Question

Use Newton's method with initial approximation x1 = 1 to find x2, the second approximation to the root of the equation x4 − x − 2 = 0.

2. Use Newton's method to approximate the indicated root of the equation correct to six decimal places. The root of x^4 − 2x^3 + 7x^2 − 9 = 0 in the interval [1, 2]

3.Suppose the line y = 4x − 1 is tangent to the curve y = f(x) when x = 2. If Newton's method is used to locate a root of the equation f(x) = 0 and the initial approximation is x1 = 2, find the second approximation x2.

4. Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. (Give your answer to four decimal places.) x^5 + 8 = 0, x1 = −1

5. Use Newton's method to find all roots of the equation correct to six decimal places. (Enter your answers as a comma-separated list.) e^x = 9 − 3x

Answer 1

Answer:

1. 3/5

2. 1.219841

3. 2 - f(2)/7

4. -1.9682

5. 1.502446

Step-by-step explanation:

1.

If we call

$\large f(x)=x^4-x-2$

then the second approximation to the root of the equation f(x)=0 would be

$\large x_2=x_1-\frac{f(x_1)}{f'(x_1)}$

$\large f(x_1)=f(1)=-2\\\\f'(x)=4x^3-1\Rightarrow f'(x_1)=f'(1)=3$

hence

$\large x_2=1-\frac{-2}{3}=1+\frac{2}{3}=\frac{5}{3}\approx1.666666$

2.

Here we have

$\large f(x)=x^4-2x^3+7x^2-9\\\\f'(x)=4x^3-6x^2+14x$

Let's start with  

$\large x_1=1$

then

$\large x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1-\frac{f(1)}{f'(1)}=1-\frac{-3}{12}=1+\frac{1}{4}=\frac{5}{4}=1.25$

$\large x_3=x_2-\frac{f(x_2)}{f'(x_2)}=1.25-\frac{f(1.25)}{f'(1.25)}=1.220343137$

$\large x_4=x_3-\frac{f(x_3)}{f'(x_3)}=1.220343137-\frac{f(1.220343137)}{f'(1.220343137)}=1.219841912$

$\large x_5=x_4-\frac{f(x_4)}{f'(x_4)}=1.219841912-\frac{f(1.219841912)}{f'(1.219841912)}=1.219841771$

Since the first 6 decimals of $\large x_4$ and $\large x_5$ are equal, the desired approximation is 1.219841

3.

If the line y = 4x − 1 is tangent to the curve y = f(x) when x = 2, then f'(2) = 4*2 - 1 = 7, so

$\large x_2=2-\frac{f(2)}{f'(2)}=2-\frac{f(2)}{7}$

4.

Here

$\large f(x)=x^5 + 8\\\\f'(x)=5x^4$

$\large x_2=-1-\frac{f(-1)}{f'(-1)}=-2.4$

$\large x_3=-2.4-\frac{f(-2.4)}{f'(-2.4)}=-1.9682$

5.

We want to find all the values x such that

$\large e^x=9-3x$

or what is the same, the x such that

$\large e^x+3x-9=0$

so, let f(x) be

$\large f(x)=e^x+3x-9$

and let's use Newton's method to find the roots of f(x).

Since

$\large f'(x)=e^x +3>0$

f is strictly increasing, and since

f(1) = e+3-9 = e - 6 < 0

and

$\large f(2)=e^2+6-9=e^2-3>0$

f has only one root in [1,2]

By using Newton's iterations starting with $\large x_1=1$

$\large x_2=1-\frac{f(1)}{f'(1)}=1.573899431$

$\large x_3=1.573899431-\frac{f(1.573899431)}{f'(1.573899431)}=1.503982961$

$\large x_4=1.503982961-\frac{f(1.503982961)}{f'(1.503982961)}=1.502446348$

$\large x_5=1.502446348-\frac{f(1.502446348)}{f'(1.502446348)}=1.50244564$

$\large x_6=1.50244564-\frac{f(1.50244564)}{f'(1.50244564)}=1.50244564$

Since $\large x_5=x_6$ then

x=1.50244564 is the desired root.

The answers as a comma-separated list would be

1,  1.573899431,  1.503982961,  1.502446348, 1.50244564

The answer rounded to six decimal places would be

1.502446