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section Q5 | 393
Polynomial and Rational Inequalities
Q.5 Polynomial and Rational Inequalities
In sections L4 and L5 , we discussed solving linear inequalities in one variable as well as solving systems of such inequalities. In this section, we examine polynomial and rational inequalities in one variable. Such inequalities can be solved using either graphical or analytic methods. Below, we discuss both types of methods with a particular interest in the analytic one.
Solving Quadratic Inequalities by Graphing
Definition 5.1 A quadratic inequality is any inequality that can be written in one of the forms 𝒂𝒂𝒙𝒙𝟐𝟐^ + 𝒃𝒃𝒙𝒙 + 𝒄𝒄 > (≥) 𝟎𝟎, or 𝒂𝒂𝒙𝒙𝟐𝟐^ + 𝒃𝒃𝒙𝒙 + 𝒄𝒄 < (≤) 𝟎𝟎, or 𝒂𝒂𝒙𝒙𝟐𝟐^ + 𝒃𝒃𝒙𝒙 + 𝒄𝒄 ≠ 𝟎𝟎 where 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐 are real numbers, with 𝑎𝑎 ≠ 0.
To solve a quadratic inequality, it is useful to solve the related quadratic equation first. For example, to solve 𝑥𝑥 2 + 2𝑥𝑥 − 3 > (≥) 0, or 𝑥𝑥 2 + 2𝑥𝑥 − 3 < (≤) 0, we may consider solving the related equation: 𝑥𝑥 2 + 2𝑥𝑥 − 3 = 0 (𝑥𝑥 + 3)(𝑥𝑥 − 1) = 0 𝑥𝑥 = − 3 or 𝑥𝑥 = 1. This helps us to sketch an approximate graph of the related function 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 2 + 𝑥𝑥 − 2 , as in Figure 1.1. The graph of function 𝑓𝑓 is a parabola that crosses the 𝑥𝑥-axis at 𝑥𝑥 = − 3 and 𝑥𝑥 = 1, and is directed upwards. Observe that the graph extends below the 𝑥𝑥-axis for 𝑥𝑥-values from the interval (−3,1) and above the 𝑥𝑥-axis for 𝑥𝑥-values from the set (−∞, −3) ∪ (1, ∞). This allows us to read solution sets of several inequalities, as listed below.
Note: If the inequalities contain equations (≥, ≤), the 𝑥𝑥-values of the intercepts are included in the solution sets. Otherwise, the 𝑥𝑥-values of the intercepts are excluded from the solution sets.
Solving Polynomial Inequalities
Definition 5.2 A polynomial inequality is any inequality that can be written in one of the forms 𝑷𝑷(𝒙𝒙) > (≥)^ 𝟎𝟎, or 𝑷𝑷(𝒙𝒙) < (≤)^ 𝟎𝟎, or 𝑷𝑷(𝒙𝒙) ≠ 𝟎𝟎 where 𝑃𝑃(𝑥𝑥) is a polynomial with real coefficients.
Inequality Solution Set 𝒙𝒙𝟐𝟐^ + 𝟐𝟐𝒙𝒙 − 𝟑𝟑 > 𝟎𝟎 (−∞,^ −^3 )^ ∪^ (^1 ,^ ∞) 𝒙𝒙𝟐𝟐^ + 𝟐𝟐𝒙𝒙 − 𝟑𝟑 ≥ 𝟎𝟎 (−∞,^ −^3 ]^ ∪^ [^1 ,^ ∞) 𝒙𝒙𝟐𝟐^ + 𝟐𝟐𝒙𝒙 − 𝟑𝟑 < 𝟎𝟎 (− 3 , 1 ) 𝒙𝒙𝟐𝟐^ + 𝟐𝟐𝒙𝒙 − 𝟑𝟑 ≤ 𝟎𝟎 [−^3 ,^1 ] 𝒙𝒙𝟐𝟐^ + 𝟐𝟐𝒙𝒙 − 𝟑𝟑 ≠ 𝟎𝟎 ℝ^ ∖^ {−^3 ,^1 }
Figure 1.
𝑥𝑥
𝒇𝒇(𝒙𝒙) = 𝒙𝒙𝟐𝟐^ + 𝟐𝟐𝒙𝒙 − 𝟑𝟑
1 1
roots of f(x)
− 3 𝑓𝑓(𝑥𝑥) < 0
𝑓𝑓(𝑥𝑥) < 0 𝑓𝑓(𝑥𝑥) > 0
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Note: Linear or quadratic inequalities are special cases of polynomial inequalities.
Polynomial inequalities can be solved graphically or analytically, without the use of a graph. The analytic method involves determining the sign of the polynomial by analysing signs of the polynomial factors for various 𝑥𝑥-values, as in the following example.
