Find the real root of xlogx=1.2 which lies between 2 and 3 by bisection method
Answers
so xlogx=1.2 which lies between 2 and 3 are 2.74218 the correct answer of xlogx=1.2 is 2.74218
Given : xlogx =1.2
To Find :root which lies between 2 and 3 Using bisection method
Solution:
f(x) = xlogx - 1.2
f(2) = -0.5979
f(3)= 0.2313
root lies between 2 and 3
f(2.5) = -0.20515
Hence root lies between 2.5 and 3
and so on
x Value
2 -0.59794
3 0.23136
2.5 -0.20515
2.75 0.00816
2.875 0.11858
2.6875 -0.04613
2.78125 0.03554
2.734375 -0.00547
2.7578125 0.01499
2.74609375 0.00475
2.740234375 -0.00036
2.743164063 0.00220
2.741699219 0.00092
2.740966797 0.00028
2.740600586 -0.00004
2.740783691 0.00012
2.740692139 0.00004
x ≈ 2.741
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