How To Find The Range Of An Inequality

How to Find the Rangeof an Inequality

Finding the range of an inequality means determining all the real numbers that make the inequality true. In other words, you are looking for the solution set, often expressed as intervals or a union of intervals. This skill is fundamental in algebra, calculus, and many applied fields because it lets you describe where a function is positive, negative, bounded, or exceeds a certain threshold. Below is a detailed, step‑by‑step guide that works for the most common types of inequalities—linear, quadratic, rational, and those involving absolute values—along with tips, pitfalls to avoid, and practice examples.

Introduction

When you see an expression like (2x-5 > 3) or (\frac{x+1}{x-2} \le 0), the goal is not just to solve for a single number but to identify the range of x‑values that satisfy the condition. The process is similar to solving an equation, but inequalities require extra attention to sign changes, especially when you multiply or divide by a negative number or when the expression involves a denominator that could be zero. By mastering a systematic approach—identifying critical points, testing intervals, and checking endpoints—you can confidently find the range for any inequality you encounter.

Step‑by‑Step Procedure

1. Rewrite the Inequality in Standard Form

Bring all terms to one side so that the other side is zero.

• For a linear inequality: (ax + b > 0) or (ax + b < 0).
• For a quadratic inequality: (ax^2 + bx + c \ge 0) or (\le 0).
• For a rational inequality: (\frac{P(x)}{Q(x)} > 0) (or (\ge, <, \le)). - For an absolute value inequality: (|f(x)| < k) or (|f(x)| \ge k).

2. Find the Critical Points

Critical points are the x‑values where the expression equals zero or is undefined.

• Set the numerator equal to zero → potential zeros.
• Set the denominator equal to zero → points of discontinuity (excluded from the solution).
• For absolute values, consider where the inside expression changes sign.

3. Create a Sign Chart (Interval Test)

Place the critical points on a number line, dividing it into open intervals. Pick a test point from each interval and substitute it into the factored form of the expression to determine whether the expression is positive or negative on that interval.

4. Determine Which Intervals Satisfy the Inequality

• If the inequality is > 0 (or ≥ 0), keep intervals where the expression is positive (or zero, if the inequality includes equality).
• If the inequality is < 0 (or ≤ 0), keep intervals where the expression is negative (or zero, if equality is allowed).

5. Include or Exclude Endpoints According to the Symbol

• Use a closed bracket ([ , ]) for (\ge) or (\le) when the point makes the expression zero and is not a denominator zero.
• Use an open parenthesis (( , )) for (>) or (<) or when the point makes the denominator zero (the expression is undefined there).

6. Write the Final Answer in Interval Notation or Set‑Builder Form

Combine the qualifying intervals using the union symbol (\cup) if needed.

Special Cases and Techniques

Linear Inequalities

A linear inequality reduces to a single critical point.
Example: Solve (3x - 7 < 5). 1. Rewrite: (3x - 12 < 0).
2. Critical point: (3x - 12 = 0 \Rightarrow x = 4).
3. Test intervals ((-\infty,4)) and ((4,\infty)).

• Choose (x=0): (3(0)-12 = -12 < 0) → satisfies.
• Choose (x=5): (3(5)-12 = 3 > 0) → does not satisfy.

4. Since the inequality is strict (<), exclude 4. Answer: ((-\infty, 4)).

Quadratic Inequalities

Factor or use the quadratic formula to find zeros. The sign of a quadratic opens upward if (a>0) and downward if (a<0).
Example: Solve (x^2 - 5x + 6 \ge 0).

1. Factor: ((x-2)(x-3) \ge 0).
2. Critical points: (x=2, x=3). 3. Sign chart:
   • ((-\infty,2)): pick (x=0) → (( -)( -) = +) → positive.
   • ((2,3)): pick (x=2.5) → ((+)(-) = -) → negative.
   • ((3,\infty)): pick (x=4) → ((+)(+)=+) → positive.
3. Because we need (\ge 0), keep the positive intervals and include the zeros.
   Answer: ((-\infty,2] \cup [3,\infty)).

Rational Inequalities

Watch for points where the denominator equals zero; these are always excluded.
Example: Solve (\frac{x+1}{x-2} \le 0).

