Question
Asked Sep 25, 2019
61. Given the fact that a root for the function f(x) = 2 - 3 lies in the interval
1, 1.5,
use the Bisection Method twice to find an interval of length 1/8 containing
the root
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61. Given the fact that a root for the function f(x) = 2 - 3 lies in the interval 1, 1.5, use the Bisection Method twice to find an interval of length 1/8 containing the root

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Expert Answer

Step 1

Refer to the question, we have to use the Bisection method and the procedure of bisection method is if f(x) is the function which has root in [a,b] then f(a)>0 and f(b)<0 then it means that the root lies between ‘a’ and ‘b’ .

Then x1=(a+b)/2 , and if  f(x1) >0 ,root is in [x1,b] and if f(x1) <0 ,root is in [a , x1].

2
>0.root is in x,.b]
and if f(x) 0.root is in [a . x]
if f(x,)
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2 >0.root is in x,.b] and if f(x) 0.root is in [a . x] if f(x,)

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Step 2

And this process is repeated until we get the required accuracy.

Here provided function is f(x)=2X-X3, in the interval [1,1.5],then follow the above procedure as,

f(x)2-x[1.1.5]
(1) 2-(1)20
f(1.5) 21.5-(1.5 = -0.5465 <0
So, root lies in the interval 1,1.5|
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f(x)2-x[1.1.5] (1) 2-(1)20 f(1.5) 21.5-(1.5 = -0.5465 <0 So, root lies in the interval 1,1.5|

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Step 3

Use the first trial of bi...

First trial
1+1.51.25
2
f(1.25) 225(1.25) 0.4252> 0
f(1.5) 215-(1.5) = -0.5465 <0
So, root lies in the interval [1.25,1.5]
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First trial 1+1.51.25 2 f(1.25) 225(1.25) 0.4252> 0 f(1.5) 215-(1.5) = -0.5465 <0 So, root lies in the interval [1.25,1.5]

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Tagged in

Math

Calculus