Bisection Method

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Bisection Method
Teacher Poon Chi Ming
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How to solve this equation?

• x3?x23x?10

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Content

• A. Locating Roots
• B. Bisection Method
• 1. Step
• 2. Example
• 3. Exercise

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A.Locating roots

• We know that x-intercepts of the graph y
  f(x) will give the roots of the equation f(x)
  0.

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Characteristic of x-intercept
f(x) changes from (ve) to(-ve)
f(x) changes from (-ve) to (ve).
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Conclusion
If f (a) lt 0 and f (b) gt 0
y f(x)
f (a)
?
b
a
f (b)
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Conclusion
If f (a) lt 0 and f (b) gt 0
or f (a) gt 0 and f (b) lt 0,
then the equation f(x) 0 has at least one root
between a and b.
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2. Example

• Show that the equation x3?x23x?10 has a root
  between 0 and 1.

Solution
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B. Bisection Method
Given that the equation x3?2x27x?50 has a
root between 0 and 1.
How to find a smaller interval containing the
root?
Lets try the mid-value x0.5 .
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y f(x)
x
0
1
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(1) Steps

• Let ? be a root f (x) 0 .
• 1. We first find an interval which brackets the
  root ( a lt ? lt b).
• 2. Reduce the bracketing interval successively by
  half until finally the root of required accuracy
  is obtained.

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(2) Example
Given f(x) x3 - 2x2 7x - 5. (a) Complete the
following table Hence find an interval
of width 1 which brackets a root ? of the
equation f(x)0. (b) Find, by the method of
bisection, the root of the equation f(x)
0 correct to 1 decimal place.
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Table Form
f(xm)
Mid-value xm
Bracketing Interval
0 lt ? lt 1
-
0.5
0.5 lt ? lt 1
-
0.75
0.75 lt ? lt 1

0.875
0.75 lt ? lt 0.875
-
0.8125
0.8125 lt ? lt 0.875

0.84375
0.8125 lt ? lt 0.84375
Hence ? 0.8 (correct to 1 d.p.)
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(3) Exercise
1. Given the equation
(a) Show that there is a root between 1 and
2. (b) Find, by the method of bisection, the
root of the equation correct to 1 d.p..
x 1.2
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2. Given the equation of x 1 3 sin x.
(a) Show that there is a root ? between 1 and
2. (b) Find ? correct to 3 sig. fig..
x 1.87
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The End
HKUST http//www.edp.ust.hk/math/
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Solution

• Let f (x) x3 ? x23x ?1
• f (0) (0)3-(0)23(0)-1 -1 lt 0
• f (1) (1)3 - (1)23(1)-1 2 gt 0
• ?There is a root between 0 and 1.