Question

A. Consider the function ?(?)=?3−?−1f(x)=x3−x−1. What is the value of x after the third iteration of the Secant Method if starting guess values are 1 and 2.

Select one:

a. 1.3372

b. 1.3285

c. 1.3230

d. 1.3229

e. 0.0

B. find the value of x at the third iteration using Newton’s Method. Starting value x=1.

Answer 1

a) option a) 1.3372

Secant Method

f(x) = x³ - x - 1

x0 = 1 and x1 = 2

First iteration:

x2 =( x0*f1 - f0*x1) / (f1 - f0)

f0 = -1

f1 = 5

x2 = 1.16667

---------------------

Second iteration:
x1 = 2 and x2 = 1.16667
Also f1 = 5 and f2 = -0.578704
x3 = 1.25311

--------------------

Third iteration:
x2 = 1.16667 and x3 = 1.25311
Also f2 = -0.578704 and f3 = -0.285363
x4 = 1.33721

So value after third iteration is 1.3372.

=================================================================================

=================================================================================

B)

f(x) = x^3 - x - 1

initially x0 = 1

Using newton method

x1 = x0 - f(x0)/f'(x0)

f'(x) = derivate of f(x) = 3x² - 1

----------------

First iteration:

x1 = 1 - [ (-1)/(2)]

x1 = 1.5

-----------------

second iteration:

x2 = 1.34783

------------------

third iteration:

x3 = 1.3252

------------------

So value after three iteration is 1.3252

======================================================

Please upvote.