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.
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"
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