# Use​ Newton s method to estimate all real solutions of the equation. Make your answers accurate to 6 decimal places.?

2x^2 -8x+8=sin x

Relevance

Let

f(x) = 2 * x^2 - 8*x + 8 - sin(x)

f'(x) = 4*x - 8 - cos(x)

f''(x) = 4 + sinx , f" > 0 => f convex (at most two roots)

Use 1.29 and 2.71 as initial root guesses to ensure no root is lost.

(because sinx between -1 and 1, ditto for the LHS quadratic, and after solving for x:

1.29 < x < 2.70 , so x way outside this interval makes little sense)

Now go to an online N-R calculator, enter the appropriate data and:

with 1.29 as x_0 we get 1.305383... in 4 iterations

with 2.71 as x_0 we get 2.534196.... in 5 iterations

• Newton's method is an iterative method for finding a zero of a function when you know an approximate value to start with. You need to rewrite your equation so it looks like

.. <something> = 0

You can do that by subtracting sin(x). Now you have a function

.. f(x) = 2x^2 -8x +8 -sin(x)

so that a value of x that makes f(x)=0 is a solution to your original equation.

The iterator used by Newton's method is of the form

.. (next guess for x) = x -f(x)/f'(x) . . . . . . where f'(x) is the derivative of f(x) with respect to x

To create this formula, you need to take the derivative of f(x).

.. f'(x) = 4x -8 -cos(x)

Now you can write the iteration function (let's call it g(x)) as

.. g(x) = x -f(x)/f'(x)

.. g(x) = x -(2x^2 -8x +8 -sin(x))/(4x -8 -cos(x))

We can express this as a rational function by performing the subtraction.

.. g(x) = ((4x^2 -8x -x*cos(x)) -(2x^2 -8x +8 -sin(x))/(4x -8 -cos(x))

.. g(x) = (2x^2 -8 -x*cos(x) +sin(x))/(4x -8 -cos(x))

Now, we need to find some suitable starting places for the iteration. A graphing calculator can help with that. It shows roots of f(x) to be near x=1.305 and x=2.534. (See the source link. Select the curve to highlight the x-intercepts; select those to see their coordinates.)

We can use the calculator at the source link or any other calculator to evaluate the iterator g(x). (Make sure the calculator is in Radian mode.) Starting with the above values, it only takes one iteration to get an answer good to 6 decimal places.

.. g(1.305) = 1.305383337

.. g(1.305383337) = 1.305383457 . . . . first 6 decimal places are the same

.. g(2.534) = 2.534196355

.. g(2.534196355) = 2.534196325 . . . . first 6 decimal places are the same

The real solutions are x = {1.305383, 2.534196}.

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Many graphing calculators have a derivative function, so you can define the iterator as we have at the source link, in terms of the original function and its derivative. This takes all the work out of using Newton's method to find solutions to full calculator accuracy.

• The first step is to graph it. There are two solutions, one near x = 1.3, and one near 2.5.

Try graphing y = 2x^2 -8x+8 - sin(x).

You also need to figure dy / dx.

Then start with x0 = 1.3 (for example), and compute y and dy/dx

The next estimate is x1 = x0 - y0 / (dy/dx(x0)). Keep repeating until the value of x doesn't change at the 6th decimal place.