Solving Polynomial Inequalities Using Sign Analysis
Solve each inequality using sign analysis.
a. (𝑥𝑥 + 1)(𝑥𝑥 − 2)𝑥𝑥 > 0 b. 2 𝑥𝑥 4 + 8 ≤ 10 𝑥𝑥 2
a. The solution set of (𝑥𝑥 + 1)(𝑥𝑥 − 2)𝑥𝑥 > 0 consists of all 𝑥𝑥-values that make the product (𝑥𝑥 + 1)(𝑥𝑥 − 2)𝑥𝑥 positive. To analyse how the sign of this product depends on the 𝑥𝑥- values, we can visualise the sign behaviour of each factor with respect to the 𝑥𝑥-value, by recording applicable signs in particular sections of a number line. For example, the expression 𝑥𝑥 + 1 changes its sign at 𝑥𝑥 = − 1. If 𝑥𝑥 > − 1 , the expression 𝑥𝑥 + 1 is positive, so we mark “+” in the interval (−1, ∞). If 𝑥𝑥 < − 1 , the expression 𝑥𝑥 + 1 is negative, so we mark “−” in the interval (−∞, −1).
So, the sign behaviour of the expression 𝑥𝑥 + 1 can be recorded on a number line, as below.
A similar analysis can be conducted for the remaining factors, 𝑥𝑥 − 2 and 𝑥𝑥. These expressions change their signs at 𝑥𝑥 = 2 and 𝑥𝑥 = 0, correspondingly. The sign behaviour of all the factors can be visualised by reserving one line of signs per each factor, as shown below.
The sign of the product (𝑥𝑥 + 1)(𝑥𝑥 − 2)𝑥𝑥 is obtained by multiplying signs in each column. The result is marked beneath the number line, as below.
Solution
Here, the zeros of polynomials are referred to as critical numbers. This is because the polynomials may change their signs at these numbers.
Remember to write the critical numbers in increasing order!
− + +^ +
− + +^ +
product 0 − 2 + solution
− + +^ +
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Quadratic Equations and Functions
its sign, it is enough to evaluate it for an easy to calculate test number from this interval. For instance, when 𝑥𝑥 = 0 ∈ (−1,1), the value of 𝑥𝑥 − 2 is negative. This means that all the values of 𝑥𝑥 − 2 are negative between − 1 and 1 .)
Finally, underneath each column, we record the sign of the product and graph the solution set to the inequality (𝑥𝑥 − 2)(𝑥𝑥 + 2)(𝑥𝑥 − 1)(𝑥𝑥 + 1)^ ≤ 0
Note: Since the inequality contains an equation, the endpoints of the resulting intervals belong to the solution set as well. Hence, they are marked by filled- in circles and notated with square brackets in interval notation.
So, the solution set is [−𝟐𝟐, −𝟏𝟏] ∪ [𝟏𝟏, 𝟐𝟐].
Solving Rational Inequalities
Definition 5.3 A rational inequality is any inequality that can be written in one of the forms 𝑷𝑷(𝒙𝒙) 𝑸𝑸(𝒙𝒙) > (≥)^ 𝟎𝟎,^ or^
𝑷𝑷(𝒙𝒙) 𝑸𝑸(𝒙𝒙) < (≤)^ 𝟎𝟎, or^
𝑷𝑷(𝒙𝒙) 𝑸𝑸(𝒙𝒙) ≠^ 𝟎𝟎 where 𝑃𝑃(𝑥𝑥) and 𝑄𝑄(𝑥𝑥) are polynomials with real coefficients.
Rational inequalities can be solved similarly as polynomial inequalities. To solve a rational inequality using the sign analysis method, we need to make sure that one side of the inequality is zero and the other side is expressed as a single algebraic fraction with a completely factored numerator and denominator.
Solving Rational Inequalities Using Sign Analysis
Solve each inequality using sign analysis.
a. (𝑥𝑥−2)𝑥𝑥 𝑥𝑥+1 ≥^0 b.^
4−𝑥𝑥 𝑥𝑥+2 ≥ 𝑥𝑥
a. The right side of the inequality (𝑥𝑥−2𝑥𝑥+1) 𝑥𝑥> 0 is zero, and the left side is a single fraction with both numerator and denominator in factored form. So, to solve this inequality, it is enough to analyse signs of the expression (𝑥𝑥−2)𝑥𝑥 𝑥𝑥+1 at different intervals of the domain.
Solution
product − 1 + − 2 solution
section Q5 | 397
Polynomial and Rational Inequalities
These intervals are determined by the critical numbers (the zeros of the numerator and denominator), which are − 1 , 0 , and 2.
