Question

Given the function

𝑥

3 − 13𝑥

2

+ 25𝑥 + 25 = 0

(a) Show that the equation has a root in the interval (− 2, 2).

(b) Use 2 iterations of Newton’s method to find an approximation for this root.

(c) Use 2 iterations of the bisection method to find an approximation for this

root.

Answer 1

Given function is-

$f(x)=x^3-13x^2+25x+25$

Then $f'(x)=3x^2-26x+25$

(a) $f(-2)=(-2)^3-13(-2)^2+25(-2)+25=-8-52-50+25=-85$

$f(2)=(2)^3-13(2)^2+25(2)+25=8-52+50+25=31$

As The value of f(-2)<0 and f(2)>0

So The root must lies in (-2,2)

(b) Let $x_o=-2,$

Then According to Newton's method-

$x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}$

at n=0,

first Iteration-

$x_1=x_o-\dfrac{f(x_0)}{f'(x_o)}=-2-\dfrac{(-2)^3-13(-2)^2+25(-2)+25}{3(-2)^2-2(-2)+25}$

$-2-(-\dfrac{85}{89})=-2+0.96=-1.04$

Second Iteration-

at n=1

$x_2=x_1-\dfrac{f(x_1)}{f'(x_1)}=-1.04-[\dfrac{(-1.04)^3-13(-1.04)^2+25(-1.04)+25}{3(-1.04)^2-26(1.04)+25}]$

$=-1.04-(-\dfrac{13.93}{55.28})=-1.04+0.2519=-0.788$

Hence Approximate root is $-0.788.$

(c) As The root lies between -2 and 2

So By bisection Method-

First Iteration-

$x_1=\dfrac{-2+2}{2}=0$

$f(0)=(0)-13(0)+25(0)+25=25>0$

So root lies in -2 and 0

So second Iteration-

$x_2=\dfrac{-2+x_o}{2}=\dfrac{-2+0}{2}=-1$