How do I solve x^3+x^2+x-1? It says to use the iterative process.?

Question

I've tried everything, but I am so confused. Please, can someone explain the process in simple terms?

I am allowed to use any one of the processes.

Answer 1

There is not just one iterative process. Newton's method, bisection, and Horner's method are all iterative processes that can be used to find roots of an equation. We need to know which method to demonstrate. Which iterative process are you supposed to use?

(In response to asker's update)

Newton's method is a calculus-based method used to find solutions to f(x) = 0 by approximating f(x) as a linear function. It requires an initial guess for the solution. It often, but not always, converges to a solution (though not necessarily the solution "close" to the initial guess).

If x0 is a number (preferably "close" to a solution of f(x)), then for x not far from x0, we have

f(x) ≈ f(x0) + f'(x0)(x - x0)

(this is the linear approximation)

or (writing "=" instead of "≈")

f(x) - f(x0) = f'(x0) (x - x0)

which has the form of the equation of a line:

y - y0 = m(x - x0)

As long as m (that is, f'(x0)) is not zero, the line intersects the x-axis. To find where, we set y to zero and solve for x.

So do the same thing with the linear approximation. Set f(x) = 0 in that approximation and solve for x:

0 = f(x0) + f'(x0) (x - x0)
-f(x0)/f'(x0) = x - x0
x0 - f(x0)/f'(x0) = x

This is the new estimate of the solution of f(x) = 0. In general it is an improved estimate, but it's not "the" answer; ordinarily you use this improved estimate as the new x0, and do the calculation again. That's what makes it an iterative process: it must be done repeatedly, and (with luck) it closes in on the solution.

That is the case with your polynomial f(x) = x³ + x² + x - 1
The iteration is

xnew = xold - (xold³ + xold² + xold -1)/(3*xold² + 2*xold + 1)

If your initial guess is xold = 1, then
xnew = 1 - (1³ + 1² + 1 - 1)/(3*1² + 2*1 + 1) = 1 - 2/6 = 2/3 ≈ .6666666667

Use this as xold to get a further improved xnew, and so on.

If you do the iteration repeatedly, you get this sequence of estimates (using 10 digits of accuracy):

1.0
0.6666666667
0.5555555556
0.5438121048
0.5436890261
0.5436890127
0.5436890127

If we kept on iterating, we would keep getting the same result; the process has converged, and you can confirm that the result does (very nearly) satisfy the cubic. (The cubic has only one real solution, by the way.)

Answer 2

ok, right here is going... So non permissible values are numbers that x can't equivalent. We additionally understand that we can't have a a fragment with 0 interior the denominator. for this reason x can't be 3 or -3 by way of fact between the 1st 2 denominators might equivalent 0 if it have been 3 or -3. We see that -3 additionally does no longer paintings for x2-9. they could be your non permissible values. to clean up this equation first discover uncomplicated denominators. We see that x2-9 is the LCM of each and all of the denominators so we multiply the 1st term by potential of x+3/x+3 to get x=3/x2-9. We then multiply the subsequent term by potential of x-3/x-3 to get x-3/x2-9. Now you have all like words! you could upload and subtract the words now. the next step could be to subtract 5/x2-9 from the two aspects of the equation. then you certainly multiply the two aspects by potential of x2-9 and you may get... x+3+x-3-5=0 then combine like words 2x-5=0 upload 5 to the two aspects 2x=5 divide the two aspects by potential of two x=5/2 good success!

Answer 3

3x^5-1