MAT 171 Precalculus Algebra Dr. Claude Moore Cape Fear Community College CHAPTER 4: Polynomial and Rational Functions 4.1 Polynomial Functions and Models 4.2 Graphing Polynomial Functions 4.3 Polynomial Division; The Remainder and Factor Theorems 4.4 Theorems about Zeros of Polynomial Functions 4.5 Rational Functions 4.6 Polynomial and Rational Inequalities 4.6 Polynomial and Rational Inequalities Solve polynomial and rational inequalities. Polynomial Inequalities A quadratic inequality can be written in the form ax 2 + bx + c > 0, where the symbol > could be replaced with either <,, or. Solve: x 2 3x 4 > 0. Let s look at the graph of f(x) = x 2 3x 4. A quadratic inequality is one type of polynomial inequality. s of polynomial inequalities: The zeros are 4 and 1. Thus the x intercepts of the graph are (0, 1) and (4, 0). 1
The zeros divide the x axis into three intervals: 1 4 The sign of the function is the same for all values of x in a given interval. We can choose a test value for x from each interval and find the sign of f(x). Test values: f( 2) = 6 f(0) = 4 f(5) = 6 + + 1 4 The solution set consists of the intervals where the sign of f(x) is positive. So the solution set is {x x < 1 or x > 4}. To Solve a Polynomial Inequality 1. Find an equivalent inequality with P(x) on one side and 0 on one side. 2. Solve the related polynomial equation; that is solve for f(x) = 0. 3. Use the solutions to divide the x axis into intervals. Then select a test value from each interval and determine the polynomial s sign on the interval. 4. Determine the intervals for which the inequality is satisfied and write interval notation or set builder notation for the solution set. Include the endpoints of the intervals in the solution set if the inequality symbol is or. Solve: 4x 3 7x 2 15x. We need to find all the zeros of the function so we solve the related equation. continued The zeros divide the x axis into four intervals. For all x values within a given interval, the sign of 4x 3 7x 2 15x 0 must be either positive or negative. To determine which, we choose a test value for x from each interval and find f(x). You may use the TI calculator as below. The zeros are 5/4, 0, and 3. Thus the x intercepts of the graph are ( 5/4, 0), (0, 0), and (3, 0). Since we are solving 4x 3 7x 2 15x 0, the solution set consists of only two of the four intervals, those in which the sign of f(x) is negative {x < x < 5/4 or 0 < x < 3}. Shade(Y1,Y2) shades the area where Y1 < Y2. 2
To Solve a Rational Inequality 1. Find an equivalent inequality with 0 on one side. 2. Change the inequality symbol to an equals sign and solve the related equation, that is, solve f (x) = 0. 3. Find the values of the variable for which the related rational function is not defined. 4. The numbers found in steps (2) and (3) are called critical values. Use the critical values to divide the x axis into intervals. Then test an x value from each interval to determine the function s sign in that interval. 5. Select the intervals for which the inequality is satisfied and write interval notation or set builder notation for the solution set. If the inequality symbol is or, then the solutions to step (2) should be included in the solution set. The x values found in step (3) are never included in the solution set. Solve f(x) =. The denominator tells us that f(x) is not defined for x = 1 and x = 1. Next, solve the related equation. NOTE: Since the fraction equals zero, the numerator must equal zero. x + 3 = 0 gives x = 3. continued The critical values are 3, 1, and 1. These values divide the x axis into four intervals. We use a test value to determine the sign of f(x) in each interval. continued f(x) = Function values are positive in the intervals ( 3, 1) and (1, ). Since 1 is not in the domain of f(x), it cannot be part of the solution set. Note that 3 does satisfy the inequality. The solution set is [ 3, 1) (1, ). Above graph is with Zoom 6. Below is [ 10, 10, 1, 1]. Shades area where Y2 Y1. x intercept ( 3,0); vertical asymptotes at x = 1 and x = 1. As shown in the graph Y2 Y1 on the intervals [ 3, 1) and (1, ). 3
