Question

Consider the function, f(x) = - x^{4} - 2x^{3} -
8x^{2} - 5x

Use parabolic interpolation (x_{0} = -2, x_{1} =
-1, x_{2}= 1, iterations = 4). Select new points
sequentially as in the secant method.

Answer #1

First iteration

X2 = X0 - f(x0)((x1-x0)/(f(x1)-f(x0)))

Now X0 = -1 and X1 = 1

so F(X1) = - (1)^{4} - 2(1)^{3} -
8(1)^{2} - 5(1) = -1-2-8-5 = -16

F(X0) = - (-1)^{4} - 2(-1)^{3} -
8(-1)^{2} - 5(-1) = -1 +2-8+5 = -2

X2 = -1 - (-2)(((1+1)/(-16-(-2)) = -1 + 2(2/-14) = -1.2857

**x2 = -1.2857**

Second Iteration

x2 = -1.2857 , f(x2) =-5.277

x1 = 1, f(x1) =-16

X3 = X1 - f(x1)((x2-x1)/f(x2)-f(x1))

if we plug the value :

**X3 =-2.410**

Third Iteration

X3 =-2.410, f(X3) =-40.1933

x2 = -1.2857 , f(x2) =-5.277

X4 = X2 - f(x2)((x3-x2)/f(x3)-f(x2))

if we plug the value :

**X4 =-1.1156**

Fourth Iteration

X4 =-1.1156, F(X4) = -3.1513

X3 =-2.410, f(X3) =-40.1933

X5 = X3 - f(x3)((x4-x3)/f(x4)-f(x3))

if we plug the value :

X5 =-1.0055

**So X5 = -1.0055 after fourth iteraitons**

Consider a function f(x) =
2x3 − 11.7x2 +
17.7x − 5.
Identify the root of the given function after the third
iteration using the secant method. Use initial guesses
x–1 = 3 and x0 = 4.
CAN YOU PLZ SHOW ALL THE WORK. THANK YOU

For the following function, determine the highest real root
of
f(x) = 2x3 – 11.7x2 + 17.7x - 5
by using (a) graphical methods, (b) fixed point iteration (three
iterations, x0 = 3) (Hint: Be certain that you develop a solution
that converges on the root), and (c) Newton-Raphson method (three
iterations, x0 = 3).
Perform an error check on each of your final root approximations
(e.g. for the last of the three iterations).

Consider the function
g (x) = 12x + 4 - cos x. Given
g (x) = 0 has a unique solution
x = b in the interval (−1/2, 0), and you can use this
without justification.
(a) Show that Newton's method of starting point
x0
= 0 gives a number sequence with
b <··· <xn+1
<xn
<··· <x1
<x0
= 0
(The word "curvature" should be included in the argument!)
(b) Calculate
x1
and x2.
Use theorem 2 in section...

Consider the function g (x) = 12x + 4 - cos x.
Given g (x) = 0 has a unique solution x =
b in the interval (−1/2, 0), and you can use this without
justification.
(a) Show that Newton's method of starting point
x0 = 0 gives a number sequence with
b <··· <xn+1 <xn <···
<x1 <x0 = 0
(The word "curvature" should be included in the argument!)
(b) Calculate x1 and x2. Use
theorem 2 in section...

Consider the function g (x) = 12x + 4 - cos x.
Given g (x) = 0 has a unique solution x =
b in the interval (−1/2, 0), and you can use this without
justification.
(a) Show that Newton's method of starting point
x0 = 0 gives a number sequence with
b <··· <xn+1 <xn <···
<x1 <x0 = 0
(The word "curvature" should be included in the argument!)
(b) Calculate x1 and x2. Use
theorem 2 in section...

Use the secant method to estimate the root of
f(x) = -56x + (612/11)*10-4 x2 -
(86/45)*10-7x3 + (3113861/55)
Start x-1= 500 and x0=900.
Perform iterations until the approximate relative error falls below
1% (Do not use any interfaces such as excel etc.)

Let
f(x)=sin(x)+x^3-2. Use the secant method to find a root of f(x)
using initial guesses x0=1 and x1=4. Continue until two consecutive
x values agree in the first 2 decimal places.

Consider the function f(x)=5x−4 and find the following:
a.) The average rate of change between the points (−1,f(−1)) and
(2,f(2)).
b.) The average rate of change between the points (a,f(a)) and
(b,f(b)).
c.) The average rate of change between the points (x,f(x)) and
(x+h,f(x+h)).

for the given functions f(x), let x0=1, x1=1.25, x2=1.6.
Construct interpolation polynomials of degree at most one and at
most two to approximate f(1.4), and find the absolute error. a.
f(x)=sin (pi x)

The seventh iteration k = 6 of the secant method in solving
the equation f(x) = x3 − 5x + 3 resulted in x8 = 1.8249 and x7 =
1.8889. Compute the next three iterations correct to 4 decimal
points

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