COLLEGE ALGEBRA: Equations and Inequalities
Section 1.7: Other Types of Inequalities
Objective: Use critical numbers to determine test intervals for a polynomial inequality.
Objective: Solve and graph a polynomial inequality.
Objective: Solve and graph a rational inequality.
Objective: Construct and use a polynomial inequality to solve an application problem.
Objective: Determine the domain of an expression involving a radical.
Finding Critical Numbers and Test Intervals
To solve a polynomial inequality (e.g., x^2 - 2x - 3 < 0):
A polynomial can change signs only at its zeros (the x-values that make the polynomial equal to zero).
Between two consecutive zeros, a polynomial is either entirely positive or entirely negative.
The real zeros of a polynomial, when arranged in order, divide the real number line into intervals where the polynomial has no sign changes.
These zeros are called critical numbers of the inequality.
The resulting intervals are called test intervals for the inequality.
Example: For x^2 - 2x - 3 < 0:
Factor the polynomial: (x + 1)(x - 3) = x^2 - 2x - 3
The zeros are x = -1 and x = 3. These are the critical numbers.
These zeros divide the real number line into three test intervals: (-\infty, -1), (-\text{1, 3}), and (3, \infty). (See Figure 1.19)
To solve the inequality, test one value from each of these test intervals.
Steps to Find Test Intervals for a Polynomial
Find Real Zeros: Find all real zeros of the polynomial and arrange them in increasing order. These are the critical numbers.
Determine Test Intervals: Use the critical numbers to define the test intervals.
Evaluate Polynomial: Choose one representative x-value in each test interval and evaluate the polynomial at that value.
If the value is negative, the polynomial is negative for all x-values in that interval.
If the value is positive, the polynomial is positive for all x-values in that interval.
Polynomial Inequalities
Example 1: Solving a Polynomial Inequality
Problem: Solve x^2 - 2x - 3 < 0
Solution:
Factor: (x + 1)(x - 3) < 0
Critical numbers: x = -1 and x = 3
Test intervals: (-\infty, -1), (-\text{1, 3}), and (3, \infty)
Test values:
Interval (-\infty, -1): Choose x = -2. ( -2 + 1)( -2 - 3) = (-1)(-5) = 5. Conclusion: Positive.
Interval (-\text{1, 3}): Choose x = 0. (0 + 1)(0 - 3) = (1)(-3) = -3. Conclusion: Negative.
Interval (3, \infty): Choose x = 4. (4 + 1)(4 - 3) = (5)(1) = 5. Conclusion: Positive.
Conclusion: The polynomial is positive on (-\infty, -1) and (3, \infty), and negative on (-\text{1, 3}).
The solution to x^2 - 2x - 3 < 0 is the interval (-\text{2, 3}) (Note: the example text has a typo, it states ( -2, 3) but the preceding interval is ( -1,3), so it should be ( -1, 3). (See Figure 1.20)
Checking solutions: Substitute x-values from the solution interval into the original inequality to confirm.
Writing in General Form
If a polynomial inequality is not in general form (polynomial on one side, zero on the other), rewrite it.
Example 2: Solving a Polynomial Inequality (not in general form)
Problem: Solve x^3 - 3x^2 > 10x
Solution:
Write original inequality: x^3 - 3x^2 > 10x
Write in general form: x^3 - 3x^2 - 10x > 0
Factor: x(x^2 - 3x - 10) > 0 \implies x(x - 5)(x + 2) > 0
Critical numbers: x = -2, x = 0, and x = 5
Test intervals: (-\infty, -2), (-\text{2, 0}), (0, 5), and (5, \infty)
Test values:
Interval (-\infty, -2): Choose x = -3. (-3)(-3-5)(-3+2) = (-3)(-8)(-1) = -24. Conclusion: Negative.
Interval (-\text{2, 0}): Choose x = -1. (-1)(-1-5)(-1+2) = (-1)(-6)(1) = 6. Conclusion: Positive.
Interval (0, 5): Choose x = 2. (2)(2-5)(2+2) = (2)(-3)(4) = -24. Conclusion: Negative.
