Recall that when you multiply or divide both sides of an inequality by a negative number, you must

Transcription

1 Unit 3, Lesson 5.3 Creating Rational Inequalities Recall that a rational equation is an equation that includes the ratio of two rational epressions, in which a variable appears in the denominator of at least one rational epression. Similarly, a rational inequality is an inequality that includes a ratio of two rational epressions, in which a variable appears in a denominator of at least one rational epression, such as. 3 The solutions to rational inequalities are written as intervals. An interval is the set of all real numbers between two given numbers. The two numbers on the ends are the endpoints. The endpoints might or might not be included in the interval depending on whether the interval is open, closed, or half-open/ half-closed. Sometimes you need to create a rational inequality and then use it to solve a real-world problem in which only positive numbers apply to the situation. Recall that when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol. Consider the inequality 8 < 1 : Multiply both sides by : Divide both sides by : 8 < 1 Original inequality 8 < 1 Original inequality 16 > Multiply both sides by > 3 Divide both sides by Reverse the inequality symbol Reverse the inequality symbol To solve an inequality means to find the complete solution set. The method of solving inequalities in this lesson involves using critical numbers. First consider polynomial inequalities. If f() is a polynomial function and an inequality is written in any of these forms: f() < 0, f() 0, f() > 0, or f() 0, then the critical numbers of the polynomial inequality are the -values that make f() = 0. Polynomial functions can change signs only at their critical numbers. The critical numbers for the polynomial function f() = 3 are 1 and 3, as shown below. Notice that between 1 and 3, the value of f() is negative, and outside of this interval, the value of f() is positive. MUnit3Lesson /3/018

2 Now consider rational inequalities. If f() is a rational function and an inequality is written in any of these forms (f() < 0, f() 0, f() > 0, or f() 0), then the critical numbers of the rational inequality are the -values that make f() = 0 or make f() undefined. Similar to polynomial functions, rational functions can change signs only at their critical numbers. The critical numbers for a rational function are the -values that make either the numerator or the denominator equal to 0. The rational function f() and its critical numbers are shown in the graph below. Note 1 that = is a critical number because it makes the numerator equal to 0, and = 1 is a critical number because it makes the denominator equal to 0. Critical numbers separate the -ais into test intervals, intervals on the -ais formed by one or more critical numbers. The sign of the function on the test interval is the same as the sign of the function value at any -value in the interval. The critical numbers and 1 separate the -ais into three test intervals, as shown in the diagram below. MUnit3Lesson5.3 9/3/018

3 Polynomial functions and rational functions can change signs only at their critical numbers. Note: Just because a function has a critical value, it doesn t necessarily change signs. If the sign does change, it will occur at a critical value. To determine the sign of all function values within a test interval, you only need to test one function value and determine its sign. Consider the function f(). 1 If you want to know the sign of the function for all -values greater than 1, test the function 3 value at any -value greater than 1, say = 3: f(3) This function value is positive, so the function is positive for all -values greater than 1. A test interval on the -ais is an eample of an interval on a number line. In this lesson, intervals are represented in three ways: number line graphs, algebraic descriptions, and interval notation. Interval notation is a way of representing an interval on a number line using a pair of parentheses, a pair of brackets, or a parenthesis and a bracket. An eample of an interval on a number line is the set of all real numbers between and 5, but not including or 5. The diagram below shows the three ways of representing this interval. Note: The interval notation (, 5) appears to be eactly the same as the ordered pair notation (, 5). The interval notation (, 5) represents the set of all real numbers between and 5, but not including or 5. The ordered pair notation (, 5) represents an ordered pair of two real numbers; this ordered pair represents a single point in the coordinate plane. The contet of the problem will always make clear which meaning is intended. Depending on the type of interval, interval notation can be written with parentheses ( ), brackets [ ], or a combination of both: [ ) or ( ]. An open interval does not include its endpoints. Eample: (, 5) is an open interval because it does not include or 5. A closed interval includes its endpoints. Eample: [, 5] is a closed interval because it includes and 5. MUnit3Lesson /3/018

