Problem 5: Use Newton’s method to find all the roots of the equation correct to eight decimal places. 𝑓(𝑥) = 𝑥 6 − 𝑥 5 − 6𝑥 4 − 𝑥 2 + 𝑥 + 10

Question

For full credit include: a) A graph of the function (which you will, hopefully, use to make your initial approximations). b) Show each approximation until you meet the correct conditions for having the ‘final answer’. c) At least one full calculation with the explanation of the steps. **Just using the Calculation feature in your calculator will result in a significant point deduction- the purpose of this problem is to verify your ability and understanding of how to use Newton’s method.

Answer 1

y = x^6 - x^5 - 6x^4 - x^2 +x+10

y' = 6x^5 - 5x^4 - 24x^3 - 2x + 1

Functions are graphed at Desmos dot com

online Calculator found at

Keisan dot casio dot com / exec / system / 1244946907

or google Newtons method calculator

Newton's Method formula:

x(N) = x(N-1) - f(x(N-1))/f'(x(N-1))

Will do 2 iterations

Per the graph, the smallest solution is close to -2,

so that will be the starting approximation.

x0 = -2

x1 = x(0) - f( x(0)) / f' (x(0))

= -2 - f(-2) / f'(-2)

= -2 - 4 / -75

= -2 + 4/75

= -1.9466666666666666666666666666667

x2 = x(1) - f(x(1))/f'(x(1))

= -1.946... - .4975/-57.46

= -1.9373418029933867037939436129481.....

Best approximation per Newton's method

-1.938228834702721447462

Here are the other solutions:

Using -1.22 as the starting approximation, the best approximation is -1.219979965004372692522

Using 1.139 as the starting approximation, the best approximation is 1.139293745967452526157

Using 3 as the starting approximation, the best approximation is 2.989841020211046396837

the other 2 solutions are complex..

Answer 2

Again, this appears to be a test question and it would not be ethical for me to work the problem.

Newton's method just uses a line in place of the function (a linearization of the function) where the slope of the line is the derivative. Consider the point slope form of a line:

y - y₁ = m(x - x₁) or y = m(x - x₁) + y₁

Think about your first "guess as the point (x₁, y₁) or (x₁, f(x₁)) then the slope is f '(x₁) so the equation becomes:

y = f '(x₁)(x - x₁) + f(x₁) and you are trying to find the value of "x" that gives a value of y = 0 (the zero) so:

0 = f '(x₁)(x - x₁) + f(x₁)

-f(x₁) = f '(x₁)(x - x₁)

-f(x₁)/f '(x₁) = x - x₁

x₁ - f(x₁)/f '(x₁) = x

x = x₁ - f(x₁)/f '(x₁)

So, plug in your "guess" for x₁ and calculate a new "guess". Then use that "new guess" in the same equation to get "another new guess", and keep going until you get an answer that doesn't change in the 8th decimal point.

Then repeat for each of the zeros.