Interval (5, \infty): Choose x = 6. (6)(6-5)(6+2) = (6)(1)(8) = 48. Conclusion: Positive.
Conclusion: The inequality is satisfied on (-\text{2, 0}) and (5, \infty). (See Figure 1.21)
Note on inequality symbols:
If the inequality includes "greater than or equal to" (\ge) or "less than or equal to" (\le), the critical numbers are included in the solution set (closed intervals/brackets).
If it's strictly "greater than" (>) or "less than" (<), critical numbers are not included (open intervals/parentheses).
Example 3: Unusual Solution Sets for Polynomial Inequalities
a. x^2 + 2x + 4 > 0
Solution: Entire set of real numbers, (-\infty, \infty).
The quadratic x^2 + 2x + 4 is positive for every real value of x.
b. x^2 + 2x + 1 \le 0
Solution: Single real number { -1 }.
The quadratic x^2 + 2x + 1 = (x + 1)^2. Its only critical number (zero) is x = -1. This is the only value that satisfies (x + 1)^2 \le 0.
c. x^2 + 3x + 5 < 0
Solution: Empty set.
The quadratic x^2 + 3x + 5 is not less than zero for any value of x.
d. x^2 - 4x + 4 > 0
Solution: All real numbers except x = 2. In interval notation: (-\infty, 2) \cup (2, \infty).
The quadratic x^2 - 4x + 4 = (x - 2)^2. It is always non-negative. It's greater than 0 for all x except when x = 2 (where it is 0).
Rational Inequalities
The concept of critical numbers and test intervals extends to rational expressions.
Rational expressions can change sign at their:
Zeros: x-values for which the numerator is zero.
Undefined values: x-values for which the denominator is zero.
These two types of numbers constitute the critical numbers of a rational inequality.
Example 4: Solving a Rational Inequality
Problem: Solve \frac{x - 2}{x - 5} \ge 0
Solution:
Write original inequality: \frac{x - 2}{x - 5} \ge 0
Identify numerator's zero: x - 2 = 0 \implies x = 2
Identify denominator's zero (undefined value): x - 5 = 0 \implies x = 5
Critical numbers: x = 2, x = 5
Test intervals: (-\infty, 2), (2, 5), (5, \infty)
Test values:
Interval (-\infty, 2): Choose x = 0. \frac{0 - 2}{0 - 5} = \frac{-2}{-5} = \frac{2}{5} (Positive). Conclusion: Satisfied.
Interval (2, 5): Choose x = 4. \frac{4 - 2}{4 - 5} = \frac{2}{-1} = -2 (Negative). Conclusion: Not satisfied.
Interval (5, \infty)$}$: Choose x = 6. \frac{6 - 2}{6 - 5} = \frac{4}{1} = 4 (Positive). Conclusion: Satisfied.
Since the inequality is \ge 0, the numerator's zero (x = 2) is included. The denominator's zero (x = 5) is always excluded.
Solution set: (-\infty, 2] \cup (5, \infty). (See Figure 1.23)
Application: Increasing the Profit for a Product
Formula: Profit = Revenue - Cost (P = R - C)
Scenario: Calculator manufacturer.
Demand equation: p = 100 - 10x, where p is price per calculator (dollars) and x is calculators sold (millions), for 0 \le x \le 10.
Revenue equation: R = xp = x(100 - 10x) (millions of dollars).
Total Cost equation: C = 10x + 2.5 (millions of dollars, includes 2.5 million development cost).
Question: What prices can the company charge to obtain a profit of at least $190,000,000?
Example 5: Increasing the Profit for a Product
Solution steps:
Verbal Model: Profit = Revenue - Cost
Equation: P = (100x - 10x^2) - (10x + 2.5) = -10x^2 + 90x - 2.5
Inequality: -10x^2 + 90x - 2.5 \ge 190 (profit of at least $190 million)
Rewrite in general form: -10x^2 + 90x - 192.5 \ge 0
Divide by -10 and reverse inequality: x^2 - 9x + 19.25 \le 0
Using techniques from this section (finding zeros, test intervals), the solution set is found to be 3.5 \le x \le 5.5. (See Figure 1.25)
Corresponding prices: Use demand equation p = 100 - 10x
When x = 3.5, p = 100 - 10(3.5) = 100 - 35 = 65
When x = 5.5, p = 100 - 10(5.5) = 100 - 55 = 45
Conclusion: The company can obtain a profit of $190,000,000 or better by charging at least $45 and at most $65 per calculator.
