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(a) Use Newton's method with $x_{1}=1$ to find the root of the equation $x^{3}-x=1$ correct to six decimal places.(b) Solve the equation in part (a) using $x_{1}=0.6$ as the initial approximation.(c) Solve the equation in part (a) using $x_{1}=0.57 .$ (You definitely need a programmable calculator for this part.)(d) Graph $f(x)=x^{3}-x-1$ and its tangent lines at $x_{1}=1,0.6,$ and 0.57 to explain why Newton's method is so sensitive to the value of the initial approximation.

(a) $x=1.324717$(b) $x=1.324717$(c) No solution after 19 iterations.(d) The change in value of the first approximation of the Newton's method to find the solution of an equation changes the tangent, and hence the second approximation gets affected by it. In turn it increases/decreases the iterations required to solve the equation.

Calculus 1 / AB

Chapter 3

Derivatives

Section 8

Linear Approximations and Taylor Polynomials

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In this problem, we are asked to use newton's method to find a root of the following equation and were asked to use different initial guesses to see the sensitivity of newton's netted to initial guesses, so we're given this function- and this is same as writing. Our f of x as x, cubed minus x, minus 1, is equal to 0 point now. Next, we're going to find the derivate of this function says we need the derivative in newton's matter and the vats x, squared minus 1. So, let's strike newton's method extent plus 1 is equal to x. Minus the function at this point, which is x, cubed and minus x, n minus 1, divided by the deetle that x, n, so 3 x, squared and is 1 point so in part a we're going to take our initial just to be 1. So if you plug 1 in this equation in this equation, to find the next approximation we find 2 to be 1.5 and pre is about 1.347826 x 4 dan is 4.3. The 520 x 5 is about 1.324718 and x. 6 is about the same as x. 5. Point so we're going to say that this is the root of this equation that we're going to take the initial guess to be .6. It now, if you plug .6 in this equation, to find next approximation that next approximation 17.9, then the third 1 is 11.9468, and we follow this procedure. We find 12 approximation, which is 1.3 g, 4718 point now in part. We'Re going to take this to initial has to be .56. Then, if we plug .56 in this equation, we find the second approximation to be negative. 54.16545 and the third approximation is negative 36.114293, and if you follow this procedure, you find root. Thirty sixth approximation- and this is 1.32471 ant now in part t- were asked to explain. This were plotting the function. So part d was plot. The function plus what function x, cubed minus x, minus 1, is 0 point. So this is what the function looks like. So this is what is x, cube minus x, minus 1 point, so this is point for it. If they take x is 1, is an initial datto draw a tangent line? You see that tangent line, our next guess x, 2 is actually pretty close to the real group. Okay. Now, let's look at what happens. 4.6. .5. 6 point now. The tangent line that x is .6 is pretty far from the real. So that's what it takes about! Paliterations and when initial guesses .57, just like in part t the tangent line is almost horizontal and x 2 in this case is really far from the real root. So that's why it takes about 36 et.

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