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

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    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 (-∞, ∞).
    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} \)

    clipboard_efc0262004238b3445893f014c353f830.png

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

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

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

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

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

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

    clipboard_e4200ac00800a4026b88723a82132bbd6.png
    -5 < x < -2

    1. \(\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}\)

    clipboard_e61db0750eee74347f74804dbd4b7221a.png

    Exercise Problems \(\PageIndex{1}\)

    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. \(2x^2 − x − 15 > 0\)
    Detail answers below
    Exercise Answers \(\PageIndex{1}\)

    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

    Exercise Answers \(\PageIndex{2}\)

    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

    Exercise \(\PageIndex{3}\)

    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

    Exercise \(\PageIndex{4}\)

    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 = 8

    Set 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.

    clipboard_edb1f455bacd71d6b73d959ccc08e1088.png

    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

    clipboard_ef7790ce0d87c838534ee0fb9f0ea6261.png

    Note the equation is negative between -2 and 8.  Therefore, the solution interval is -2 < x < 8 or (-2, 8) in Interval notation.

    Exercise \(\PageIndex{5}\)

    5. \(2x^2 − x − 15 > 0\)

    Answer

    Using the box method we can factor the \(2x^2 − x − 15 > 0\).

    clipboard_e2e60a5120c720d9371fdcadae57a17d5.png

    \(2x^2 − x − 15 \) = (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 2

    2x/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 = 3

    Divide the number line into three regions 

    clipboard_e81c14c03554786144fca5db8f97d96a4.png

    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 clipboard_e78f122915a8a905242245cddb54e3f93.png.

    Graph the quadratic equation

    clipboard_ed71206ccaa5060cf33c7b5542bb121c6.png

    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 clipboard_e417c294ab81bceee35a058feb125b915.png.

    This page titled 2.2: Solving and Graphing Inequalities, and Writing Answers in Interval Notation is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Victoria Dominguez, Cristian Martinez, & Sanaa Saykali (ASCCC Open Educational Resources Initiative) .