2.2: Solving and Graphing Inequalities, and Writing Answers in Interval Notation
Interval Notation
- Symbol to not include a value are ( or ).
- Symbol to include a value are [ or ].
- Use the Symbol ( or ) when using infinity (-∞, ∞).
- x < -4 The region is from negative infinity (-∞) up to and not including -4. Therefore, the Interval Notation is ( -∞ , -4) .
- x < -4 The region is from negative infinity (-∞) up to and including -4. There for , the Interval Notation is ( -∞, -4].
- -3 < x < 2 The region is from -3 up to and not including 2. Therefore, the Interval Notation is (-3, 2).
- -3 < x < 2 The region is including -3 and up to and including 2. Therefore, the Interval Notation is [-3, 2].
- x > 5 The region is greater than and not including 5. Therefore, the Interval Notation is (5, ∞) .
- x > 5 The region is greater than and including 5. Therefore, the Interval Notation is [5, ∞).
- -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].
- 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:
- Solve the inequality using the Properties of Inequalities from the previous section.
- Graph the solution set on a number line.
- Write the solution set in interval notation.
Solve the inequality, graph the solution set on a number line and show the solution set in interval notation:
- \(−1 ≤ 2x − 5 < 7\)
- \(x^2 + 7x + 10 < 0\)
- \(−6 < x − 2 < 4\)
Solution
- \(\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} \)
- \(\begin{array} &&x^2 + 7x + 10 < 0 &\text{Example problem} \\ &(x + 5)(x + 2) < 0 &\text{Factor the polynomial.} \\ &(x + 5)(x + 2) < 0 &\text{Set each factor equal to 0 and solve for x. x = -5 and x = -2.} \\ \end{array}\)
\(\begin{array} &\text{Since the inequality is a strict inequality(< or >), -5 and -2 is not included in the solution set.} \end{array}\)
\(\begin{array} &\text{Pick a point less than -5, x = -6. Evaluate the expression (-6+5)(-6+2) = (-1)(-4) = 4. It is positive, x < -5 is not in the solution set.} \end{array}\)
\(\begin{array} &\text{Pick a point between -5 and x = -2. Evaluate the expression (-4 + 5)(-4+2) = (1)(-2) = -2. It is negative, -5 < x < -2 is in the solution set.} \end{array}\)
\(\begin{array} &\text{Finally, pick a point greater than -2, x = 0. Evaluate the expression (0 + 5)(0 + 2) = 5(2) = 10. It is positive, x > -2 is not the solution set.} \end{array}\)
\(\begin{array} &\text{Interval notation: (-5, -2) } \end{array}\)
-5 < x < -2
- \(\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}\)
Solve the inequalities, graph the solution sets on a number line and show the solution sets in interval notation:
- \(0 ≤ x + 1 ≤ 4\)
- \(0 < 2(x − 1) ≤ 4\)
- \(6 < 2(x − 1) < 12\)
- \(x^2 − 6x − 16 < 0\)
- \(2x^2 − x − 15 > 0\)
1. \(0 ≤ x + 1 ≤ 4\)
- Answer
-
\(0 ≤ x + 1 ≤ 4\)
subtract 1 from all three sides
\(0 -1 ≤ x + 1 - 1 ≤ 4 - 1\)
\(-1 ≤ x + 0≤ 3\)
\(-1 ≤ x ≤ 3\)
[-1, 3] Interval Notation
2. \(0 < 2(x − 1) ≤ 4\)
- Answer
-
Multiply the 2 through the parenthesis in the middle section.
\(0 < 2x − 2 ≤ 4\)
Add 2 to all three sides
\(0 + 2 < 2x − 2 + 2 ≤ 4 + 2\)
\(2 < 2x + 0 ≤ 6\)
\(2 < 2x ≤ 6\)
Divide all three sides by 2
\(2/2 < 2x/2 ≤ 6/2\)
\(1 < x ≤ 3\)
(1, 3] Interval Notation
3. \(6 < 2(x − 1) < 12\)
- Answer
-
Add texts here. Do not delete this text first.
Multiply the 2 through the parenthesis in the middle section.
\(6 < 2x − 2 < 12\)
Add 2 to all three sides
\(6 + 2 < 2x − 2 + 2 < 12 + 2\)
\(8 < 2x + 0 < 14\)
\(8 < 2x < 14\)
Divide all sides by 2
\(8/2 < 2x/2 < 14/2\)
\(4 < x < 7\)
(4, 7) Interval Notation
4. \(x^2 − 6x − 16 < 0\)
- Answer
-
Factor \(x^2 − 6x − 16\) to (x - 8)(x + 2).
Set x - 8 = 0 and solve for x
x - 8 = 0 Add 8 to both sides
x - 8 + 8 = 0 + 8
x + 0 = 8
x = 8Set x + 2 = 0 and solve for x
x + 2 = 0 Subtract 2 from both sides
x + 2 - 2 = 0 - 2
x + 0 = -2
x = -2
Create a number line with the two points selected. Since it is a strict inequality, <, the points will not be included.
Select a point to the left of x = -2, -3. Substitute it in the factored form of the quadratic equation.
(-3 - 8)(-3 +2) = (-11)(-1) = 11 > 0. For values to the left of x = -2 the function is positive.
Select a point between x = -2 and x = 8, 0. Substitute it in the factored form of the quadratic equation.
(0 -8)(0 + 2) = (-8)(2) = -16 < 0. For values between -2 and 8, the function is negative.
Select a point to the right of x = 8, 9. Substitute the value in the factored form of the quadratic equation.
(9 - 8)(9 + 2) = (1)(11) = 11 > 0. For values to the right of x = 8 the function is positive.
The solution is in the interval -2 < x < 8, or (-2, 8) Interval notation.
Graph of quadratic equation
Note the equation is negative between -2 and 8. Therefore, the solution interval is -2 < x < 8 or (-2, 8) in Interval notation.
5. \(2x^2 − x − 15 > 0\)
- Answer
-
Using the box method we can factor the \(2x^2 − x − 15 > 0\).
\(2x^2 − x − 15 \) = (2x + 5)(x - 3)
Set each factor equal to 0 and solve for x
2x + 5 = 0 subtract 5 from both sides
2x + 5 - 5 = 0 - 5
2x + 0 = -5
2x = -5 divide both sides by 22x/2 = -5/2
x = -5/2 or -2.5
x - 3 = 0 add 3 to both sides
x - 3 + 3 = 0 + 3
x + 0 = 3
x = 3Divide the number line into three regions
Select a number to the left of -2.5, -3. Substitute the value in the factored form of the quadratic equation.
(2(-3) + 5)(-3 - 3) = (-6 + 5)(-3 - 3) = (-1)(-6) = 6 > 0. The region to the left of -2.5 is positive.
Select a number between -2.5 and 3, 0. Substitute the value in the factored form of the quadratic equation.
(2(0) + 5)(0 - 3) = (0 + 5)(-3) = (5)(-3) = -15 < 0. The region between -2.5 and 3 is negative.
Select a number to the right of 3, 4. Substitute the value in the factored form of the quadratic equation.
(2(4) + 5)(5 - 3) = (8 + 5)(2) = (13)(2) = 26 > 0. The region to the right of 3 is positive.
The solution is .
Graph the quadratic equation
You can see the graph is positive to the left of -2.5 and to the right of 3. The solution in interval notation is .