368/2. For the function f(x) = x 2 + 2x 15, solve f(x) < 0. 368/9. For the function f(x) = (x 2) / (x + 4), solve g(x) 0. Function values are negative only in the interval 5, 3. The solution interval is 5, 3. The critical value 4 is not included because it is not in the domain of the function. The solution set is, 4 [2,. 368/14. For the function h(x) = 7x / ((x 1)(x + 5)), solve h(x) > 0. 369/18. For the function g(x) = x 5 9x 3, solve g(x) 0. Set the denominator equal to zero to find x = 5 and x = 1 as vertical asymptotes. Set the numerator equal to zero to find x = 0 and get 0, 0 as x intercept. Set x = 0 to get y = 0. So 0, 0 is y intercept. Put open circles at V.A. x = 5 and x = 1. Since h x > 0, put an open circle at x intercept 0, 0. If we had h x 0, we would put a closed circle at (0, 0). Identify the sections of the graph above the x axis because h x > 0. 5 0 1 The solution set is 5, 0 1,. The solution set is, 3] [0, 3]. 4
369/24. A related function is graphed. Solve the given inequality: 8 / (x 2 4) < 0. 369/30. Solve x 2 + 4x + 7 5x + 9. The solution set is 2, 2. The solution set is, 1 ] [2, ]. 369/40. Solve 2x 3 x 2 < 5x. Rewrite to get zero on one side of the inequality. 2x 3 x 2 5x < 0 Graph Y = 2x 3 x 2 5x and find where the graph is below the x axis. 369/46. Solve x 5 + 24 > 3x 3 + 8x 2. Rewrite to get zero on one side of the inequality. x 5 3x 3 8x 2 + 24 > 0 Graph Y = x 5 3x 3 8x 2 + 24 and find where the graph is above the x axis. This can be factored as x 3 x 2 3 8 x 2 3 = 0 to get x 3 8 x 2 3 = 0 and solve for x = 2, x = 3, x = 3. The solution set is approximately (, 1.3508) (0, 1.8508). The solution set is ( 3, 3) (2, ). 5
369/54. Solve 1 / (x 3) 0. 370/60. Solve (x + 1) / (x 2) + (x 3) / (x 1) < 0. The solution set is (, 3). The solution set is (1, 2). 370/68. Solve 3 / (x 2 4) 5 / (x 2 + 7x + 10). Get zero on one side of inequality: 3 / (x 2 4) - 5 / (x 2 + 7x + 10) 0. Graph the new function and find where the graph is below or on the x axis. From the first fraction, we find x = 2 and x = 2 as V.A. From the second fraction, we find another V.A. is x = 5. Set the two fraction equal and solve to get x = 2 or x = 12.5. However, x = 2 is a V.A. and can not be used. The only x intercept is (12.5, 0) which the calculator does not show without changing the WINDOW setting. The graph crossing the H.A. and comes back up from beneath it. The solution set is (, 5) ( 2, 2) [12.5, ). 370/72. Solve 3 / (x 2 + 3) > 3 / (5 + 4x 2 ) The solution set is (, ). 6
370/78. Population Growth. The population P, in thousands, of Lordsburg is given by P(t) = 500t / (2t 2 + 9), where t is the time, in months. Find the interval on which the population was 40,000 or greater. (See Exercise 81 in Exercise Set 4.5.) NOTE: 40,000 would be represented by 40 because P is in thousands. The solution set is [0.830, 5.420]. 80t 2 500t + 360 0 8t 2 50t + 36 0 Use the quadratic formula to get solutions of x = 0.8303 or x = 5.4196 rounded to four decimals. Check to the left of x = 0.8303; between the two values; and to the right of x = 5.4196. For x = 0, we get 0 40 false. For x = 1, we get 500/11 40 true. For x = 10, we get 5000/209 40 false. So, the solution is [0.8303, 5.4196]. 371/82. Number of Handshakes. If there are n people in a room, the number N of possible handshakes by all the people in the room is given by the function N(n) = n(n 1) / 2. For what number n of people is 66 N 300? The solution set of the original inequality is {n n 12 and 0 < n 25}, or in simplified form the solution set is {n 12 n 25}. The solution set of the original inequality is {n n 12 and 0 < n 25}, or in simplified form the solution set is {n 12 n 25}. 7