Question

Consider the following real 3rd order polynomial

f (x)= x^3− 5.5 x^2− 5x+ 37.5

A) Use the bisection method to determine one of the roots, employing initial guesses of xl = - 10, xu = -1, and a stopping criterion εs=12% .

B) Use the false position method to determine a root, employing initial guesses of xl = - 1, xu = 4, and a stopping criterion εs=3%. Was this method the best for these initial guesses?

C) Use the fixed point iteration to determine a root, employing an initial guess x0 = 7 and a stopping criterion εs=5%

D) Use the Newton-Raphson to determine a root, employing an initial guess x0 = 7 and a stopping criterion εs=5%

Please please show all steps

Thank you

Answer 1

A.  bisection method

Let f(x)=x3-5.5x2-5x+37.5


1st iteration :

Here f(-10)=-1462.5<0 and f(-1)=36>0

∴ Now, Root lies between -10 and -1

x0=-10+(-1)2=-5.5

f(x0)=f(-5.5)=-5.53-(5.5-5.52)-(5-5.5)+37.5=-267.75<0


2nd iteration :

Here f(-5.5)=-267.75<0 and f(-1)=36>0

∴ Now, Root lies between -5.5 and -1

x1=-5.5+(-1)2=-3.25

f(x1)=f(-3.25)=-3.253-(5.5-3.252)-(5-3.25)+37.5=-38.671875<0


3rd iteration :

Here f(-3.25)=-38.671875<0 and f(-1)=36>0

∴ Now, Root lies between -3.25 and -1

x2=-3.25+(-1)2=-2.125

f(x2)=f(-2.125)=-2.1253-(5.5-2.1252)-(5-2.125)+37.5=13.69335938>0


4th iteration :

Here f(-3.25)=-38.671875<0 and f(-2.125)=13.69335938>0

∴ Now, Root lies between -3.25 and -2.125

x3=-3.25+(-2.125)2=-2.6875

f(x3)=f(-2.6875)=-2.68753-(5.5-2.68752)-(5-2.6875)+37.5=-8.19799805<0


5th iteration :

Here f(-2.6875)=-8.19799805<0 and f(-2.125)=13.69335938>0

∴ Now, Root lies between -2.6875 and -2.125

x4=-2.6875+(-2.125)2=-2.40625

root is x=-2.40625

error is 3%

B. false position method

Find a root of an equation f(x)=x3-5.5x2-5x+37.5 between -4 and -1, using False Position method (regula falsi method)

Solution:
Here x3-5.5x2-5x+37.5=0

Let f(x)=x3-5.5x2-5x+37.5


1st iteration :

Here f(-4)=-94.5<0 and f(-1)=36>0

∴ Now, Root lies between x0=-4 and x1=-1

x2=x0-f(x0)⋅x1-x0f(x1)-f(x0)

x2=-4-(-94.5)⋅-1-(-4)36-(-94.5)

x2=-1.8276

f(x2)=f(-1.8276)=-1.82763-(5.5-1.82762)-(5-1.8276)+37.5=22.1633>0


2nd iteration :

Here f(-4)=-94.5<0 and f(-1.8276)=22.1633>0

∴ Now, Root lies between x0=-4 and x1=-1.8276

x3=x0-f(x0)⋅x1-x0f(x1)-f(x0)

x3=-4-(-94.5)⋅-1.8276-(-4)22.1633-(-94.5)

x3=-2.2403

f(x3)=f(-2.2403)=-2.24033-(5.5-2.24032)-(5-2.2403)+37.5=9.8536>0


3rd iteration :

Here f(-4)=-94.5<0 and f(-2.2403)=9.8536>0

∴ Now, Root lies between x0=-4 and x1=-2.2403

x4=x0-f(x0)⋅x1-x0f(x1)-f(x0)

x4=-4-(-94.5)⋅-2.2403-(-4)9.8536-(-94.5)

x4=-2.4065

f(x4)=f(-2.4065)=-2.40653-(5.5-2.40652)-(5-2.4065)+37.5=3.7458>0


4th iteration :

Here f(-4)=-94.5<0 and f(-2.4065)=3.7458>0

∴ Now, Root lies between x0=-4 and x1=-2.4065

x5=x0-f(x0)⋅x1-x0f(x1)-f(x0)

x5=-4-(-94.5)⋅-2.4065-(-4)3.7458-(-94.5)

x5=-2.4672

root is x= -2.4672

error is 1.3%

C. it is very large so i am skip this if you want it then comment im send to you

D. Find a root of an equation f(x)=x3-5.5x2-5x+37.5 initial solution x0=7, using Newton Raphson method

Solution:
Here x3-5.5x2-5x+37.5=0

Let f(x)=x3-5.5x2-5x+37.5

∴f′(x)=3x2-11x-5

x0=7


1st iteration :

f(x0)=f(7)=73-5.5⋅72-5⋅7+37.5=76

f′(x0)=f′(7)=3⋅72-11⋅7-5=65

x1=x0-f(x0)f′(x0)

x1=7-7665

x1=5.83077


2nd iteration :

f(x1)=f(5.83077)=5.830773-5.5⋅5.830772-5⋅5.83077+37.5=19.5916

f′(x1)=f′(5.83077)=3⋅5.830772-11⋅5.83077-5=32.85515

x2=x1-f(x1)f′(x1)

x2=5.83077-19.591632.85515

x2=5.23447

root is 5.23447 (by NR methode)

and actual root is 5

error is 4%