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In Exercises \(15-18,\) use Newton's method to estimate all real solutions of the equation. Make your answers accurate to 6 decimal places. \(x^{3}+x-1=0\)

Short Answer

Expert verified
The real solution of the equation \(x^{3}+x-1=0\) accurate up to six decimal places using Newton's method is approximatly \(0.682328\)

Step by step solution

01

Define the Function and its Derivative

We are given the equation \(x^{3}+x-1=0\). Let's denote the left hand side of the equation by \(f(x)\) so we have \(f(x) = x^{3}+x-1\). We need to find the derivative of \(f(x)\), \(f'(x) = 3x^{2}+1\).
02

Choose an Initial Approximation

We have to choose a seed or an initial approximation. A good initial approximation is essential for Newton's method to converge. A reasonable starting point for our problem could be \(x_{0} = 1\).
03

Apply Newton's Method with x0

To start the iteration, plug \(x_{0}\) into the Newton's method formula. Thus, we get: \(x_{1} = x_{0} - \frac{f(x_{0})}{f'(x_{0})}\). Substitute \(x_{0} = 1\), \(f(x_{0}) = 1^{3}+1-1 = 1\), and \(f'(x_{0}) = 3*1^{2}+1 = 4\) into the equation you get: \(x_{1} = 1 - \frac{1}{4} = 0.75\).
04

Iteratively Apply Newton's Method

Repeat this process, each time plugging in the previous iteration's \(x\) value, until the result does not change, or changes very minimally (i.e., up to six decimal places). Continue until accuracy is reached, with the final value being the root of the equation.
05

Check Against the Original Equation

Once the root has been found, it is always a good practice to check if this is indeed a root by substituting it back into the original equation. Should the left and right side of the equation be equal (within a small tolerance), we can be sure that it is a root of the equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Numerical Solutions of Equations
When we encounter equations involving polynomials, trigonometry, or other complex functions, finding exact solutions can be quite challenging and sometimes impossible with algebra alone. This is where numerical solutions come into play. These methods allow us to approximate the solutions of an equation to a desired degree of accuracy.

Numerical approaches, like Newton's method, which is a root-finding technique, give us a practical way to tackle equations too tough for algebra. It is particularly useful because it uses the concept of iterations to hone in on a more precise answer with each step. However, to make sure these iterations lead to the correct answer, we start by carefully selecting an initial guess, then progressively improve our estimate.

Consider the equation from the exercise, \(x^{3} + x - 1 = 0\). The exact solution is not obvious, but through numerical methods, we can still arrive at a very precise approximation. Depending on the complexity of the equation and the initial guess, the number of iterations required can vary. Through this steadfast process, we can solve equations which are otherwise analytically unsolvable.
Derivative of a Function
The concept of the derivative is at the heart of calculus and plays a critical role in many numerical methods, including Newton's method. A derivative essentially measures how a function's output changes as its input changes. In technical terms, it represents the rate of change or the slope of the function at a particular point.

In the context of the given exercise, the function \( f(x) = x^{3} + x - 1 \) has a derivative \( f'(x) = 3x^{2} + 1 \). This derivative is crucial because, in Newton's method, it helps us understand how to adjust our current approximation to get closer to the actual root. When applying Newton's method, the value of \( f'(x) \) tells us the 'steepness' of the curve at our current guess, which influences the size and direction of the next step we take toward the root.

Understanding derivatives allows us to predict the behavior of functions and is the foundation for correctly applying iterative root-finding methods. Without this concept, methods like Newton's would not be possible.
Iteration for Root Finding
The process of iteration is a cornerstone of numerical methodologies for root finding. Iteration involves repeating a set of operations, with each repetition using the outcome of the previous one to move closer to the desired result. In our case, the mission is to identify a root of the equation—the value for which the function equals zero. Using Newton's method, we iterate by continually updating our guess of the root using a specific formula.

Each step of iteration involves two main activities: calculating the function value at the current guess, and evaluating its derivative. With these in hand, we apply the Newton's formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \), systematically nudging our guess closer to the actual root with each cycle. The beauty of this method lies in its speed and efficiency; typically, it converges to a highly accurate solution within just a few iterations.

