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
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2b9bf898-9a72-4e11-a33c-691d270253f8%2Fcd725763-6f6e-458a-9af4-48f95e9247a6%2Ftd67sf_processed.jpeg&w=3840&q=75)
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.
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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