Domain of a Radical Expression
The domain of an expression is the set of all x-values for which the expression is defined.
Inequalities can be used to find the domain of expressions involving radicals, especially even roots.
Example 6: Finding the Domain of an Expression
Problem: Find the domain of the expression \sqrt{64 - 4x^2}.
Solution:
For the expression to have real values, the radicand must be non-negative: 64 - 4x^2 \ge 0
Write in general form: 64 - 4x^2 \ge 0
Divide each side by 4: 16 - x^2 \ge 0
Factor: (4 - x)(4 + x) \ge 0
Critical numbers: x = -4 and x = 4
Test intervals: (-\infty, -4), (-\text{4, 4}), (4, \infty)
Test values:
Interval (-\infty, -4): Choose x = -5. (4 - (-5))(4 + (-5)) = (9)(-1) = -9 (Negative).
Interval (-\text{4, 4}): Choose x = 0. (4 - 0)(4 + 0) = (4)(4) = 16 (Positive).
Interval (4, \infty): Choose x = 5. (4 - 5)(4 + 5) = (-1)(9) = -9 (Negative).
The expression is greater than or equal to 0 in the closed interval [-4, 4].
Conclusion: The domain of \sqrt{64 - 4x^2} is the interval [-4, 4]. (See Figure 1.26)
Section 2.1: Graphs of Equations
Objective: Plot points in the Cartesian plane.
Objective: Use the Distance Formula to find the distance between two points in the coordinate plane, and use the Midpoint Formula to find the midpoint of a line segment joining two points.
Objective: Determine whether a point is a solution of an equation, and sketch the graph of an equation.
Objective: Find the x- and y-intercepts of the graph of an equation.
Objective: Determine the symmetry of a graph.
Objective: Write the equation of a circle in standard form.
The Cartesian Plane
Definition: Also called the rectangular coordinate system, named after René Descartes.
Formation: Formed by two real number lines intersecting at right angles.
x-axis: Horizontal real number line.
y-axis: Vertical real number line.
Origin: Point of intersection of the two axes.
Quadrants: The two axes divide the plane into four parts.
Coordinates: Each point in the plane corresponds to an ordered pair (x, y) of real numbers.
x-coordinate: Represents the directed distance from the y-axis to the point.
y-coordinate: Represents the directed distance from the x-axis to the point.
Notation: (x, y) can denote a point in the plane or an open interval on the real number line; context clarifies meaning.
Example 1: Plotting Points in the Cartesian Plane
Method: To plot (-1, 2), imagine a vertical line through -1 on the x-axis and a horizontal line through 2 on the y-axis. The intersection is the point.
Other points shown (Figure 2.3): (3, 4), (0, 0), (3, 0), and (-2, -3).
The Distance and Midpoint Formulas
Pythagorean Theorem: For a right triangle with hypotenuse length c and legs lengths a and b, a^2 + b^2 = c^2.
The converse is also true: if a^2 + b^2 = c^2, the triangle is a right triangle.
Derivation of Distance Formula:
Consider two points (x1, y1) and (x2, y2).
A right triangle can be formed with vertical side length |y2 - y1| and horizontal side length |x2 - x1|.
By Pythagorean Theorem: d^2 = (x2 - x1)^2 + (y2 - y1)^2 (Absolute values are not needed due to squaring).
Distance Formula: d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} (Choose positive square root).
Usage: The order of assigning points as (x1, y1) and (x2, y2) does not affect the result.
Midpoint Formula: The midpoint of the line segment joining (x1, y1) and (x2, y2) is:
Midpoint = \left( \frac{x1 + x2}{2}, \frac{y1 + y2}{2} \right)
Example 2: Using the Distance and Midpoint Formulas
Problem: Find (a) the distance between, and (b) the midpoint of the line segment joining, (-2, 1) and (3, 4).
