Question

Each equation has one real root. Use Newton’s Method to approximate the root to eight correct decimal places. (a) x5 + x = 1 (b) sin x = 6x + 5 (c) ln x + x2 = 3

**MUST BE DONE IN MATLABE AND SHOW CODE

Answer #1

**(a)**

**root = 0.754877666246693**

Matlab code:

format long;

x=5;

h = (x^5 + x -1) /(5*(x^4) + 1 );

i = 1;

while(abs(h) >= 0.00000001)

h = (x^5 + x -1) /(5*(x^4) + 1 );

x = x - h;

i=i+1;

end

x

**(b) root = -0.970898923504256**

**Matlab code:**

format long;

x=5;

h = (sin(x) - 6*x - 5) /(cos(x) -6 );

i = 1;

while(abs(h) >= 0.00000001)

h = (sin(x) - 6*x - 5) /(cos(x) -6 );

x = x - h;

i=i+1;

end

x

**c) root = 1.592142937058094**

**Matlab code:**

format long;

x=5;

h = (log(x) + x^2 - 3) /((1/x) + 2*x );

i = 1;

while(abs(h) >= 0.00000001)

h = (log(x) + x^2 - 3) /((1/x) + 2*x );

x = x - h;

i=i+1;

end

x

Each equation has one root. Use Newton’s Method to approximate
the root to eight correct
decimal places. (a) x3 = 2x + 2 (b) ex + x = 7 (c) ex + sin x =
4
**MUST BE DONE IN MATLAB AND NEED CODE

Use
Newton's method to approximate the root of the equation to four
decimal places. Start with x 0 =-1 , and show all work
f(x) = x ^ 5 + 10x + 3
Sketch a picture to illustrate one situation where Newton's
method would fail . Assume the function is non-constant
differentiable , and defined for all real numbers

Use Newton's method to find all the roots of the equation
correct to eight decimal places. Start by drawing a graph to find
initial approximations.
3 sin(x2) = 2x

: Consider f(x) = 3 sin(x2) − x.
1. Use Newton’s Method and initial value x0 = −2 to approximate
a negative root of f(x) up to 4 decimal places.
2. Consider the region bounded by f(x) and the x-axis over the
the interval [r, 0] where r is the answer in the previous part.
Find the volume of the solid obtain by rotating the region about
the y-axis. Round to 4 decimal places.

Use Newton's method to find all solutions of the equation
correct to eight decimal places. Start by drawing a graph to find
initial approximations. (Enter your answers as a comma-separated
list.) x x2 + 1 = 1 − x

Use Newton’s method to find all solutions of the equation
correct to six decimal places: ?^2 − ? = √? + 1

Show that the equation
x+sin(x/3)−8=0
has exactly one real root. Justify your answer.

Part A.
Consider the nonlinear equation
x5-x=15
Attempt to find a root of this equation with Newton's method
(also known as Newton iteration).
Use a starting value of x0=4 and apply Newton's
method once to find x1
Enter your answer in the box below correct to four
decimal places.
Part B.
Using the value for x1 obtained in Part A, apply
Newton's method again to find x2
Note you should not round x1 when computing
x2

The given equation has one real solution. Approximate it by
Newton's method. (x^3)+2x-2

(a) Show that the equation x = 14 \ln x has at least one real
solution.
(b) This is an example of what sort of equation? (Professor said
the answer is Ans. fixed point but I don't know why.)
(c) By using an iterative method of your choice, find a
numerical approximation of
the solution valid to 5 decimal
places.

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