Question #110546
Taking the endpoints of the last interval obtained in f(x) = x^3−5x^2 +1 = 0 as the initial
approximations, perform two iterations of the secant method to approximate the
root.
1
Expert's answer
2020-04-21T16:26:37-0400

According to the secant method, the root can be found by the following equation:


xn=xn2f(xn1)xn1f(xn2)f(xn1)f(xn2)x_n=\frac{x_{n-2}f(x_{n-1})-x_{n-1}f(x_{n-2})}{f(x_{n-1})-f(x_{n-2})}



Take the last interval of [xn2;xn1]=[0.469;0.5][x_{n-2};x_{n-1}]=[0.469;0.5]. The first iteration gives


xn=0.469f(0.5)0.5f(0.469)f(0.5)f(0.469)=0.495.x_n=\frac{0.469f(0.5)-0.5f(0.469)}{f(0.5)-f(0.469)}=0.495.


Choose points for the second iteration. The signs of the function at the interval ends must be different:


f(0.495)f(0.5)>0,f(0.469)f(0.495)<0.f(0.495)f(0.5)>0,\\ f(0.469)f(0.495)<0.

Therefore, we choose [0.469;0.495].

The second iteration gives


xn=0.469f(0.495)0.495f(0.469)f(0.495)f(0.469)=0.470.x_n=\frac{0.469f(0.495)-0.495f(0.469)}{f(0.495)-f(0.469)}=0.470.

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