Solution: Let (x1, y1) = (-2, 1) and (x2, y2) = (3, 4).
a. Distance:
d = \sqrt{(3 - (-2))^2 + (4 - 1)^2}
d = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} \approx 5.83 (See Figure 2.6)
b. Midpoint:
M = \left( \frac{-2 + 3}{2}, \frac{1 + 4}{2} \right)
M = \left( \frac{1}{2}, \frac{5}{2} \right) = (0.5, 2.5) (See Figure 2.7)
The Graph of an Equation
Solution of an equation: For an equation in variables x and y, a point (a, b) is a solution if substituting x = a and y = b satisfies the equation.
Example 3: Solution of an Equation
Problem: Determine whether (-1, 0) is a solution of y = 2x^2 - 4x - 6.
Solution:
Substitute x = -1 and y = 0:
0 = 2(-1)^2 - 4(-1) - 6
0 = 2(1) + 4 - 6
0 = 2 + 4 - 6
0 = 0
Both sides are equivalent, so (-1, 0) is a solution.
Graph of an equation: The set of all points that are solutions of the equation.
Point-Plotting Method of Graphing:
If possible, isolate one of the variables.
Construct a table of values (solution points).
Plot these points in a rectangular coordinate system.
Connect the points with a smooth curve or line.
Caution: Too few points can misrepresent the graph. Include negative values, zero, and positive values for x.
Example 4: Sketching the Graph of an Equation (Linear)
Problem: Sketch 3x + y = 5.
Solution:
Isolate y: y = 5 - 3x
Table of values:
| x | y = 5 - 3x | Solution Point |
| --- | ---------- | -------------- |
| -1 | 8 | (-1, 8) |
| 0 | 5 | (0, 5) |
| 1 | 2 | (1, 2) |
| 2 | -1 | (2, -1) |
| 3 | -4 | (3, -4) |Plot points and connect. The graph is a line (Figure 2.10).
Example 5: Sketching the Graph of an Equation (Quadratic)
Problem: Sketch y = x^2 - 2.
Solution:
Table of values:
| x | y = x^2 - 2 | Solution Point |
| --- | ----------- | -------------- |
| -2 | 2 | (-2, 2) |
| -1 | -1 | (-1, -1) |
| 0 | -2 | (0, -2) |
| 1 | -1 | (1, -1) |
| 2 | 2 | (2, 2) |Plot points and connect with a smooth curve (Figure 2.11).
Intercepts of a Graph
Definition of Intercepts:
x-intercepts: Points where the graph intersects the x-axis (y = 0).
To find: Let y = 0 and solve for x. (Points are (a, 0))
y-intercepts: Points where the graph intersects the y-axis (x = 0).
To find: Let x = 0 and solve for y. (Points are (0, b))
Note: An intercept may refer to the point or coordinate. A graph can have no intercepts, one, or several (Figure 2.12).
Example 6: Finding x- and y-Intercepts
Problem: Find the x- and y-intercepts of x = y^2 - 3.
Solution:
x-intercept: Let y = 0
x = 0^2 - 3 \implies x = -3
x-intercept: (-3, 0)
y-intercepts: Let x = 0
0 = y^2 - 3 \implies y^2 = 3 \implies y = \pm \sqrt{3}
y-intercepts: (0, \sqrt{3}) and (0, -\sqrt{3}) (See Figure 2.13)
Symmetry
Knowing symmetry helps sketch graphs (fewer points needed).
Types of Symmetry (Figure 2.14):
x-Axis symmetry: Graph is a mirror image across the x-axis. If (x, y) is on graph, (x, -y) is also on graph.
y-Axis symmetry: Graph is a mirror image across the y-axis. If (x, y) is on graph, (-x, y) is also on graph.
Origin symmetry: Graph looks the same after a 180° rotation about the origin. If (x, y) is on graph, (-x, -y) is also on graph.
Tests for Symmetry:
x-axis: Replace y with -y. If equivalent equation, symmetric with x-axis.
y-axis: Replace x with -x. If equivalent equation, symmetric with y-axis.
