Answered: cos(x? – x) = x*

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter4: Calculating The Derivative

Section4.3: The Chain Rule

Problem 4YT: YOUR TURN Find dy/dx if y=5x2-6x-2.

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Question

Use Newton’s method to find all the roots of the
equation correct to eight decimal places. Start by drawing a
graph to find initial approximations.

Expert Solution

Step 1

Newton's method: This method is also known as Newton Raphson method and it is a technique used to

find the approximation of the root of the function.

It is given that $\cos (x^{2} - x) = x^{4}$ . Rewriting it, we get, $x^{4} - \cos (x^{2} - x) = 0$ . Let us define a function $f (x)$

such that $f (x) = x^{4} - \cos (x^{2} - x)$ . Then, the derivative of $f (x)$ is $f' (x) = 4 x^{3} + (2 x - 1) \sin (x^{2} - x)$ .

Step 2

The graph of the function $f (x) = x^{4} - \cos (x^{2} - x)$ is given below:

Since the curve intersects the x-axis at two points, there are two roots of the equation $f (x) = 0$ . Let

us consider the initial approximations $x_{0} = - 0.5$ and $x_{0} = 1.2$ .

Step 3

First, we will use the initial approximation $x_{0} = - 0.5$ .

First iteration:

We have $f (x_{0}) = - 0.66918887$ and $f' (x_{0}) = - 1.86327752$ . So we get,

$\begin{array}{rcl} x_{1} & = & x_{0} - \frac{f (x_{0})}{f' (x_{0})} \\ = & - 0.5 - \frac{- 0.66918887}{- 1.86327752} \\ = & - 0.85914611 \end{array}$

Hence, $x_{1} = - 0.85914611$ .

Second iteration:

We have $f (x_{1}) = 0.57131762$ and $f' (x_{1}) = - 5.25399218$ . So we get,

$\begin{array}{rcl} x_{2} & = & x_{1} - \frac{f (x_{1})}{f' (x_{1})} \\ = & - 0.85914611 - \frac{0.57131762}{- 5.25399218} \\ = & - 0.75040639 \end{array}$

Hence, $x_{2} = - 0.75040639$ .

Step 4

Third iteration:

We have $f (x_{2}) = 0.06264139$ and $f' (x_{2}) = - 4.10874469$ . So we get,

$\begin{array}{rcl} x_{3} & = & x_{2} - \frac{f (x_{2})}{f' (x_{2})} \\ = & - 0.75040639 - \frac{0.06264139}{- 4.10874469} \\ = & - 0.73516052 \end{array}$

Hence, $x_{3} = - 0.73516052$ .

Fourth iteration:

We have $f (x_{3}) = 0.00119097$ and $f' (x_{3}) = - 3.9527852$ . So we get,

$\begin{array}{rcl} x_{4} & = & x_{3} - \frac{f (x_{3})}{f' (x_{3})} \\ = & - 0.73516052 - \frac{0.00119097}{- 3.9527852} \\ = & - 0.73485922 \end{array}$

Hence, $x_{4} = - 0.73485922$ .

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