💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

## Educators

### Problem 1

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.

### Problem 2

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

Yuki H.

### Problem 3

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.

### Problem 4

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.

### Problem 5

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.

### Problem 6

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.

### Problem 7

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.

### Problem 8

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.

### Problem 9

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.

### Problem 10

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.

### Problem 11

Use Newton's method to approximate the given number correct to eight decimal places.
$\sqrt[4]{75}$

Oswaldo J.

### Problem 12

Use Newton's method to approximate the given number correct to eight decimal places.
$\sqrt[8]{500}$

Amrita B.

### Problem 13

(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.

### Problem 14

(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.

### Problem 15

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.

### Problem 16

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.

### Problem 17

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

$3\cos x = x + 1$

Amrita B.

### Problem 18

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

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

Amrita B.

### Problem 19

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

$2^x = 2 - x^2$

Amrita B.

### Problem 20

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

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

Amrita B.

### Problem 21

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

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

Amrita B.

### Problem 22

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

$\sin x = x^2 - 2$

Amrita B.

### Problem 23

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.

### Problem 24

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.

### Problem 25

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.

### Problem 26

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.

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

Sherrie F.

### Problem 27

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.

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

Sherrie F.

### Problem 28

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.

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

Sherrie F.

### Problem 29

(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.
(a) Apply Newton's method to the equation $1/x - a = 0$ to derive the following reciprocal algorithm:
$$x_{x + 1} = 2x_n - ax{_n}^2$$