Use the Bisection method to find solutions accurate to within 10-2 for x³ – 7x2 + 14x – 6 = 0 on [0,1]. Solution: Let f(x) = x³ – 7x² + 14x – 6 = 0. Note that f (0) = –6 < 0 and f(1) = 2 > 0, therefore, based on the Intermediate Value Theorem, since f is continuous, there is p E (0, 1) such that f(p) = 0. Let ao = 0, bo = 1, with f(ao) < 0, ƒ(bo) > 0. Let po = ao + sign as f(ao), therefore a1 = Po = 0.5, bị = bọ = 1 and repeat: pi = 0.75, .. This yields the following results for p, and f(Pn): - %3D bo-an 2 0.5, and we have f(po) = –0.6250 < 0 (the same 0.75, ... %3D %3D f (Pn) n Pn 0 0.5 -0.6250000
Use the Bisection method to find solutions accurate to within 10-2 for x³ – 7x2 + 14x – 6 = 0 on [0,1]. Solution: Let f(x) = x³ – 7x² + 14x – 6 = 0. Note that f (0) = –6 < 0 and f(1) = 2 > 0, therefore, based on the Intermediate Value Theorem, since f is continuous, there is p E (0, 1) such that f(p) = 0. Let ao = 0, bo = 1, with f(ao) < 0, ƒ(bo) > 0. Let po = ao + sign as f(ao), therefore a1 = Po = 0.5, bị = bọ = 1 and repeat: pi = 0.75, .. This yields the following results for p, and f(Pn): - %3D bo-an 2 0.5, and we have f(po) = –0.6250 < 0 (the same 0.75, ... %3D %3D f (Pn) n Pn 0 0.5 -0.6250000
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![3a
• Use the Bisection method to find solutions accurate to within 10-2 for
x³ – 7x2 + 14x – 6 = 0 on [0,1].
Solution: Let f(x) = x³ – 7x² + 14x – 6 = 0. Note that f (0) = -6 < 0
and f(1) = 2 > 0, therefore, based on the Intermediate Value Theorem, since
f is continuous, there is p e (0, 1) such that f(p) = 0.
Let ao = 0, bo = 1, with f(ao) < 0, ƒ(bo) > 0.
Let po = ao + bn-an
sign as f(ao), therefore a1 = Po = 0.5, b1 =
This yields the following results for p, and f(pn):
0.5, and we have f(po) = -0.6250 < 0 (the same
bo = 1 and repeat: P1
||
2
= 0.75, ...
Pn
f(Pn)
0.5
-0.6250000
1
0.75000000 +0.9843750
2
0.62500000
+0.2597656
3
0.56250000 -0.1618652
4
0.59375000 +0.0540466
5
0.57812500 -0.0526237
6 0.58593750
+0.0010313](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbd2bf6e5-548f-4d46-9a16-2efd3e333581%2Fea988a26-0255-4ba5-a06f-24dee704cccb%2F20ifli_processed.png&w=3840&q=75)
Transcribed Image Text:3a
• Use the Bisection method to find solutions accurate to within 10-2 for
x³ – 7x2 + 14x – 6 = 0 on [0,1].
Solution: Let f(x) = x³ – 7x² + 14x – 6 = 0. Note that f (0) = -6 < 0
and f(1) = 2 > 0, therefore, based on the Intermediate Value Theorem, since
f is continuous, there is p e (0, 1) such that f(p) = 0.
Let ao = 0, bo = 1, with f(ao) < 0, ƒ(bo) > 0.
Let po = ao + bn-an
sign as f(ao), therefore a1 = Po = 0.5, b1 =
This yields the following results for p, and f(pn):
0.5, and we have f(po) = -0.6250 < 0 (the same
bo = 1 and repeat: P1
||
2
= 0.75, ...
Pn
f(Pn)
0.5
-0.6250000
1
0.75000000 +0.9843750
2
0.62500000
+0.2597656
3
0.56250000 -0.1618652
4
0.59375000 +0.0540466
5
0.57812500 -0.0526237
6 0.58593750
+0.0010313
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