As indicated in the above table of signs, the solution set to the inequality (𝑥𝑥−2𝑥𝑥+1) 𝑥𝑥≥ 0 contains numbers between − 1 and 0 and numbers higher than 2. In addition, since the inequality includes an equation, 𝑥𝑥 = 0 and 𝑥𝑥 = 2 are also solutions. However, 𝑥𝑥 = − 1 is not a solution because − 1 does not belong to the domain of the expression (𝑥𝑥−2)𝑥𝑥 𝑥𝑥+ since it would make the denominator 0. So, the solution set is (−𝟏𝟏,𝟎𝟎]^ ∪ [𝟐𝟐, ∞).
Attention: Solutions to a rational inequality must belong to the domain of the inequality. This means that any number that makes the denominator 0 must be excluded from the solution set.
b. To solve 4−𝑥𝑥 𝑥𝑥+2 ≥ 𝑥𝑥^ by the sign analysis method, first, we would like to keep the right side equal to zero. So, we rearrange the inequality as below. 4 − 𝑥𝑥 𝑥𝑥 + 2
When multiplying by a negative number, remember to reverse the inequality sign!
When working with inequalities, avoid multiplying by the denominator as it can be positive or negative for different x-values!
− + +^ +
product 0 − 2 + solution
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Polynomial and Rational Inequalities
a. First, notice that the left side of the inequality (3𝑥𝑥 + 2)^2 > − 1 is a perfect square and as such, it assumes a nonnegative value for any input 𝑥𝑥. Since a nonnegative quantity is always bigger than − 1 , the inequality is satisfied by any real number 𝑥𝑥. So, the solution set is ℝ.
Note: The solution set of an inequality that is always true is the set of all real numbers, ℝ. For example, inequalities that take one of the following forms
𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒂𝒂𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏 > 𝒏𝒏𝒏𝒏𝒏𝒏𝒂𝒂𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏
𝒑𝒑𝒏𝒏𝒑𝒑𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏 ≥ 𝒏𝒏𝒏𝒏𝒏𝒏𝒂𝒂𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏
𝒑𝒑𝒏𝒏𝒑𝒑𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏 > 𝒏𝒏𝒏𝒏𝒏𝒏𝒑𝒑𝒏𝒏𝒑𝒑𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏
𝒑𝒑𝒏𝒏𝒑𝒑𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏 > 𝟎𝟎
𝒏𝒏𝒏𝒏𝒏𝒏𝒂𝒂𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏 < 𝟎𝟎
are always true. So their solution set is ℝ.
b. Since the left side of the inequality (𝑥𝑥−4)^2 𝑥𝑥 2 ≤^0 is a perfect square, it is bigger or equal to zero for all 𝑥𝑥-values. So, we have
(𝑥𝑥 − 4)^2
which can be true only if (𝑥𝑥−4)^2 𝑥𝑥 2 = 0. Since a fraction equals to zero only when its numerator equals to zero, the solution to the last equation is 𝑥𝑥 = 4. Thus, the solution set for the original inequality is {𝟒𝟒}.
Observation: Notice that the inequality (𝑥𝑥−4)^2 𝑥𝑥 2 < 0^ has no solution as a perfect square is never negative.
Note: The solution set of an inequality that is never true is the empty set, ∅. For example, inequalities that take one of the following forms
𝒑𝒑𝒏𝒏𝒑𝒑𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏 (𝒏𝒏𝒐𝒐 𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒂𝒂𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏) ≤ 𝒏𝒏𝒏𝒏𝒏𝒏𝒂𝒂𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏
𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒂𝒂𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏 < 𝒏𝒏𝒏𝒏𝒏𝒏𝒑𝒑𝒏𝒏𝒑𝒑𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏
𝒑𝒑𝒏𝒏𝒑𝒑𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏 ≤ 𝒏𝒏𝒏𝒏𝒏𝒏𝒑𝒑𝒏𝒏𝒑𝒑𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏
𝒑𝒑𝒏𝒏𝒑𝒑𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏 ≤ 𝟎𝟎 or 𝒏𝒏𝒏𝒏𝒏𝒏𝒂𝒂𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏 ≥ 𝟎𝟎
𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒂𝒂𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏 < 𝟎𝟎 or 𝒏𝒏𝒏𝒏𝒏𝒏𝒑𝒑𝒏𝒏𝒑𝒑𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏 > 𝟎𝟎
are never true. So, their solution sets are ∅.
Solution
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Quadratic Equations and Functions
Polynomial and Rational Inequalities in Application Problems
Some application problems involve solving polynomial or rational inequalities.
