find a real root of x^3-2x-5=0
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Answer:
Here, x
3
−2x−5=0
Let f(x)=x
3
−2x−5
f
′
(x)=3x
2
−2
Here f(2)=−1<0 and f(3)=16>0
Root lies between 2 and 3.
x
0
=
2
2+3
=2.5
First Iteration:
f(x
0
)=f(2.5)=5.625
f
′
(x
0
)=f
′
(2.5)=16.75
x
1
=x
0
−
f
′
(x
0
)
f(x
0
)
=2.5−
16.75
5.625
=2.16418
Second Iteration:
f(x
1
)=f(2.16418)=0.80795
f
′
(x
1
)=f
′
(2.16418)=12.05101
x
2
=x
1
−
f
′
(x
1
)
f(x
1
)
=2.16418−
12.05101
0.80795
=2.09714
Third Iteration:
f(x
2
)=f(2.09714)=0.02888
f
′
(x
2
)=f
′
(2.09714)=11.19393
x
3
=x
2
−
f
′
(x
2
)
f(x
2
)
=2.09714−
11.19393
0.02888
=2.09456
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