x^3-4x-9=0 find the bisection method
Answers
Answer:
Hence the required root is x_4=2.6875.
Step-by-step explanation:
Given:
Let f(x) =x^3- 4x - 9f(x)=x
3
−4x−9
Then f(1) =(1)^3- 4(1) - 9(1)
3
−4(1)−9 = -12
f(2)=(2)^3- 4(2) - 9(2)
3
−4(2)−9 = 8-8-9= -9
f(3)=(3)^3- 4(3) - 9(3)
3
−4(3)−9 =27-12-9= 6
Here f(2) is -ve and f(3) is positive. Therefore root lies in (2,3)
x_1=\frac{2+3}{2} =2.5x
1
=
2
2+3
=2.5
Hence the first approximation to the root is x_1 =2.5x
1
=2.5
f( x_1) =2.5^3- 4\times2.5 - 9f(x
1
)=2.5
3
−4×2.5−9
f( x_1) =15.625-10 - 9f(x
1
)=15.625−10−9
= -3.375
Root lies between 2.5 and 3
x_2=\frac{2.5+3}{2} =2.75x
2
=
2
2.5+3
=2.75
x_2=2.75x
2
=2.75
f( x2) =2.75^3- 4\times2.75 - 9f(x2)=2.75
3
−4×2.75−9
f( x_2) =20.79- 11 - 9f(x
2
)=20.79−11−9
f( x_2) =0.79f(x
2
)=0.79
Root lies between 2.5 and 2.75.
x_3=\frac{2.5+2.75}{2} =2.625x
3
=
2
2.5+2.75
=2.625
x_3=2.625x
3
=2.625
f( x_3) =2.625^3- 4\times2.625 - 9f(x
3
)=2.625
3
−4×2.625−9
f(x_3)=18.09-10.5-9f(x
3
)=18.09−10.5−9
f(x_3)=-1.41f(x
3
)=−1.41
Root lies between 2.625 and 2.75.
x_4=\frac{2.625+2.75}{2}=2.6875x
4
=
2
2.625+2.75
=2.6875
x_4=2.6875x
4
=2.6875
Hence the required root is x_4=2.6875x
4
=2.6875 .