The figure shows the graph of a function $ f $. Suppose that Newton's method is used to approximately the root $ s $ of the equation $ f(x) = 0 $ with initial approximaton $ x_1 = 6 $.

(a) Draw the tangent lines that are used to find $ x_2 $ and $ x_3 $, and estimate the numerical values of $ x_2 $ and $ x_3 $.

(b) Would $ x_1 = 8 $ be a better first approximation? Explain.

Amrita B.

Numerade Educator

Follow the instructions for Exercise 1(a) but use $ x_1 = 1 $ as the starting approximation for finding the root $ r $.

Yuki H.

Numerade Educator

Suppose the tangent line to the curve $ y = f(x) $ at the point $ (2, 5) $ has the equation $ y = 9 - 2x $. If Newton's method is used to locate a root of the equation $ f(x) = 0 $ and the initial approximation is $ x_1 = 2 $, find the second approximation $ x_2 $.

Yuki H.

Numerade Educator

For each initial approximation, determine graphically what happens if Newton's method is used for the function whose graph is shown.

(a) $ x_1 = 0 $ (b) $ x_1 = 1 $ (c) $ x_1 = 3 $

(d) $ x_1 = 4 $ (e) $ x_1 = 5 $

Sherrie F.

Numerade Educator

For which of the initial approximations $ x_1 = a $, $ b $, $ c $, and $ d $ do you thing Newton's method will work and lead to the root of the equation $ f(x) = 0 $?

Amrita B.

Numerade Educator

Use Newton's method with the specified initial approximation $ x_1 $ to find $ x_3 $, the third approximation to the root of the given equation. (Give your answer to four decimal places.)

$ 2x^3 - 3x^2 + 2 = 0 $, $ x_1 = -1 $

Amrita B.

Numerade Educator

Use Newton's method with the specified initial approximation $ x_1 $ to find $ x_3 $, the third approximation to the root of the given equation. (Give your answer to four decimal places.)

$ \dfrac{2}{x} - x^2 + 1 = 0 $, $ x_1 = 2 $

Amrita B.

Numerade Educator

Use Newton's method with the specified initial approximation $ x_1 $ to find $ x_3 $, the third approximation to the root of the given equation. (Give your answer to four decimal places.)

$ x^7 + 4 = 0 $, $ x_1 = -1 $

Amrita B.

Numerade Educator

Use Newton's method with initial approximation $ x_1 = -1 $ to find $ x_2 $, the second approximation to the root of the equation $ x^3 + x + 3 = 0 $. Explain how the method works by first graphing the function and its tangent line at $ (-1, 1) $.

Amrita B.

Numerade Educator

Use Newton's method with initial approximation $ x_1 = 1 $ to find $ x_2 $, the second approximation to the root of the equation $ x^4 - x - 1 = 0 $. Explain how the method works by first graphing the function and its tangent line at $ (1, -1) $.

Amrita B.

Numerade Educator

Use Newton's method to approximate the given number correct to eight decimal places.

$ \sqrt[4]{75} $

Oswaldo J.

Numerade Educator

Use Newton's method to approximate the given number correct to eight decimal places.

$ \sqrt[8]{500} $

Amrita B.

Numerade Educator

(a) Explain how we know that the given equation must have a root in the given interval. (b) Use Newton's method to approximate the root correct to six decimal places.

$ 3x^4 - 8x^3 + 2 = 0 $, $[2, 3] $

Amrita B.

Numerade Educator

(a) Explain how we know that the given equation must have a root in the given interval. (b) Use Newton's method to approximate the root correct to six decimal places.

$ -2x^5 + 9x^4 - 7x^3 - 11x = 0 $, $[3, 4] $

Amrita B.

Numerade Educator

Use Newton's method to approximate the indicated root of the equation correct to six decimal places.

The negative root of $ e^x = 4 - x^2 $

Amrita B.

Numerade Educator

Use Newton's method to approximate the indicated root of the equation correct to six decimal places.

The positive root of $ 3\sin x = x $

Amrita B.

Numerade Educator

Use Newton's method to find all solutions of the equation correct to six decimal places.

$ 3\cos x = x + 1 $

Amrita B.

Numerade Educator

Use Newton's method to find all solutions of the equation correct to six decimal places.

$ \sqrt{x + 1} = x^2 - x $

Amrita B.

Numerade Educator

Use Newton's method to find all solutions of the equation correct to six decimal places.

$ 2^x = 2 - x^2 $

Amrita B.

Numerade Educator

Use Newton's method to find all solutions of the equation correct to six decimal places.

$ \ln x = \dfrac{1}{x - 3} $

Amrita B.

Numerade Educator

Use Newton's method to find all solutions of the equation correct to six decimal places.

$ x^3 = \tan^{-1} x $

Amrita B.

Numerade Educator

Use Newton's method to find all solutions of the equation correct to six decimal places.

$ \sin x = x^2 - 2 $

Amrita B.

Numerade Educator

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

$ -2x^7 - 5x^4 + 9x^3 + 5 = 0 $

Amrita B.

Numerade Educator

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

$ x^5 - 3x^4 + x^3 - x^2 + 6 = 0 $

Sherrie F.

Numerade Educator

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

$ \dfrac{x}{x^2 + 1} = \sqrt{1 - x} $

Sherrie F.

Numerade Educator

$ \cos (x^2 - x) = x^4 $

Sherrie F.

Numerade Educator

$ 4e^{-x^2} \sin x = x^2 - x + 1 $

Sherrie F.

Numerade Educator

$ \ln (x^2 + 2) = \dfrac{3x}{\sqrt{x^2 + 1}} $

Sherrie F.