Origin: Replace x with -x AND y with -y. If equivalent equation, symmetric with origin.
Example (y-axis symmetry): For y = x^2 - 1
Replace x with -x: y = (-x)^2 - 1 \implies y = x^2 - 1
Equation is equivalent, so symmetric with y-axis (Figure 2.15).
Example 7: Testing for Symmetry
Problem: Test y = 2x^3 for symmetry with respect to both axes and the origin.
Solution:
x-axis: Replace y with -y: -y = 2x^3 \implies y = -2x^3. Not equivalent.
y-axis: Replace x with -x: y = 2(-x)^3 \implies y = -2x^3. Not equivalent.
Origin: Replace x with -x and y with -y: -y = 2(-x)^3 \implies -y = -2x^3 \implies y = 2x^3. Equivalent.
Conclusion: Only origin symmetry (Figure 2.16).
Example 8: Using Symmetry as a Sketching Aid
Problem: Use symmetry to sketch the graph of x - y^2 = 1.
Solution:
Test for x-axis symmetry: Replace y with -y: x - (-y)^2 = 1 \implies x - y^2 = 1. Equivalent.
Conclusion: Symmetric with x-axis.
Sketch points above x-axis (e.g., y = \sqrt{x-1} for y \ge 0) and use symmetry to complete (Figure 2.17).
The Equation of a Circle
Equation recognition: An additional graphing aid.
Definition of a circle: All points (x, y) that are a given positive distance r (radius) from a fixed point (h, k) (center).
Derivation (from Distance Formula):
d = \sqrt{(x - h)^2 + (y - k)^2} becomes r = \sqrt{(x - h)^2 + (y - k)^2}
Square both sides:
Standard Form of the Equation of a Circle: (x - h)^2 + (y - k)^2 = r^2
(h, k) is the center, r is the radius.
Special case: Center at origin (h = 0, k = 0): x^2 + y^2 = r^2.
Example 9: Finding the Equation of a Circle
Problem: Point (3, 4) lies on a circle whose center is (-1, 2).
Solution:
1. Find radius (r): Distance between (-1, 2) and (3, 4).
r = \sqrt{(3 - (-1))^2 + (4 - 2)^2}
r = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20}
2. Write equation: Using (h, k) = (-1, 2) and r = \sqrt{20}.
(x - (-1))^2 + (y - 2)^2 = (\sqrt{20})^2
(x + 1)^2 + (y - 2)^2 = 20 (Standard form) (Figure 2.19)
General form of a circle: Expanding (x + 1)^2 + (y - 2)^2 = 20
(x^2 + 2x + 1) + (y^2 - 4y + 4) = 20
x^2 + y^2 + 2x - 4y - 15 = 0 (General form)
General form: Ax^2 + Ay^2 + Dx + Ey + F = 0, where A \ne 0.
Graphing from general form: Use completing the square to convert to standard form.
Example 10: Completing the Square to Sketch a Circle
Problem: Sketch the circle given by 4x^2 + 4y^2 + 20x - 16y + 37 = 0.
Solution:
1. Divide by A (coefficient of x^2 and y^2):
x^2 + y^2 + 5x - 4y + \frac{37}{4} = 0
2. Group x-terms and y-terms:
(x^2 + 5x) + (y^2 - 4y) = -\frac{37}{4}
3. Complete the square for x and y:
For x: add (\frac{5}{2})^2 = \frac{25}{4}.
For y: add (\frac{-4}{2})^2 = (-2)^2 = 4.
(x^2 + 5x + \frac{25}{4}) + (y^2 - 4y + 4) = -\frac{37}{4} + \frac{25}{4} + 4
4. Factor and simplify:
(x + \frac{5}{2})^2 + (y - 2)^2 = -\frac{12}{4} + 4 = -3 + 4 = 1
Standard form: (x + \frac{5}{2})^2 + (y - 2)^2 = 1^2
Center: (-\frac{5}{2}, 2) = (-2.5, 2)
Radius: r = 1$$
Use this information to sketch the circle (Figure 2.20).