4 A half-open (or half-closed) interval includes one endpoint but not the other. Eample: [, 5) is a half-open (or half-closed) interval because it includes, but not 5. Eample: (, 5] is a half-open (or half-closed) interval because it includes 5, but not. The diagram below shows these intervals. An interval that etends to infinity goes on without bound. Infinity is represented by the symbol. (negative infinity) and (positive infinity) are used when representing intervals The symbols that are unbounded on either one side or both sides. These symbols do not represent any particular real number because given any real number, you can always find a lesser or greater number. When an infinity symbol appears in interval notation, it is always written with a parenthesis to indicate an open interval. The diagram below shows intervals using the infinity symbols. The first four intervals are each unbounded on one side; the last interval, all real numbers, is unbounded on both sides. A previous diagram showing critical numbers and test intervals is repeated below, this time including interval notation for each test interval. MUnit3Lesson5.3 9/3/018

5 Interval notation can be used to represent a union of two intervals. The symbol for union is. The diagram below shows a union of two intervals. Common Errors/Misconceptions confusing open interval notation with ordered pair notation neglecting to distribute the negative sign when multiplying a binomial by a negative number forgetting to change the direction of the inequality symbol when multiplying or dividing by a negative number confusing closed and open interval notation Eample 1: Solve Find the critical numbers and use them to identify the test intervals. The critical numbers are the -values that make either the numerator or the denominator equal to 0. 0 Set the numerator equal to 0 Subtract from both sides 1 Divide both sides by 0 Set the denominator equal to 0 Add to both sides The critical numbers are 1 and. For the rational inequality 0, the related rational function is Verify that 1 and are critical numbers. (1) () f(1) and f() , which is undefined f(). MUnit3Lesson /3/018

6 Eample 1 (continued): The critical numbers 1 and form the test intervals (,1), (1,), and (, ).. Determine the sign of the related rational function in each test interval. The related rational function is f(). The sign of f() does not change within a test interval. Choose any -value in each test interval and determine the sign of f() at that value; the function will have that sign on the entire test interval. MUnit3Lesson /3/018

7 Eample 1 (continued): The following summary diagram shows the signs of f() in the test intervals and the values of f() at the critical numbers. 3. Use the summary diagram from step to identify the solution set of the inequality. The diagram shows that greater than. 0 at all -values less than 1, at 1, and at all -values The solution set of the inequality is given in three different but equivalent forms below.. Connect the solution set of the rational inequality to the graph of its related rational function. The following diagram shows the graph of the related function f(). The parts of the graph that correspond to the solution set of 0 are emphasized. MUnit3Lesson /3/018

8 Eample : Solve Write the inequality in an equivalent form with 0 on one side and the related rational function on the other side. 3 Original inequality 3 0 Subtract from both sides 3 ( ) 0 Rewrite the subtraction as an addition 3 ( ) 0 Multiply by a fraction equal to 1 whose denominator is the LCD 3 ( ) 0 Simplify 3 0 Distribute 3 0 Add the fractions 7 0 Simplify The inequality 7 0 is equivalent to the original inequality. 7 The related rational function is f().. Find the critical numbers and use them to identify the test intervals. The critical numbers are the -values that make either the numerator or the denominator equal to Set the numerator equal to 0 0 Set the denominator equal to 0 7 Subtract 7 from both sides Add to both sides 7 Divide both sides by 7 7 The critical numbers are and. Verify that and are critical numbers f () 7 and f() , which is undefined The critical numbers 7 and form the test intervals 7,, 7,, and (, ). MUnit3Lesson /3/018

9 Eample (continued): 3. Determine the sign of the related rational function in each test interval. 7 The related rational function is f(). The sign of f() does not change within a test interval. Choose any -value in each test interval and determine the sign of f() at that value; the function will have that sign on the entire test interval. The following summary diagram shows the signs of f() in the test intervals and the values of f() at the critical numbers.. Use the summary diagram in step 3 to identify the solution set. The diagram shows that 7 0 at all -values between Note: Between means not including the end values. 7 and. The inequality 7 0 is equivalent to the original inequality 3, so the two inequalities have the same solution set. MUnit3Lesson /3/018

10 Eample (continued): The solution set is given in three different but equivalent forms below. 5. Connect the solution set of + 7 <0 to the graph of its related rational function. - 7 The following diagram shows the graph of f(), the related function of the inequality 7 0. The part of the graph that corresponds to the solution set of 7 0 is emphasized. 6. The solution set shown in step is the solution set of both + 7 <0 - inequality + 3 <. Connect the solution set to the graph of the function g() =. - and the original The following diagram shows the graph of 3 g(). MUnit3Lesson /3/018

