For this graph, if we use Newton's method with initial point
𝑥
1
=
0
, what will happen?

Question

For this graph, if we use Newton's method with initial point 
𝑥
1
=
0
, what will happen?

BlackTom AI · Accepted Answer

To analyze what happens when we apply Newton's method starting at x1 = 0 for the given graph, we first restate the available option: We obtain a sequence of points converging to the root at x = 1.

Consider the behavior at the initial point x = 0. From the graph, the curve appears to cross the x-axis somewhere near a positive x-value (the right-hand root). The Newton update is x_{n+1} = x_n - f(x_n)/f'(x_n). If we plug in x_0 = 0, the next iterate is determined by the value of f(0) and the slope f'(0).

- If f(0) is nonzero and f'(0) is not zero, the Newton step will move from x = 0 toward the root that is in the direction indicated by the tangent line at x = 0. On this graph, the tangent at x = 0 typically intersects the x-axis near the right-hand root, which is around x = 1. This makes it plausible that x_2 (or the next iterate after x_0) lands close to or at x ≈ 1, starting the convergence toward the root at x = 1.
- If the derivative at 0 were extremely small or if f(0) and f'(0) had unfavorable signs, the method could stall or move toward a different root. However, the plotted curve shows a smooth, monotone behavior near 0 toward the right-hand side, with the immediate nearby root being at x ≈ 1, making convergence to that root the most likely outcome.

Therefore, given the provided option, the reasoning aligns with the scenario where the Newton iteration from x = 0 produces a sequence that converges to the root at x = 1. The key ideas are: the next Newton step from 0 tends to cross the x-axis near the right-hand root due to the slope and position of f(0) on the graph, and subsequent iterations then refine toward that same root, assuming f'(x) remains nonzero near the root and there are no anomalous Newton-step cycles.

If one were to consider potential pitfalls, such as f'(0) = 0 or the presence of a nearby horizontal tangent causing a poor first step, those would be noted as cases where convergence might fail or head elsewhere. But with the given graph and typical smooth behavior around x = 0, the path toward the root at x = 1 is the expected outcome.

MATH-112-301-001 Unproctored Midcourse Exam 3 Practice Exam 1

For this graph, if we use Newton's method with initial point 𝑥 1 = 0 , what will happen?

View Explanation

Log in for full answers

Similar Questions

Use Newton's method to approximate all the intersection points of the pair of curves. Some preliminary graphing or analysis may help in choosing good initial approximations. Round to six decimal places. y = ex and y = x2 + 5

Use Newton's method to approximate all the intersection points of the pair of curves. Some preliminary graphing or analysis may help in choosing good initial approximations. Round to six decimal places. y = ex and y = x2 + 3

Use a calculator to compute the first 10 iterations of Newton's method when applied to the function with the given initial approximation. Make a table for the values. Round to six decimal places. f(x) = 3x - cos x; x0 = 1

使用牛顿方法的两次迭代和初始近似来找到第三个近似方程。（给出你的答案到小数点后位）。提示：牛顿法 .

Consider Newton's Method. What happens when the gradient is small, and how may you optimise in this regime?

Using Newton's method to approximate the zero of 𝑓 ( 𝑥 ) = 3 𝑥 2 − 3 𝑥 + 1 , determine the second approximation if the initial one is 𝑥 1 = 2 .

Let 𝑓 ( 𝑥 ) = 3 𝑥 2 − 2 𝑥 − 9 . Using Newton's method, and an initial guess of 𝑥 1 = 2 , find the next approximation, 𝑥 2 , for a root of this function.

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler