/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Find the slope \(m\) of the line... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the slope \(m\) of the line joining \((a, f(a))\) and \((b, f(b))\). Then use the Newton-Raphson method to estimate the values of \(c\) for which \(f^{\prime}(c)=m .\) Continue the process until successive iterations obtained by the calculator are identical. $$ f(x)=x^{2}-2 / x ; a=1, b=2 $$

Short Answer

Expert verified
The slope \(m\) is 4. Use Newton-Raphson with \(c_0 = 1.5\) to estimate \(c\) where \(f'(c) = 4\). Iterate until convergence.

Step by step solution

01

Calculate the Function Values

First, we need to calculate the values of the function \(f(x) = x^2 - \frac{2}{x}\) at \(x = a\) and \(x = b\). \( f(1) = 1^2 - \frac{2}{1} = 1 - 2 = -1 \) \( f(2) = 2^2 - \frac{2}{2} = 4 - 1 = 3 \)Hence, we have the points \((1, -1)\) and \((2, 3)\).
02

Calculate the Slope (m)

Now, calculate the slope \(m\) of the line connecting the points \((1, -1)\) and \((2, 3)\):\[ m = \frac{f(b) - f(a)}{b - a} = \frac{3 - (-1)}{2 - 1} = \frac{3 + 1}{1} = 4\]Thus, the slope \(m = 4\).
03

Find the Derivative f'(x)

Find the derivative \(f'(x)\) of the function \(f(x) = x^2 - \frac{2}{x}\):\[ f'(x) = \frac{d}{dx} \left( x^2 - \frac{2}{x} \right) = 2x + \frac{2}{x^2}\]
04

Set f'(x) = m and Use Newton-Raphson Method

We need to find \(c\) such that \(f'(c) = m = 4\). Thus, we solve the equation:\[ 2c + \frac{2}{c^2} = 4\]For the Newton-Raphson method, the formula is:\[ c_{n+1} = c_n - \frac{f'(c_n) - 4}{f''(c_n)}\]
05

Calculate Second Derivative for Newton-Raphson

Calculate the second derivative \(f''(x)\):\[ f''(x) = \frac{d}{dx} \left( 2x + \frac{2}{x^2} \right) = 2 - \frac{4}{x^3}\]
06

Initial Guess and Iterations

Make an initial guess for \(c\); typically, a value between \(a\) and \(b\), say \(c_0 = 1.5\).Use the Newton-Raphson formula to iterate:1. Calculate \(f'(c_0) - 4\) and substitute \(c_0\) into \(f''(x)\).2. Compute \(c_1 = c_0 - \frac{f'(c_0) - 4}{f''(c_0)}\).3. Repeat until \(c_{n+1} = c_n\).
07

Verify Convergence

Continue to iterate the Newton-Raphson step until the values \(c_n\) converge to a stable value, where the successive iterations are the same to the desired precision. The result will give the value of \(c\) where \(f'(c) = m\).

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Key Concepts

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

Derivative Calculation
Calculating a derivative is a crucial aspect of understanding how functions change at any given point. In this exercise, we begin by dealing with the function \(f(x) = x^2 - \frac{2}{x}\). To find its derivative, \(f'(x)\), we perform differentiation, which gives us a new function expressing how \(f(x)\) changes as \(x\) changes.
In our function, differentiating \(x^2\) results in \(2x\), whereas the derivative of \(-\frac{2}{x}\) is \(\frac{2}{x^2}\) because of the power rule. Combining these, we're left with:
\[f'(x) = 2x + \frac{2}{x^2}\]
This derivative tells us the rate at which our original function's value is changing at any specific \(x\) value. Derivatives are a fundamental tool in calculus. They help us understand the behavior of graphs and optimize functions in various fields like physics and engineering.
Slope of a Line
The slope of a line represents how steep the line is, indicating the rate of change between two points. In this scenario, we seek the slope \(m\) of the line connecting the points \((1, -1)\) and \((2, 3)\).
To compute \(m\), we use the formula:
\[m = \frac{f(b) - f(a)}{b - a} = \frac{3 - (-1)}{2 - 1} = 4\]
Here's what happens:
  • Subtract the y-values, which provides the change in the y-direction.
  • Subtract the x-values for the x-direction change.
  • Divide these results to find the slope \(m = 4\).
Understanding slope is key when interpreting graphs, as it determines how the function behaves. Knowing the slope helps predict future values and understand past behavior.
Convergence Criteria
Convergence criteria are essential when employing iterative methods, like the Newton-Raphson method, to ensure we reach a stable solution. In our case, we are finding a \(c\)-value such that \(f'(c) = 4\).
Using the Newton-Raphson formula:
\[c_{n+1} = c_n - \frac{f'(c_n) - 4}{f''(c_n)}\]
The idea is to start with an initial guess \(c_0\), and update it iteratively until the sequence \(c_n\) converges. Convergence implies that successive values of \(c_n\) become practically identical, indicating stability and accuracy. To check for convergence, observe if \(c_{n+1} - c_n\) is less than a chosen small value (like 0.001).
This small threshold ensures high precision in our results. Iterative methods without convergence criteria might yield incorrect conclusions due to non-stability.
Second Derivative
The second derivative, \(f''(x)\), offers more insight into the geometry of the function graph, such as concavity and inflection points. For our function \(f(x) = x^2 - \frac{2}{x}\), we've already calculated \(f'(x) = 2x + \frac{2}{x^2}\).
To find \(f''(x)\), we differentiate \(f'(x)\):
\[f''(x) = \frac{d}{dx}(2x) + \frac{d}{dx}(\frac{2}{x^2}) = 2 - \frac{4}{x^3}\]
Here, the second derivative tells us about the acceleration of the slope. It's particularly useful in the Newton-Raphson method to adjust the slope for iteration, helping us refine our approximation of \(c\). Understanding \(f''(x)\) is vital in various applications, from determining forces in physics to predicting price changes in economics.

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Most popular questions from this chapter

The Spice-of-Life Company is preparing to create shipping crates. The company wishes the volume of each crate to be 6 cubic feet, with the crate's base to be square between 1 and 2 feet on a side. Assume that the material for the bottom costs \(\$ 5\), the sides \(\$ 2\), and the top \(\$ 1\) per square foot. Find the dimensions that yield the minimum cost.

Let \(h(t)\) denote the height at time \(t\) of an object moving in a vertical direction above the ground. Assume that the object is subject only to the force of gravity. a. Show that the graph of \(h\) is concave downward. b. In terms of the behavior of the velocity \(v\), tell why the result of (a) is true.

Suppose the current \(I(t)\) flowing in an electrical circuit at time \(t\) is given by $$ I(t)=\frac{100}{1+t^{2}}+3 \sin \frac{30 t}{\pi} \text { for } t \geq 0 $$ Show that $$ \lim _{t \rightarrow \infty}\left(I(t)-3 \sin \frac{30 t}{\pi}\right)=0 $$ Thus for large values of \(t, I(t)\) is very nearly equal to \(3 \sin (30 t / \pi)\). The expression \(3 \sin (30 t / \pi)\) is called the steady-state current, and the expression \(100 /\left(1+t^{2}\right)\) is the transient current (since it is significant only for small values of \(t\) ).

Plot the graph of \(f^{\prime}\), and then use the Newton-Raphson method to approximate all values of \(c\) for which \((c, f(c))\) is an inflection point. Continue until the output of the calculator does not change. $$ f(x)=e^{x}-\sin x \text { for }-e \leq x<\infty $$

Find all inflection points (if any) of the graph of the function. Then sketch the graph of the function. $$ f(x)=x^{3}+3 $$

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