11 Eample (continued): The part of the graph that corresponds to the solution set is emphasized. Eample 3: Ashrita makes and sells leather goods. One of her new products is a belt. Her startup cost for tools and equipment to make the belts is $160 and it cost her $10 to make each belt. She sells each belt for $18. Ashrita wants to earn an average profit of at least $5 per belt. How many belts could she make and sell to earn an average profit of at least $5 per belt? What will be her minimum profit if she achieves her goal? 1. Define an average profit function to represent the situation. First, define a revenue function R() in dollars, a cost function C() in dollars, and a profit function P() in dollars. R() 18 is the revenue taken in if belts are sold. C() is the cost to make belts. P() R() C() is the total profit from making and selling belts, because profit revenue cost. Now define an average profit function P A() in dollars per belt. P() P A() An average is the total divided by the number of items R() C() P A() Substitute the function rule for P() [(18) (160 10)] P A() Substitute the function rules for R() and C() P A() Simplify The average profit function is P A() is the number of belts made and sold., where P A() is dollars of profit per belt and MUnit3Lesson /3/018

12 Eample 3 (continued):. Using the average profit function, write an inequality that can be used to solve the problem. Ashrita wants to earn an average profit of at least $5 per belt, so the inequality is Write the inequality 5 in an equivalent form with 0 on one side and the related rational function on the other side of the inequality Original inequality Subtract 5 from both sides Multiply 5 by a fraction equal to 1 whose denominator is the LCD Combine the fractions Simplify The inequality is equivalent to the original inequality. The related rational function is f() To solve the inequality 5, you need to solve the inequality 0. Begin by finding the critical numbers and using them to identify the test intervals Inequaltiy Set the numerator equal to Add 160 to both sides Divide both sides by 3 and simplify Set the denominator equal to 0; this is already solved The critical numbers are 0 and 1 0,53 3, and 1 53, ; they form the test intervals (,0), 3 MUnit3Lesson /3/018

13 Eample 3 (continued): 5. Determine the sign of the related rational function in each test interval The related rational function is f(). Choose any -value in each test interval and determine the sign of f() at that value; the function will have that sign on the entire test interval. The following summary diagram shows the signs of f() in the test intervals and the values of f() at the critical numbers. 6. Use the summary diagram in step 5 to identify the solution set of the inequality. The diagram shows that related rational function is at all -values less than 0 and 1 at all -values greater than or equal to The solution set, shown in three forms below, is the solution set of both and because these two inequalities are equivalent. 7. Use the solution set of the inequality to identify the answers to the problem. You need to find how many belts Ashrita could make and sell to earn an average profit of at least $5 per belt. Because represents a number of belts, neither negative numbers nor mied numbers make sense in this situation. MUnit3Lesson /3/018

14 Eample 3 (continued): Based on the positive solutions of the inequality, it seems that the number of belts is a whole 1 number greater than 53 ; that is, any number in the set {5,55,56,57, } Check the apparent answers that you arrived at in step 7. For 53 belts, the cost is (53) , the revenue is 18(53) 95, 690 the profit is , and the average profit per belt is 6.98, which is 53 less than 5. For 5 belts, the cost is (5) , the revenue is 18(5) 97, 700 the profit is , and the average profit per belt is 7 5.0, which is 5 greater than 5. For 55 belts, the cost is (55) , the revenue is 18(55) 990, 710 the profit is , and the average profit per belt is , which is 55 greater than 5. The checks support the conclusion that to earn an average profit of at least $5 per belt, 53 belts is not enough, and 5 or more belts is enough. 9. Summarize your conclusions. To earn an average profit of at least $5 per belt, Ashrita could make and sell any whole 1 number of belts greater than 53 ; that is, any number in the set {5,55,56,57, }. 3 Recall from step 1 that the total profit from making and selling belts is given by the following: P() R() C() 18 (160 10) If she makes and sells 5 belts, her profit will be as follows: P(5) 8(5) If she achieves her goal, her minimum profit will be $7. MUnit3Lesson /3/018

15 Eample 3 (continued): 10. Make an observation about using a rational inequality for a problem situation. If you are using a rational inequality to solve a real-world problem in which only positive numbers apply, you might be able to multiply or divide both sides by an epression containing a variable, keeping in mind that you will not be finding the complete solution set of the inequality. To solve the problem about Ashrita and her belts, you need only find all the positive solutions of the inequality, because represents a number of belts. So, the following method can be used, with the understanding that it is valid only for positive values of Write the inequality Multiply both sides by. The inequality symbol stays the same because is positive Subtract 5 from both sides Add 160 to both sides 160 Divide both sides by Eample : Note that this method should not be use if you need to find the complete solution set of the inequality. The following diagram represents an electrical circuit with an E-volt battery and two resistors of resistances R 1 and R, connected in parallel. The total resistance R in the circuit satisfies the equation. R R R 1 Find the value of R that will make the total resistance at least 3 ohms. 1. Develop a rational inequality that can be solved for R. R 3 The total resistance R must be at least 3 ohms 3 1 R Divide both sides by R. The inequality stays the same because R is positive 1 1 Divide both sides by 3 3 R MUnit3Lesson /3/018