1. Critical points: numerator zero at (x=-1); denominator zero at (x=2).
2. Intervals: ((-\infty,-1)), ((-1,2)), ((2,\infty)).
3. Test signs:
   • (x=-2): (\frac{-1}{-4}=+) → positive.
   • (x=0): (\frac{1}{-2}=- ) → negative.
   • (x=3): (\frac{4}{1}=+) → positive. 4. Inequality is (\le 0): keep negative interval and include the zero at (x=-1) (since it makes the fraction zero). Exclude (x=2).
     Answer: ([-1,2)).

Absolute Value Inequalities

Convert to a compound inequality.

7. Absolute Value Inequalities

An inequality that contains (|,x-a,|) can be rewritten as a pair of linear bounds.
The rule is:

[ |,x-a,| ;<; k \quad\Longleftrightarrow\quad a-k ;<; x ;<; a+k, ] [ |,x-a,| ;>; k \quad\Longleftrightarrow\quad x ;<; a-k ;;\text{or};; x ;>; a+k, ]

where (k\ge 0). If the right‑hand side is negative, the inequality has no solution for “(<)” and is automatically true for “(>)” (except at points where the expression is undefined).

Example 1 – Strict Inequality

Solve (|,2x-5,| ;<; 3).

1. Isolate the absolute value (already isolated).
2. Apply the rule with (a=\tfrac{5}{2}) and (k=3):
   [ \tfrac{5}{2}-3 ;<; x ;<; \tfrac{5}{2}+3. ]
3. Compute the bounds: (\tfrac{5}{2}-3 = -\tfrac{1}{2}) and (\tfrac{5}{2}+3 = \tfrac{11}{2}).
4. Because the inequality is strict, the endpoints are excluded.

Solution set: (\displaystyle\left(-\tfrac12,;\tfrac{11}{2}\right)).

Example 2 – Non‑strict Inequality

Solve (|,x+4,| ;\ge; 2).

1. Write the compound form for “(\ge)”:
   [ x+4 \le -2 \quad\text{or}\quad x+4 \ge 2. ]
2. Solve each linear inequality:
   [ x \le -6 \quad\text{or}\quad x \ge -2. ]
3. Union the two intervals and note that the endpoints are included (the inequality is non‑strict).

Solution set: ((-\infty,-6] \cup [-2,\infty)).

Example 3 – No Solution Case Solve (|,3x-1,| ;<; -4).

The right‑hand side is negative, and an absolute value is always non‑negative. Hence there is no real number that satisfies the inequality.

Solution set: (\varnothing).

Example 4 – Rational Expression Inside the Absolute Value

Solve (\displaystyle\frac{|,x-3,|}{x+1} \le 2).

1. First note the restriction (x\neq -1).
2. Multiply both sides by (|x+1|) (which is always non‑negative) to avoid sign flips:
   [ |,x-3,| \le 2|x+1|. ]
3. Remove the outer absolute values by considering the sign of the inner expressions. It is easier to square both sides (preserving the inequality because both sides are non‑negative):
   [ (x-3)^2 \le 4(x+1)^2. ]
4. Expand and bring everything to one side:
   [ x^2-6x+9 \le 4x^2+8x+4 \ 0 \le 3x^2+14x-5. ]
5. Solve the quadratic inequality (3x^2+14x-5 \ge 0). Its zeros are found with the quadratic formula:
   [ x = \frac{-14 \pm \sqrt{14^2-4\cdot3\cdot(-5)}}{2\cdot3} = \frac{-14 \pm \sqrt{196+60}}{6} = \frac{-14 \pm \sqrt{256}}{6} = \frac{-14 \pm 16}{6}. ] Hence (x_1 = \frac{2}{6}= \frac13) and (x_2 = \frac{-30}{6}= -5).
6. Because the leading coefficient (3>0), the parabola opens upward, so the expression is non‑negative outside the interval between the roots:
   [ x \le -5 \quad\text{or}\quad x \ge \frac13. ]
7. Re‑impose the original restriction (x\neq -1). The point (-1) lies in the interval ((-5,\frac13)), which we have already excluded, so no further adjustment is needed.

Solution set: ((-\infty,-5] \cup [\tfrac13,\infty)), with the understanding that (x=-1) is not allowed (but (-1) is not in the final set anyway).