2.2: Solving and Graphing Inequalities, and Writing Answers in Interval Notation

Interval Notation

Definition: How to write inequalities in Interval Notation

• Symbol to not include a value are ( or ).
• Symbol to include a value are [ or ].
• Use the Symbol ( or ) when using infinity (-∞, ∞).

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1. x < -4           The region is from negative infinity (-∞) up to and not including -4.  Therefore, the Interval Notation is (-∞, -4).
2. x < -4            The region is from negative infinity (-∞) up to and including -4.  There for , the Interval Notation is (-∞, -4].
3. -3 < x < 2     The region is from -3 up to and not including 2.  Therefore, the Interval Notation is (-3, 2).
4. -3 < x < 2     The region is including -3 and up to and including 2.  Therefore, the Interval Notation is [-3, 2].
5. x > 5              The region is greater than and not including 5.  Therefore, the Interval Notation is (5, ∞).
6. x > 5              The region is greater than and including 5.  Therefore, the Interval Notation is [5, ∞).
7. -3 < x < 10   The region is greater than and not including 3 but less than or equal to 10.  Therefore, the Interval Notation is (-3, 10].
8. 4 < x < 7       The region is greater than and including 4 and less than not including 7.  Therefore, the Interval Notation is [4, 7).

To solve and graph inequalities:

1. Solve the inequality using the Properties of Inequalities from the previous section.
2. Graph the solution set on a number line.
3. Write the solution set in interval notation.

Example 2.2.1

Solve the inequality, graph the solution set on a number line and show the solution set in interval notation:

1. \(−1 ≤ 2x − 5 < 7\)
2. \(x^2 + 7x + 10 < 0\)
3. \(−6 < x − 2 < 4\)

Solution

1. \(\begin{array} &&−1 ≤ 2x − 5 < 7 &\text{Example problem} \\ &−1 + 5 ≤ 2x − 5 + 5 < 7 + 5 &\text{The goal is to isolate the variable \(x\), so start by adding \(5\) to all three regions in the inequality.} \\ &4 ≤ 2x < 12 &\text{Simplify.} \\ &\dfrac{4}{2} ≤ 2x < \dfrac{4}{2} &\text{Divide all by \(2\) to isolate the variable \(x\).} \\ &2 ≤ x < 6 &\text{Final answer written in inequality/solution set form.} \\ &[2, 6) &\text{Final answer written in interval notation (see section on Interval Notation for more details)} \end{array} \)

2. \(\begin{array} &&x^2 + 7x + 10 < 0 &\text{Example problem} \\ &(x + 5)(x + 2) < 0 &\text{Factor the polynomial.} \\ &(x + 5)(x + 2) < 0 &\text{Set each factor equal to 0 and solve for x.  x = -5 and x = -2.} \\ \end{array}\)

\(\begin{array} &\text{Since the inequality is a strict inequality(< or >), -5 and -2 is not included in the solution set.} \end{array}\)

\(\begin{array} &\text{Pick a point less than -5, x = -6.  Evaluate the expression (-6+5)(-6+2) = (-1)(-4) = 4.  It is positive, x < -5 is not in the solution set.} \end{array}\)

\(\begin{array} &\text{Pick a point between -5 and x = -2.  Evaluate the expression (-4 + 5)(-4+2) = (1)(-2) = -2.  It is negative, -5 < x < -2 is in the solution set.} \end{array}\) 

\(\begin{array} &\text{Finally, pick a point greater than -2, x = 0.  Evaluate the expression (0 + 5)(0 + 2) = 5(2) = 10.  It is positive, x > -2 is not the solution set.} \end{array}\)

\(\begin{array} &\text{Interval notation: (-5, -2) } \end{array}\)

-5 < x < -2

3. \(\begin{array}&&−6 < x − 2 ≤ 4 &\text{Example problem} \\ &−6 + 2 < x − 2 + 2 ≤ 4 + 2 &\text{The goal is to isolate the variable \(x\), so start by adding \(2\) to all three regions in the inequality.} \\ &−4 < x ≤ 6 &\text{Final answer written in inequality/solution set form.} \\ &(−4, 6] &\text{Final answer written in interval notation (see section on Interval Notation for more details).} \end{array}\)

Exercise 2.2.1

Solve the inequalities, graph the solution sets on a number line and show the solution sets in interval notation:

1. \(0 ≤ x + 1 ≤ 4\)
2. \(0 < 2(x − 1) ≤ 4\)
3. \(6 < 2(x − 1) < 12\)
4. \(x^2 − 6x − 16 < 0\)
5. \(2x^2 − x − 15 > 0\)