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The bisection method is applied to compute a zero of the function f(x) = x4 – x3 – x2 – 4 in the
interval [1,9]. The method converges to a solution after ––––– iterations.
interval [1,9]. The method converges to a solution after ––––– iterations.
- a)1
- b)3
- c)5
- d)7
Correct answer is option 'B'. Can you explain this answer?
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The bisection method is applied to compute a zero of the function f(x)...
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The bisection method is applied to compute a zero of the function f(x)...
- 3x + 1 in the interval [0,1].
1. First, we need to check if f(0) and f(1) have opposite signs, which is a necessary condition for the bisection method to work.
f(0) = (0)^4 - 3(0) + 1 = 1
f(1) = (1)^4 - 3(1) + 1 = -1
Since f(0) and f(1) have opposite signs, we can proceed with the bisection method.
2. We start by finding the midpoint of the interval [0,1]:
c = (0 + 1)/2 = 0.5
3. We evaluate f(c):
f(0.5) = (0.5)^4 - 3(0.5) + 1 = 0.0625 - 1.5 + 1 = -0.4375
4. Since f(0.5) and f(1) have opposite signs, we discard the left half of the interval [0,1] and consider the right half [0.5,1].
5. We repeat the process by finding the midpoint of [0.5,1]:
c = (0.5 + 1)/2 = 0.75
6. We evaluate f(c):
f(0.75) = (0.75)^4 - 3(0.75) + 1 = 0.3164
7. Since f(0.75) and f(0.5) have the same sign, we discard the right half of the interval [0.5,1] and consider the left half [0.5,0.75].
8. We repeat the process by finding the midpoint of [0.5,0.75]:
c = (0.5 + 0.75)/2 = 0.625
9. We evaluate f(c):
f(0.625) = (0.625)^4 - 3(0.625) + 1 = -0.0762
10. Since f(0.625) and f(0.5) have opposite signs, we discard the right half of the interval [0.5,0.75] and consider the left half [0.5,0.625].
11. We repeat the process by finding the midpoint of [0.5,0.625]:
c = (0.5 + 0.625)/2 = 0.5625
12. We evaluate f(c):
f(0.5625) = (0.5625)^4 - 3(0.5625) + 1 = -0.0156
13. Since f(0.5625) and f(0.5) have opposite signs, we discard the right half of the interval [0.5,0.625] and consider the left half [0.5,0.5625].
14. We repeat the process by finding the midpoint of [0.5,0.5625]:
c = (0.5 + 0.5625)/2 = 0.53125
15. We evaluate f(c):
f(0.
1. First, we need to check if f(0) and f(1) have opposite signs, which is a necessary condition for the bisection method to work.
f(0) = (0)^4 - 3(0) + 1 = 1
f(1) = (1)^4 - 3(1) + 1 = -1
Since f(0) and f(1) have opposite signs, we can proceed with the bisection method.
2. We start by finding the midpoint of the interval [0,1]:
c = (0 + 1)/2 = 0.5
3. We evaluate f(c):
f(0.5) = (0.5)^4 - 3(0.5) + 1 = 0.0625 - 1.5 + 1 = -0.4375
4. Since f(0.5) and f(1) have opposite signs, we discard the left half of the interval [0,1] and consider the right half [0.5,1].
5. We repeat the process by finding the midpoint of [0.5,1]:
c = (0.5 + 1)/2 = 0.75
6. We evaluate f(c):
f(0.75) = (0.75)^4 - 3(0.75) + 1 = 0.3164
7. Since f(0.75) and f(0.5) have the same sign, we discard the right half of the interval [0.5,1] and consider the left half [0.5,0.75].
8. We repeat the process by finding the midpoint of [0.5,0.75]:
c = (0.5 + 0.75)/2 = 0.625
9. We evaluate f(c):
f(0.625) = (0.625)^4 - 3(0.625) + 1 = -0.0762
10. Since f(0.625) and f(0.5) have opposite signs, we discard the right half of the interval [0.5,0.75] and consider the left half [0.5,0.625].
11. We repeat the process by finding the midpoint of [0.5,0.625]:
c = (0.5 + 0.625)/2 = 0.5625
12. We evaluate f(c):
f(0.5625) = (0.5625)^4 - 3(0.5625) + 1 = -0.0156
13. Since f(0.5625) and f(0.5) have opposite signs, we discard the right half of the interval [0.5,0.625] and consider the left half [0.5,0.5625].
14. We repeat the process by finding the midpoint of [0.5,0.5625]:
c = (0.5 + 0.5625)/2 = 0.53125
15. We evaluate f(c):
f(0.
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The bisection method is applied to compute a zero of the function f(x) = x4 – x3 – x2 – 4 in theinterval [1,9]. The method converges to a solution after ––––– iterations.a)1b)3c)5d)7Correct answer is option 'B'. Can you explain this answer?
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