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In Problems 85-88, each equation has a solution r in the interval indicated. Use the method of Example 10 to approximate this solution correct to two decimal places. 85. &r - 2r + 5x -1- 0;0 srsl 87. 2r + 6r- 8x + 2- 0; -5srs-4 86. x + 8r - x+ 2 - 0; -1 srs0 88. 3x - 10x + 9 -0; -3 srs-2 In Problems 89-92, each polynomial function has exactly one positive real zero. Use the method of Example 10 to approximate the zero correct to two decimal places. 89. f(x) -x' +x +x - 4 91. f(x) - 2r - 3x - 4x - 8 90. f(x) - 2 +-1 92. f(x) - 3x - 2r - 20

In Problems 85-88, each equation has a solution r in the interval indicated. Use the method of Example 10 to approximate this solution correct to two decimal places. 85. &r - 2r + 5x -1- 0;0 srsl 87. 2r + 6r- 8x + 2- 0; -5srs-4 86. x + 8r - x+ 2 - 0; -1 srs0 88. 3x - 10x + 9 -0; -3 srs-2 In Problems 89-92, each polynomial function has exactly one positive real zero. Use the method of Example 10 to approximate the zero correct to two decimal places. 89. f(x) -x' +x +x - 4 91. f(x) - 2r - 3x - 4x - 8 90. f(x) - 2 +-1 92. f(x) - 3x - 2r - 20

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Please help me with 85,91 please step by step
wone or S STEP 5: Continue with Steps 3 and 4 until the desired accuracy is achieved. Note: If at Step 3 the value of f equals 0, the process ends since that value is a zero. EXAMPLE 10 Approximating a Real Zero of a Polynomial Function bvice o Find the positive real zero of f(x) = x – x³ – 1 correct to two decimal places. 2me Solution From Example 9 we know that the positive real zero is between 1 and 2. Divide the interval [1,2] into 10 equal subintervals: [1, 1.1], [1.1, 1.2], [1.2, 1.3], [1.3, 1.4], [1.4, 1.5], [1.5, 1.6], [1.6, 1.7], [1.7, 1.8], [1.8, 1.9], [1.9,2]. Now find the value of f(x) = x³ – x³ – 1 at each endpoint until the Intermediate Value Theorem applies. f(1.0) = -1 f(1.1) = -0.72049 f(1.2) = -0.23968 f(1.3) = 0.51593 We can stop here and conclude that the zero is between 1.2 and 1.3. Now divide the interval [ 1.2, 1.3] into 10 equal subintervals and evaluate f at each endpoint. nal Zeros f(1.20) = -0.23968 f(1.21) ~ -0.1778185 f(1.22) ~ -0.1131398 f(1.23) - -0.0455613 f(1.24) ~ 0.025001 The zero lies between 1.23 and 1.24, and so, correct to two decimal places, the zero is 1.23. NORMAL FLOAT AUTO REAL RADIAN MP CALC ZERO Y1=X^5-X3-1 Exploration dei We examine the polynomial function f given in Example 10. The Theorem on Bounds of Zeros tells us that every real zero is between -1 and 2. Graphing f using -1 < x < 2 (see Figure 50), we see fei that f has exactly one x-intercept. Using ZERO or ROOT, we find this zero to be 1.23 correct to two decimal places. Zero X-1.2365057 Y=0 Figure 50 TI-84 Plus C - Now Work PROBLEM 91 COMMENT The TABLE feature of a graphing calculator makes the computations in the solution to Example 10 a lot easier. ( There are many other numerical techniques for approximating a real zero of a polynomial. The one outlined in Example 10 (a variation of the bisection method) has the advantages that it always works, it can be programmed on a computer, and each time it is used, another decimal place of accuracy is achieved. See Problem 118 for the bisection method, which places the zero in a succession of intervals, with each new interval being half the length of the preceding one.
Transcribed Image Text:wone or S STEP 5: Continue with Steps 3 and 4 until the desired accuracy is achieved. Note: If at Step 3 the value of f equals 0, the process ends since that value is a zero. EXAMPLE 10 Approximating a Real Zero of a Polynomial Function bvice o Find the positive real zero of f(x) = x – x³ – 1 correct to two decimal places. 2me Solution From Example 9 we know that the positive real zero is between 1 and 2. Divide the interval [1,2] into 10 equal subintervals: [1, 1.1], [1.1, 1.2], [1.2, 1.3], [1.3, 1.4], [1.4, 1.5], [1.5, 1.6], [1.6, 1.7], [1.7, 1.8], [1.8, 1.9], [1.9,2]. Now find the value of f(x) = x³ – x³ – 1 at each endpoint until the Intermediate Value Theorem applies. f(1.0) = -1 f(1.1) = -0.72049 f(1.2) = -0.23968 f(1.3) = 0.51593 We can stop here and conclude that the zero is between 1.2 and 1.3. Now divide the interval [ 1.2, 1.3] into 10 equal subintervals and evaluate f at each endpoint. nal Zeros f(1.20) = -0.23968 f(1.21) ~ -0.1778185 f(1.22) ~ -0.1131398 f(1.23) - -0.0455613 f(1.24) ~ 0.025001 The zero lies between 1.23 and 1.24, and so, correct to two decimal places, the zero is 1.23. NORMAL FLOAT AUTO REAL RADIAN MP CALC ZERO Y1=X^5-X3-1 Exploration dei We examine the polynomial function f given in Example 10. The Theorem on Bounds of Zeros tells us that every real zero is between -1 and 2. Graphing f using -1 < x < 2 (see Figure 50), we see fei that f has exactly one x-intercept. Using ZERO or ROOT, we find this zero to be 1.23 correct to two decimal places. Zero X-1.2365057 Y=0 Figure 50 TI-84 Plus C - Now Work PROBLEM 91 COMMENT The TABLE feature of a graphing calculator makes the computations in the solution to Example 10 a lot easier. ( There are many other numerical techniques for approximating a real zero of a polynomial. The one outlined in Example 10 (a variation of the bisection method) has the advantages that it always works, it can be programmed on a computer, and each time it is used, another decimal place of accuracy is achieved. See Problem 118 for the bisection method, which places the zero in a succession of intervals, with each new interval being half the length of the preceding one.
In Problems 85-88, each equation has a solution r in the interval indicated. Use the method of Example 10 to approximate this solution correct to two decimal places. 85. & - 2x + 5x - 1- 0;0 srs1 87. 2r + 6x? - &r + 2 = 0; -5 srs -4 86. x + &r' - x+2 = 0; -1 srs0 88. 3x - 10x + 9 = 0; -3 srs -2 In Problems 89-92, each polynomial function has exactly one positive real zero. Use the method of Example 10 to approximate the zero correct to two decimal places. 89. f(x) -x + x + x - 4 \91. f(x) - 2r - 3x - 4x - 8 90. f(x) - 2 + x - 1 92. f(x) - 3x - 2r - 20
Transcribed Image Text:In Problems 85-88, each equation has a solution r in the interval indicated. Use the method of Example 10 to approximate this solution correct to two decimal places. 85. & - 2x + 5x - 1- 0;0 srs1 87. 2r + 6x? - &r + 2 = 0; -5 srs -4 86. x + &r' - x+2 = 0; -1 srs0 88. 3x - 10x + 9 = 0; -3 srs -2 In Problems 89-92, each polynomial function has exactly one positive real zero. Use the method of Example 10 to approximate the zero correct to two decimal places. 89. f(x) -x + x + x - 4 \91. f(x) - 2r - 3x - 4x - 8 90. f(x) - 2 + x - 1 92. f(x) - 3x - 2r - 20
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