3.5 Polynomial & Rational Inequalities
December 01, 2008
3.5 Polynomial & Rational Inequalities
Objectives:
>Solve polynomial inequalities algebraically & graphically
>Solve rational inequalities algebraically & graphically
Nov 25­11:29 AM
1
3.5 Polynomial & Rational Inequalities
December 01, 2008
Polynomial Inequalities
Example #1: x2 < ­7x ­ 12 Solve and graph the solution set.
Hint: We are trying to find where the parabola is equal to or below the line.
Step 1: Write the inequality so that the polynomial is on the left and zero is on the right.
x2 < ­7x ­ 12
x2 + 7x + 12 < 0 (Therefore, we want to know when the function is negative.)
Step 2: Determine the numbers at which the expression f on the left side equals zero. Hint: FACTOR
x2 + 7x + 12 = 0
(x + 3)(x + 4) = 0
x = ­3 x = ­4
x = ­4 x = ­3 (We will need these in this order in the next step.)
Step 3: Use the numbers found in Step 2 to separate the real number line into intervals.
x = ­4 x = ­3
(­∞, ­4) (­4, ­3) (­3, ∞)
Nov 25­11:31 AM
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3.5 Polynomial & Rational Inequalities
December 01, 2008
Example #1: x2 < ­7x ­ 12 Solve and graph the solution set.
Step 4: Select a number in each interval and evaluate f at the number.
a) If f is positive, then f(x)>0 for all numbers x in the interval.
b) If f is negative, then f(x)<0 for all numbers x in the interval.
Remember: f(x) = x2 + 7x + 12
Interval
(­∞, ­4) (­4, ­3) Number
­5
­3.5
Value of f f(­5) = 2 f(­3.5) = ­0.25
Conclusion Positive
Negative
(­3, ∞)
­2
f(­2) = 2
Positive
Since we want to know where f(x) is negative (remember: x2 + 7x + 12 < 0), we conclude that f(x)<0 for all x such that ­4 < x < ­3.
ANSWER: [­4, ­3] ­5
­4
­3
­2
­1
0
1
2
3
4
5
Nov 25­2:04 PM
3
3.5 Polynomial & Rational Inequalities
December 01, 2008
Example #2: x4 > 1 Solve and graph the solution set.
Step 1: Write the inequality so that the polynomial is on the left and zero is on the right.
x4 > 1
x4 ­ 1 > 0 (Therefore, we want to know when the function is positive.)
Step 2: Determine the numbers at which the expression f on the left side equals zero. x4 ­ 1 = 0
(x2 + 1)(x2 ­ 1) = 0
(x2 + 1)(x ­ 1)(x + 1) = 0
x = 1 x = ­1
Step 3: Use the numbers found in Step 2 to separate the real number line into intervals.
x = 1 x = ­1
(­∞, ­1) (­1, 1) (1, ∞)
Nov 25­11:33 AM
4
3.5 Polynomial & Rational Inequalities
December 01, 2008
Example #2: x4 > 1 Solve and graph the solution set.
Step 4: Select a number in each interval and evaluate f at the number.
Remember: f(x) = x4 ­ 1
Interval
(­∞, ­1) Number
­2
Value of f f(­2) = 15
Conclusion Positive
(­1, 1) 0
f(0) = ­1
Negative
(1, ∞)
2
f(2) = 15
Positive
Since we want to know where f(x) is positive (remember: x4 ­ 1 > 0), we conclude that f(x)>0 for all x such that x < ­1 or x > 1.
ANSWER: (­∞, ­1) or (1, ∞)
­5
­4
­3
­2
­1
0
1
2
3
4
5
Nov 25­2:18 PM
5
3.5 Polynomial & Rational Inequalities
December 01, 2008
Rational Inequalities
Example #3: Solve and graph the solution set.
Step 1: Write the inequality so that the rational is on the left and zero is on the right. Be sure to write the left side as a single quotient.
Step 2: Determine the numbers at which the expression f on the left side equals zero and the numbers at which the rational is undefined.
zero of f
f is undefined
­x ­ 2 = 0
2x ­ 2 = 0
­x = 2
2x = 2
x = ­2
x = 1
Step 3: Use the numbers found in Step 2 to separate the real number line into intervals.
x = ­2 x = 1
(­∞, ­2) (­2, 1) (1, ∞)
Nov 25­11:35 AM
6
3.5 Polynomial & Rational Inequalities
December 01, 2008
Solve and graph the solution set.
Example #3: Step 4: Select a number in each interval and evaluate f at the number.
a) If f is positive, then f(x)>0 for all numbers x in the interval.
b) If f is negative, then f(x)<0 for all numbers x in the interval.
Interval
(­∞, ­2) Number
­3
Value of f f(­3) = ­1/8
Conclusion Negative
(­2, 1) 0
f(0) = 1
Positive (1, ∞)
2
f(2) = ­2
Negative
Since we want to know where f(x) is negative, we conclude that f(x)<0 for all x such that x < ­2 or x > 1.
ANSWER: (­∞, ­2] or [1, ∞) ­5
­4
­3
­2
­1
0
1
2
3
4
5
Nov 28­10:44 AM
7
3.5 Polynomial & Rational Inequalities
December 01, 2008
HW: p. 217/3­54 mult. of 3 Nov 25­11:36 AM
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