8.3 Inequalities, Absolute Value & Function Basics

Inequalities, Absolute Value & Function Basics

Inequalities, absolute values, and function relationships test a candidate's understanding of real number order, distance dynamics on a 1D line, and mathematical input-output mapping. This section covers linear inequalities, compound conditions, absolute value equations and inequalities, interval notation, and foundational function operations.

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1. Linear Inequalities & Order Properties

An inequality connects two algebraic expressions using one of four inequality symbols: $<$ (less than), $>$ (greater than), $\le$ (less than or equal to), or $\ge$ (greater than or equal to).

Fundamental Axioms of Order

For any real numbers $a, b, c$:

1. Addition/Subtraction Property: If $a < b$, then $a + c < b + c$ and $a - c < b - c$.
2. Positive Scalar Multiplication/Division: If $a < b$ and $c > 0$, then $ac < bc$ and $\frac{a}{c} < \frac{b}{c}$.
3. Negative Scalar Multiplication/Division (CRITICAL RULE): If $a < b$ and $c < 0$, then $ac > bc$ and $\frac{a}{c} > \frac{b}{c}$.

Exam Caution: Whenever you multiply or divide an inequality by a negative number, you MUST flip the direction of the inequality symbol ($<$ becomes $>$, $\le$ becomes $\ge$).

Worked Example: Inequality Sign Reversal

$\text{Solve for } x: \quad 5 - 3(2x - 4) \ge 29$

1. Distribute $-3$: $5 - 6x + 12 \ge 29 \implies 17 - 6x \ge 29$.
2. Subtract 17 from both sides: $-6x \ge 12$.
3. Divide by $-6$ and flip the inequality sign: $x \le \frac{12}{-6} \implies x \le -5$.
4. Interval Notation: $(-\infty, -5]$.

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2. Compound Inequalities & Interval Notation

A compound inequality combines two inequalities using logical AND (intersection) or logical OR (union).

Conjunctions ("AND" Inequalities)

Expressed as double inequalities $a < x < b$. The solution contains all values satisfying both inequalities simultaneously.

Worked Example: Double Inequality

$\text{Solve: } -5 \le 3x + 4 < 19$

1. Subtract 4 from all three parts: $-5 - 4 \le 3x < 19 - 4 \implies -9 \le 3x < 15$.
2. Divide all three parts by 3: $-3 \le x < 5$.
3. Interval Notation: $[-3, 5)$. (Square bracket includes $-3$; round parenthesis excludes $5$).

Disjunctions ("OR" Inequalities)

Expressed as two separate statements joined by OR. The solution set is the union ($,\cup,$) of values satisfying at least one condition.

Worked Example: Disjunction

$\text{Solve: } 2x + 1 < -7 \quad \text{OR} \quad 3x - 4 \ge 11$

1. Solve left part: $2x < -8 \implies x < -4$.
2. Solve right part: $3x \ge 15 \implies x \ge 5$.
3. Combined Union: $(-\infty, -4) \cup [5, \infty)$.

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3. Absolute Value Equations

The absolute value $|x|$ represents the non-negative distance of $x$ from $0$ on the number line: $|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$

Solving $|ax + b| = c$

• If $c < 0$: No real solution.
• If $c = 0$: $ax + b = 0 \implies x = -\frac{b}{a}$.
• If $c > 0$: Split into two separate linear equations: $ax + b = c \quad \text{OR} \quad ax + b = -c$

Worked Example: Absolute Value Equation

$\text{Solve: } |3x + 6| = 15$

1. Case 1: $3x + 6 = 15 \implies 3x = 9 \implies x = 3$.
2. Case 2: $3x + 6 = -15 \implies 3x = -21 \implies x = -7$.
3. Sum of Solutions: $3 + (-7) = -4$.

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4. Absolute Value Inequalities

Absolute value inequalities transform into bounded intervals or two-tailed outer rays:

Worked Example: Bounded Inequality $|x - a| \le b$

$\text{Solve: } |2x - 3| \le 7$

1. Unfold into double inequality: $-7 \le 2x - 3 \le 7$.
2. Add 3 to all parts: $-4 \le 2x \le 10$.
3. Divide by 2: $-2 \le x \le 5$.
4. Integer Solutions Count: The integers in $[-2, 5]$ are $-2, -1, 0, 1, 2, 3, 4, 5$ (total of 8 integers).

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5. Function Basics for NTS GAT

A function $f$ is a rule that assigns to each element $x$ in a set called the domain exactly one element $f(x)$ in a set called the range.

Key Function Concepts

• Domain Restrictions:
  1. Denominators cannot equal zero (e.g., $f(x) = \frac{1}{x-3} \implies \text{Domain: } x \neq 3$).
  2. Even radicals cannot contain negative radicands (e.g., $g(x) = \sqrt{2x - 6} \implies 2x - 6 \ge 0 \implies x \ge 3$).
• Composite Functions ($f \circ g$): Defined as $(f \circ g)(x) = f(g(x))$. Evaluate the inner function $g(x)$ first, then substitute the result into the outer function $f$.

Worked Example: Composite Function Evaluation

$\text{If } f(x) = 2x^2 - 3 \text{ and } g(x) = 3x + 1, \text{ evaluate } f(g(2)).$

1. Evaluate inner function $g(2)$: $g(2) = 3(2) + 1 = 7$.
2. Evaluate outer function $f(7)$: $f(7) = 2(7)^2 - 3 = 2(49) - 3 = 98 - 3 = 95$.