is the name of the function rule, $x$ is the independent variable (input), and $f(x)$ is the dependent variable (output). It is essential to recognize that $f(x)$ represents a single output value, not the multiplication of $f$ by $x$.

Evaluating Functions at Numerical and Algebraic Inputs

To evaluate a function $f(x)$ at an argument, substitute the entire argument—enclosed in parentheses—into every instance of $x$ in the algebraic rule.

Worked Example: Multi-Argument Evaluation

Given $f(x) = 3x^2 - 4x + 5$, evaluate $f(-3)$, $f(2a)$, and $f(x + h)$.

1. Numerical Evaluation $f(-3)$: $f(-3) = 3(-3)^2 - 4(-3) + 5 = 3(9) + 12 + 5 = 27 + 12 + 5 = 44$
2. Algebraic Monomial Evaluation $f(2a)$: $f(2a) = 3(2a)^2 - 4(2a) + 5 = 3(4a^2) - 8a + 5 = 12a^2 - 8a + 5$
3. Algebraic Binomial Evaluation $f(x + h)$: $f(x + h) = 3(x + h)^2 - 4(x + h) + 5$ $= 3(x^2 + 2xh + h^2) - 4x - 4h + 5 = 3x^2 + 6xh + 3h^2 - 4x - 4h + 5$

The Difference Quotient

In calculus, the difference quotient represents the secant line slope connecting points $(x, f(x))$ and $(x+h, f(x+h))$ on the graph of $f$, measuring the average rate of change over an interval of length $h$:

$\text{Difference Quotient} = \frac{f(x+h) - f(x)}{h}, \quad h \ne 0$

Step-by-Step Difference Quotient Evaluation

For $f(x) = 2x^2 - 5x + 3$:

1. Expand $f(x+h)$: $f(x+h) = 2(x+h)^2 - 5(x+h) + 3 = 2(x^2 + 2xh + h^2) - 5x - 5h + 3 = 2x^2 + 4xh + 2h^2 - 5x - 5h + 3$
2. Subtract $f(x)$: $f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 - 5x - 5h + 3) - (2x^2 - 5x + 3)$ $= 4xh + 2h^2 - 5h$
3. Divide by $h$ and factor: $\frac{f(x+h) - f(x)}{h} = \frac{h(4x + 2h - 5)}{h} = 4x + 2h - 5$

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3. Systematic Algebraic Rules for Determining Domain

The implicit domain (or natural domain) of a real-valued function is the set of all real numbers $x$ for which $f(x)$ produces a real output. Unless specified otherwise, domain analysis requires identifying algebraic operations that produce undefined outputs or imaginary quantities.

Master Domain Restriction Matrix

Combining Multiple Domain Restrictions

When a function contains multiple algebraic constraints, solve each inequality or equation independently and find the intersection of all valid sets.

Worked Example: Advanced Multi-Restriction Domain

Find the domain of $f(x) = \frac{\sqrt{3x + 12}}{x^2 - 16} + \ln(9 - x)$.

1. Even Radical Condition (Numerator): $3x + 12 \ge 0 \implies 3x \ge -12 \implies x \ge -4$
2. Rational Condition (Denominator): $x^2 - 16 \ne 0 \implies (x - 4)(x + 4) \ne 0 \implies x \ne 4 \text{ and } x \ne -4$
3. Logarithmic Condition: $9 - x > 0 \implies x < 9$
4. Intersect All Conditions:
   • Radical requires $x \ge -4$.
   • Denominator excludes $x = -4$ and $x = 4$.
   • Logarithm requires $x < 9$. Combining these: $x \in (-4, 4) \cup (4, 9)$. $\text{Domain in Interval Notation: } (-4, 4) \cup (4, 9)$

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4. Comprehensive Strategies for Determining Range

The range of a function is the set of all possible output values $y = f(x)$ generated as $x$ spans the entire domain.

Analytical Range Determination Techniques

1. Graphical Projection: Sketch or analyze the function to identify absolute vertical bounds (minimum $y_{min}$ and maximum $y_{max}$).
2. Algebraic Inversion Method: Set $y = f(x)$, solve the resulting equation for $x$ in terms of $y$, and determine the values of $y$ that yield real, valid $x$ values within the function's domain.
3. Non-Negativity Inequality Bounds: Utilize known lower bounds of parent functions ($x^2 \ge 0$, $|x| \ge 0$, $\sqrt{g(x)} \ge 0$, $e^x > 0$).

Worked Example: Algebraic Inversion for Range

Find the range of $g(x) = \frac{2x + 1}{x - 3}$.

1. Domain: $x \ne 3$, so Domain = $(-\infty, 3) \cup (3, \infty)$.
2. Set $y = g(x)$ and solve for $x$: $y = \frac{2x + 1}{x - 3} \implies y(x - 3) = 2x + 1 \implies xy - 3y = 2x + 1$ Collect $x$ terms on one side: $xy - 2x = 3y + 1 \implies x(y - 2) = 3y + 1 \implies x = \frac{3y + 1}{y - 2}$
3. Identify Restrictions on $y$: For $x$ to be a valid real number, the denominator $y - 2$ cannot equal zero $\implies y \ne 2$. $\text{Range in Interval Notation: } (-\infty, 2) \cup (2, \infty)$