16 Eample (continued): Substitute for 1 3 R1 R R1 R R Substitute for R 1. This is a rational inequality that can be solve for R 3 R. Find the positive values of R that are solutions of the rational inequality Write the rational inequality developed in step 1 3 R R 1R 3 R Multiply both sides by 1R, the LCD The inequality symbol stays the same because 1R is positive R 3R 1 Distribute and simplify R 1 Subtract 3R from both sides 3. Identify answers to the problem. Based on the result in step, any value of R greater than or equal to 1 ohms will make the total resistance R at least 3 ohms.. Check your answers and then write your final answer to the problem. Substitute values for R into the equation to check. Begin by choosing a value less than 1: try If R 11, then R R R ; and if R, then 1 R , which is less than 3 ohms Substitute 1 for R If R 1, then R R R ; and if 1 1, then R 3, which is at least 3 ohms 3 R 3 MUnit3Lesson /3/018

17 Eample (continued): Substitute a value greater than 1: try If R 13, then R R R ; and if R, then 1 R , which is at least 3 ohms The checks support the conclusion that any value of R greater than or equal to 1 ohms will make the value of R at least 3 ohms. Eample 5: Adam has 100 meters of fencing. He wants to fence in a rectangular region in which he can have a garden that is protected from deer. He wants the region to have an area of 50 square meters. What are all the possible widths that Adam can use? Describe two rectangles that Adam can use and eplain why they satisfy the requirements. 1. Write the relevant formulas for a rectangle and draw a sketch. For a rectangle with length l and width w: P l w represents the perimeter A l w represents the area w l. Develop a rational inequality that can be used to find the values of the width w that satisfy the requirements. You know: A 50 The area must be equal to 50 square meters P 100 The perimeter must be less than or equal to 100 meters because Adam only has 100 meters of fencing Use the equation to obtain an epression for l in terms of w. A 50 Given l w 50 Substitute l w for A 50 w l Divide both sides by w to solve for l in terms of w MUnit3Lesson /3/018

18 Eample 5 (continued): Use the inequality to develop a rational inequality. P 100 Given l w 100 Substitute l w for P 50 w 100 w Substitute 50 w for l 50 w 50 Divide both sides by to simplify w 50 w 50 is a rational inequality that represents the situation. It can be used to find the w values of w that satisfy the requirements of the problem. 3. Find the positive solutions of the rational inequality. 50 w 50 w Write the rational inequality developed in step 50 w 50w Multiply both sides by w to eliminate the denominator w 50w 50 0 Subtract 50w from both sides and write the epression on the left in descending order Now you have a polynomial inequality, w 50w 50 0 To solve it, find the critical numbers and then use test intervals. w 50w 50 0 Set the polynomial equal to 0 ( 50) ( 50) (1)(50) w Use the quadratic formula to determine w (1) (50) w Simplify w 50 0 w w w 5 w 105 w w or w w 35.6 or w w 35.5 or w 1.75 The quadratic polynomial function f(w) w 50w 50 has the value 0 at w 1.75 and w These values are the critical numbers. MUnit3Lesson /3/018

19 Eample 5 (continued): The following summary diagram shows a sample function value within each test interval, the sign of f(w) in each interval, and the value of f(w) at the critical numbers. The diagram shows that w-values between and Write the answers to the problem. w 50w 50 0 at w 5 105, at w 5 105, and at all Adam can use any width w in meters that satisfies w The corresponding approimate values are given by 1.75 w Two possible rectangles that Adam can use are shown below; one has a width of 0 meters and the other has a width of 5 meters l l w w l l 0 5 l 6 l 0.8 The requirements are that the area must be 50 square meters and the perimeter must be less than or equal to 100 meters so that Adam has enough fence. A rectangle with a width of 0 meters and a length of 6 meters has an area of 50 square meters and a perimeter of 9 meters. A rectangle with a width of 5 meters and a length of 0.8 meters has an area of 50 square meters and a perimeter of 91.6 meters. Therefore, both rectangles satisfy the requirements. MUnit3Lesson /3/018