Answer to Question #305611 in Differential Geometry | Topology for Junior

Question

use the Bisection method with 3 iterations to find solutions for f(x) = x ^ 3 + x - 4 on interval [1, 4] .

Answer 1

$f(x)=x^3+x-4\\ \left[1,4\right]\\ a=1, b=4,a<b\\ f(a)=f(1)=1+1-4=-2\\ f(b)=f(4)=4^3+4-4=64\\ f(a)f(b)=-2\cdot64<0\\ x_0=\frac{a+b}{2}=\frac{1+4}{2}=2.5\\ f(x_0)=f(2.5)=2.5^3+2.5-4=14.125\\ f(a)f(x_0)=-2\cdot14.125<0, b=x_0=2.5\\ x_1=\frac{1+2.5}{2}=1.75\\ f(x_1)=f(1.75)=1.75^3+1.75-4=3.109\\ f(a)f(x_1)=-2\cdot3.109<0, b=1.75\\ x_2=\frac{1+1.75}{2}=1.375\\ f(x_2)=1.375^3+1.375-4=-0.025\\ x=1.375$