Numerade Educator

(a) Apply Newton's method to the equation $ x^2 - a = 0 $ to derive the following square-root algorithm (used by the ancient Babylonians to compute $ \sqrt{a} $):

$$ x_{n + 1} = \dfrac{1}{2} \left( x_n + \dfrac{a}{x_n} \right) $$

(b) Use part (a) to compute $ \sqrt{1000} $ correct to six decimal places.

Sherrie F.

Numerade Educator

(a) Apply Newton's method to the equation $ 1/x - a = 0 $ to derive the following reciprocal algorithm:

$$ x_{x + 1} = 2x_n - ax{_n}^2 $$

(This algorithm enables a computer to find reciprocals without actually dividing.)

(b) Use part (a) to compute $ 1/1.6984 $$ correct to six decimal places.

Sherrie F.

Numerade Educator

Explain why Newton's method doesn't work for finding the root of the equation

$$ x^3 - 3x + 6 = 0 $$

if the initial approximation is chosen to be $ x_1 = 1 $.

Tyler G.

Numerade Educator

(a) Use Newton's method with $ x_1 = 1 $ to find the roof of the equation $ x^3 - x = 1 $ correct to six decimal places.

(b) Solve the equation in part (a) using $ x_1 == 0.6 $ as the initial approximation.

(c) Solve the equation in part (a) using $ x_1 = 0.57 $.

(You definitely need a programmable calculator for this part.)

(d) Graph $ f(x) = x^3 - x - 1 $ and its tangent lines at $ x_1 = 1 $, $ 0.6 $, and $ 0.57 $ to explain why Newton's method is so sensitive to the value of the initial approximation.

Sherrie F.

Numerade Educator

Explain why Newton's method fails when applied to the equation $ \sqrt[3]{x} = 0 $ with any initial approximation $ x_1 \not= 0 $. Illustrate your explanation with a sketch.

Sherrie F.

Numerade Educator

If

$ f(x) = \left\{

\begin{array}{ll}

\sqrt{x} & \mbox{if} x \geqslant 0\\

-\sqrt{-x} & \mbox{if} x < 0\\

\end{array} \right. $

then the root of the equation $ f(x) = 0 $ is $ x = 0 $. Explain why Newton's method fails to find the root no matter which initial approximation $ x_1 \not= 0 $ is used. Illustrate your explanation with a sketch.

Sherrie F.

Numerade Educator

(a) Use Newton's method to find the critical numbers of the function

$$ f(x) = x^+ - x^4 + 3x^3 - 2x $$

correct to six decimal places.

(b) Find the absolute minimum value of $ f $ correct to four decimal places.

Sherrie F.

Numerade Educator

Use Newton's method to find the absolute maximum value of the function $ f(x) = x\cos x $, $ 0 \leqslant 0 \leqslant \pi $, correct to six decimal places.

Sherrie F.

Numerade Educator

Use Newton's method to find the coordinates of the inflection point of the curve $ y = x^2 \sin x $, $ 0 \leqslant x \leqslant \pi $, correct to six decimal places.

Sherrie F.

Numerade Educator

Of the infinitely many lines that are tangent to the curve $ y = -\sin x $ and pass through the origin, there is one that has the largest slope. Use Newton's method to find the slope of that line correct to six decimal places.

Sherrie F.

Numerade Educator

Use Newton's method to find the coordinates, correct to six decimal places, of the point on the parabola $ y = (x - 1)^2 $ that is closest to the origin.

Sherrie F.

Numerade Educator

In the figure, the length of the chord $ AB $ is $ 4 $ cm and the length of the arc is $ AB $ is $ 5 $ cm. Find the central angle $ \theta $, in radians, correct to four decimal places. Then give the answer to the nearest degree.

Sherrie F.

Numerade Educator

A car dealer sells a new car for $ \$ 18,000 $. He also offers to sell the same car for payments of $ \$ 375 $ per month for five years. What monthly interest rate is this dealer charging?

To solve this problem you will need to use the formular for the present value $ A $ of an annuity consisting of $ n $ equal payments of size $ R $ with interest rate $ i $ per time period:

$$ A = \dfrac{R}{i} [1 - (1 + i)^{-n}] $$

Replacing $ i $ by $ x $, show that

$$ 48x(1 + x)^{60} - (1 + x)^{60} + 1 = 0 $$

Use Newton's method to solve this equation.

Sherrie F.

Numerade Educator

The figure shows the sun located at the origin and the earth at the point $ (1, 0) $. (The unit here is the distance between the centers of the earth and the sun, called an astronomical unit: $ 1 AU \approx 1.496 x 10^8 $ km.) There are five locations $ L_1 $, $ L_2 $, $ L_3 $, $ L_4 $, and $ L_5 $ in this plane of rotation of the earth about the sun where a satellite remains motionless with respect to the earth because the forces acting on the satellite (including the gravitational attractions of the earth and the sun) balance each other. These locations are called libration points. (A solar research satellite has been placed at one of these libration points.) If $ m_1 $ is the mass of the sun, $ m_2 $ is the mass of the earth, and $ r = m_2/(m_1 + m_2) $, it turns out that the $ x $-coordinate of $ L_1 $ is the unique root of the fifth-degree equation

$$

\begin{equation}

p(x) = x^5 - (2 + r)x^4 + (1 + 2r)x^3 - (1 - r)x^2 \\

+ 2(1 - r)x + r - 1 = 0

\end{equation}

$$

and the $ x $-coordinate of $ L_2 $ is the root of the equation

$$ p(x) - 2rx^2 = 0 $$

Using the value of $ r = 3.04042\: X\: 10^{-6} $, find the locations of the libration points of (a) $ L_1 $ and (b) $ L_2 $.

Sherrie F.

Numerade Educator