Finding the Range of Values Satisfying the Given Condition
An outerwear manufacturer determines that its weekly revenue, 𝑅𝑅, for selling a rain parka at a price of 𝑝𝑝 dollars is modeled by the function 𝑅𝑅(𝑝𝑝) = 150𝑝𝑝 − 𝑝𝑝 2. What range of prices for the parka will generate a weekly revenue of at least $5000?
Since the revenue must be at least $5000, we can set up the inequality
𝑅𝑅(𝑝𝑝) = 150𝑝𝑝 − 𝑝𝑝 2 ≥ 5000,
and solve it for 𝑝𝑝. So, we have
−𝑝𝑝 2 + 150𝑝𝑝 − 5000 ≥ 0
𝑝𝑝 2 − 150 𝑝𝑝 + 5000 ≤ 0
(𝑝𝑝 − 50)(𝑝𝑝 − 100) ≤ 0
Since the left-hand side expression represents a directed upwards parabola with roots at 𝑝𝑝 = 50 and 𝑝𝑝 = 100, its graph looks like in the accompanying figure. The graph extends below the 𝑝𝑝-axis for 𝑝𝑝-values between 50 and 100. So, to generate weekly revenue of at least $5000, the price 𝑝𝑝 of a parka must take a value within the interval [50,100].
Q.5 Exercises
Concept Check True or False.
1. To determine if the value of an expression is greater than or less than 0 in a given interval, a test number can be used. 2. If the solution to the inequality 𝑃𝑃(𝑥𝑥) ≥ 0 , where 𝑃𝑃(𝑥𝑥) is a polynomial with real coefficients, is [2, 5), then the solution to the inequality 𝑃𝑃(𝑥𝑥) < 0 is (−∞, 2) ∪ [5, ∞). 3. The inequalities (𝑥𝑥 − 1)(𝑥𝑥 + 3) ≤ 0 and ((𝑥𝑥−1𝑥𝑥+3)) ≤ 0 have the same solutions. 4. The solution set of the inequality (𝑥𝑥 − 1)^2 > 0 is the set of all real numbers. 5. The inequality 𝑥𝑥 2 + 1 ≤ 0 has no solution. 6. The solution set of the inequality (𝑥𝑥−1)^2 (𝑥𝑥+1)^2 ≥^0 is the set of all real numbers.
Solution
(^50 100) 𝑝𝑝
𝑅𝑅
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Quadratic Equations and Functions
(𝑥𝑥+2)(𝑥𝑥−1) (𝑥𝑥+4)^2 ≥^0 53.^
𝑥𝑥 2 −2𝑥𝑥− 𝑥𝑥 2 +10𝑥𝑥+25 > 0^ 54.^
𝑥𝑥 2 −4𝑥𝑥 𝑥𝑥 2 −𝑥𝑥−6 ≤^0
Solve each inequality.
55. (4 − 3 𝑥𝑥)^2 ≥ − 2 56. (5 + 2𝑥𝑥)^2 < − 1 57. (1−2𝑥𝑥)
2 2𝑥𝑥 4 ≤^0
58. (1−2𝑥𝑥)^2 (𝑥𝑥+2)^2 >^ −^3 59.^
−2𝑥𝑥 2 (𝑥𝑥+2)^2 ≥^0 60.^
−𝑥𝑥 2 (𝑥𝑥−3)^2 < 0
Analytic Skills Solve each problem.
60. A penny is tossed upwards with an initial velocity of 48 ft/sec. The height, ℎ, of the penny relative to the point of release 𝑡𝑡 seconds after it is tossed is modeled by the function ℎ(𝑡𝑡) = − 16 𝑡𝑡 2 + 48𝑡𝑡. For what interval of time is the penny above the point of release? 61. A furniture maker determines that the weekly cost, 𝐶𝐶, for producing 𝑥𝑥 accent tables is given by the function 𝐶𝐶(𝑥𝑥) = 2𝑥𝑥 2 − 60 𝑥𝑥 + 900. How many accent tables can be produced to keep the weekly cost under $500? 62. A gardener wants to enclose a rectangular flower bed using 90 ft of fencing. For what range of lengths will the area exceed 450 ft 2? 63. A company determines that its average cost 𝐶𝐶, in dollars, for selling 𝑥𝑥 units of a product is modeled by the function 𝐶𝐶(𝑥𝑥) = 870+2𝑥𝑥𝑥𝑥. For what number of units 𝑥𝑥 will the average cost be less than $8? 64. A publishing company had a revenue of $18 million last year. The company’s financial analyst uses the formula 𝑃𝑃(𝑅𝑅) = 100𝑅𝑅−1800𝑅𝑅 to determine the percent growth, 𝑃𝑃, in the company’s revenue this year over last year, where 𝑅𝑅 is in millions of dollars. For what revenues 𝑅𝑅 will the company’s revenue grow by more than 10%?
