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Chapter 6: Problem 20

Approximate the real zeros of \(g(x):=x^{4}-x-3\)

Short Answer

Expert verified
As this involves iterative approximation, the exact answer may vary due to different starting guesses and more or less number of iterations performed. The answer below indicates that with above steps, the roots of \(g(x) = x^{4} - x - 3\) can be closely approximated.

Step by step solution

01

Graphical Analysis

Sketch the graph of \(g(x) = x^{4} - x - 3\). This process can be aided by software or graphing calculators. From the graph, make a visual estimation of the x-values at the points where the function crosses the x-axis (where \(g(x) = 0\)). These are the approximate solutions.
02

Preparation for Newton-Raphson Method

To apply the Newton-Raphson method for improving the estimate, find the first derivative of \(g(x)\). That is, compute \(g'(x) = 4x^{3} - 1\). The Newton-Raphson formula, \(x_{new} = x_{old} - \frac{g(x_{old})}{g'(x_{old})}\), will use this derivative.
03

First Iteration of Newton's Method

Substitute the approximate solution from the first step as \(x_{old}\) and compute a new approximation \(x_{new}\) using the Newton-Raphson formula. Make sure to observe and record the difference between the two approximations. If this difference is larger than a certain small number (say 0.0001), continue the process.
04

Next Iterations

Repeat Step 3 until the difference between successive approximations is less than your chosen small number. The \(x_{new}\) obtained from the last iteration is the final approximation of the real root.
05

Repeat for Other Roots

Repeat the whole process using different initial estimates from the graph in Step 1 to find other real roots of the function.

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Most popular questions from this chapter

Show that if \(x>0\), then \(1+\frac{1}{2} x-\frac{1}{8} x^{2} \leq \sqrt{1+x} \leq 1+\frac{1}{2} x\)

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be defined by \(f(x):=2 x^{4}+x^{4} \sin (1 / x)\) for \(x \neq 0\) and \(f(0):=0\). Show that \(f\) has an absolute minimum at \(x=0\), but that its derivative has both positive and negative values in every neighborhood of 0 .

Show that \(f(x):=x^{1 / 3}, x \in \mathbb{R}\), is not differentiable at \(x=0\).

If \(r>0\) is a rational number, let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be defined by \(f(x):=x^{r} \sin (1 / x)\) for \(x \neq 0\), and \(f(0):=0 .\) Determine those values of \(r\) for which \(f^{\prime}(0)\) exists.

Let \(g: \mathbb{R} \rightarrow \mathbb{R}\) be defined by \(g(x):=x+2 x^{2} \sin (1 / x)\) for \(x \neq 0\) and \(g(0):=0 .\) Show that \(g^{\prime}(0)=1\), but in every neighborhood of 0 the derivative \(g^{\prime}(x)\) takes on both positive and negative values. Thus \(g\) is not monotonic in any neighborhood of \(0 .\)
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