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Solving Inequalities

Core Definition

An inequality expresses an order relationship between two expressions using <<, ≤\le, >>, or ≥\ge. Solving an inequality determines the set of all real numbers that satisfy the condition, typically expressed in interval notation. The fundamental algebraic rule is that multiplying or dividing both sides by a negative number reverses the direction of the inequality sign.

Concept Explanation

1. Single Linear Inequalities

For single inequalities involving one relational operator (such as ax+b<cax + b < c), isolate the variable using standard linear operations while tracking sign flips:

  • Addition/Subtraction: Adding or subtracting any real number preserves the inequality direction:
    x−5≥3  ⟹  x≥8  ⟺  [8,∞)x - 5 \ge 3 \implies x \ge 8 \quad \iff \quad [8, \infty)
  • Multiplication/Division by Positive Numbers: Preserves direction:
    3x<12  ⟹  x<4  ⟺  (−∞,4)3x < 12 \implies x < 4 \quad \iff \quad (-\infty, 4)
  • Multiplication/Division by Negative Numbers: Reverses the inequality:
    −2x≤10  ⟹  x≥10−2  ⟹  x≥−5  ⟺  [−5,∞)-2x \le 10 \implies x \ge \frac{10}{-2} \implies x \ge -5 \quad \iff \quad [-5, \infty)

2. Double (Compound / Sandwich) Inequalities

A double inequality of the form a<f(x)<ba < f(x) < b represents an intersection ("AND") condition: f(x)>af(x) > a and f(x)<bf(x) < b.

When solving a linear double inequality, apply algebraic operations simultaneously to all three parts:

1≤3x+4<131 \le 3x + 4 < 13

  1. Subtract 44 across all three segments:
    1−4≤3x<13−4  ⟹  −3≤3x<91 - 4 \le 3x < 13 - 4 \implies -3 \le 3x < 9
  2. Divide all three segments by 33:
    −33≤x<93  ⟹  −1≤x<3  ⟺  [−1,3)\frac{-3}{3} \le x < \frac{9}{3} \implies -1 \le x < 3 \quad \iff \quad [-1, 3)

If dividing all three segments by a negative number, every inequality sign reverses:

−6≤−2x<8  ⟹  −6−2≥x>8−2  ⟹  3≥x>−4  ⟺  (−4,3]-6 \le -2x < 8 \implies \frac{-6}{-2} \ge x > \frac{8}{-2} \implies 3 \ge x > -4 \quad \iff \quad (-4, 3]

3. Non-Linear Inequalities & Sign Charts

For polynomials and rational expressions (e.g., x2−4>0x^2 - 4 > 0 or x−1x+3≤0\frac{x-1}{x+3} \le 0), solve by factoring and analyzing interval signs across critical boundary points (roots and vertical asymptotes):

  1. Set the inequality to zero: P(x)≥0P(x) \ge 0 or P(x)Q(x)≤0\frac{P(x)}{Q(x)} \le 0.
  2. Identify all critical points where the expression equals zero or is undefined.
  3. Test a sample point in each sub-interval on the real number line to determine whether the expression is positive (+)(+) or negative (−)(-).

Engineering Applications

Inequalities define operational bounds, structural tolerances, and sensor envelopes:

  • Structural Safety Envelopes: Civil engineers specify stress bounds σmin≤σ(t)≤σallowable\sigma_{\text{min}} \le \sigma(t) \le \sigma_{\text{allowable}} to prevent bridge beam buckling.
  • Control Signal Clamping: Electrical engineers design operational thresholds Vlow<Vsensor(t)<VhighV_{\text{low}} < V_{\text{sensor}}(t) < V_{\text{high}} to filter voltage noise and trigger emergency shutdown clamps.
  • Squeeze Theorem Bounding: When mathematical functions oscillate wildly, bounding them between two solvable envelope curves g(x)≤f(x)≤h(x)g(x) \le f(x) \le h(x) allows engineers to prove stability as time t→0t \to 0 or t→∞t \to \infty.

Example Problems

Example 1: Solving a Double Inequality

Solve the double inequality for xx and express the solution in interval notation:

−5<2−3x≤11-5 < 2 - 3x \le 11

Solution

Step 1: Subtract 22 from all three parts:

−5−2<−3x≤11−2-5 - 2 < -3x \le 11 - 2
−7<−3x≤9-7 < -3x \le 9

Step 2: Divide all three parts by −3-3 (and flip the inequality signs):

−7−3>x≥9−3\frac{-7}{-3} > x \ge \frac{9}{-3}
73>x≥−3  ⟺  −3≤x<73\frac{7}{3} > x \ge -3 \quad \iff \quad -3 \le x < \frac{7}{3}

Step 3: State the interval notation:

x∈[−3,73)x \in \left[-3, \frac{7}{3}\right)

Example 2: Bounding Inequality for the Squeeze Theorem

Given −1≤sin⁡(1x)≤1-1 \le \sin\left(\frac{1}{x}\right) \le 1 for all x≠0x \neq 0, construct a sandwich inequality for f(x)=x2sin⁡(1x)f(x) = x^2 \sin\left(\frac{1}{x}\right) and determine whether the inequality direction changes for x<0x < 0.

Solution

Step 1: Analyze the multiplier x2x^2: Since x2>0x^2 > 0 for all real numbers x≠0x \neq 0, multiplying across the inequality preserves the direction for both positive and negative xx.

Step 2: Multiply each term by x2x^2:

−1⋅x2≤x2sin⁡(1x)≤1⋅x2-1 \cdot x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le 1 \cdot x^2
−x2≤x2sin⁡(1x)≤x2-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2

Step 3: Conclusion for Squeeze Theorem: Because lim⁡x→0(−x2)=0\lim_{x \to 0} (-x^2) = 0 and lim⁡x→0(x2)=0\lim_{x \to 0} (x^2) = 0, the trapped center term lim⁡x→0x2sin⁡(1/x)=0\lim_{x \to 0} x^2 \sin(1/x) = 0.


Connections & References