Answer to Question #187328 in Statistics and Probability for Khalilah Spencer

Question #187328

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.


1
Expert's answer
2021-05-07T10:25:02-0400

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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