However, the success of the iteration depends on the right choice of starting value and the nature of the function. Some functions may require more careful handling to ensure convergence. In the end, with a correct approach, iteration becomes a powerful tool in our mathematical arsenal to tackle equations that defy simpler, algebraic solutions.

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Most popular questions from this chapter

Writing to Learn You have been asked to determine whether the function \(f(x)=3+4 \cos x+\cos 2 x\) is ever negative. (a) Explain why you need to consider values of \(x\) only in the interval \([0,2 \pi] . \quad\) (b) Is f ever negative? Explain.

Quadratic Approximations (a) Let \(Q(x)=b_{0}+b_{1}(x-a)+b_{2}(x-a)^{2}\) be a quadratic approximation to \(f(x)\) at \(x=a\) with the properties: \(\begin{aligned} \text { i. } Q(a) &=f(a) \\ \text { ii. } Q^{\prime}(a) &=f^{\prime}(a) \\ \text { ii. } & Q^{\prime \prime}(a)=f^{\prime \prime}(a) \end{aligned}\) Determine the coefficients \(b_{0}, b_{1},\) and \(b_{2}\) (b) Find the quadratic approximation to \(f(x)=1 /(1-x)\) at \(x=0 .\) (c) Graph \(f(x)=1 /(1-x)\) and its quadratic approximation at \(x=0 .\) Then zoom in on the two graphs at the point \((0,1) .\) Comment on what you see. (d) Find the quadratic approximation to \(g(x)=1 / x\) at \(x=1\) Graph \(g\) and its quadratic approximation together. Comment on what you see. (e) Find the quadratic approximation to \(h(x)=\sqrt{1+x}\) at \(x=0 .\) Graph \(h\) and its quadratic approximation together. Comment on what you see. (f) What are the linearizations of \(f, g,\) and \(h\) at the respective points in parts \((b),(d),\) and \((e) ?\)

Multiple Choice If the volume of a cube is increasing at 24 \(\mathrm{in}^{3} / \mathrm{min}\) and the surface area of the cube is increasing at 12 \(\mathrm{in}^{2} / \mathrm{min}\) , what is the length of each edge of the cube? \(\mathrm{}\) \(\begin{array}{lll}{\text { (A) } 2 \text { in. }} & {\text { (B) } 2 \sqrt{2} \text { in. (C) } \sqrt[3]{12} \text { in. (D) } 4 \text { in. }}\end{array}\)

The Effect of Flight Maneuvers on the Heart The amount of work done by the heart's main pumping chamber, the left ventricle, is given by the equation $$W=P V+\frac{V \delta v^{2}}{2 g}$$ where \(W\) is the work per unit time, \(P\) is the average blood pressure, \(V\) is the volume of blood pumped out during the unit of time, \(\delta("\) delta") is the density of the blood, \(v\) is the average velocity of the exiting blood, and \(g\) is the acceleration of gravity. When \(P, V, \delta,\) and \(v\) remain constant, \(W\) becomes a function of \(g,\) and the equation takes the simplified form $$W=a+\frac{b}{g}(a, b\( constant \))$$ As a member of NASA's medical team, you want to know how sensitive \(W\) is to apparent changes in \(g\) caused by flight maneuvers, and this depends on the initial value of \(g\) . As part of your investigation, you decide to compare the effect on \(W\) of a given change \(d g\) on the moon, where \(g=5.2 \mathrm{ft} / \mathrm{sec}^{2},\) with the effect the same change \(d g\) would have on Earth, where \(g=32\) \(\mathrm{ft} / \mathrm{sec}^{2} .\) Use the simplified equation above to find the ratio of \(d W_{\mathrm{moon}}\) to \(d W_{\mathrm{Earth}}\)

$$ \begin{array}{l}{\text { Multiple Choice Which of the following functions is an }} \\ {\text { antiderivative of } \frac{1}{\sqrt{x}} ? \quad \mathrm{}}\end{array} $$ $$ (\mathbf{A})-\frac{1}{\sqrt{2 x^{3}}}(\mathbf{B})-\frac{2}{\sqrt{x}} \quad(\mathbf{C}) \frac{\sqrt{x}}{2}(\mathbf{D}) \sqrt{x}+5(\mathbf{E}) 2 \sqrt{x}-10 $$

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