Chapter 1.

1.1 Problem Statement

{2,6,7}
round(pow($a,-0.25),6)
round($ans,1)
pow(0.5,-4) - $a
pow(1,-4) - $a
pow($x0,-4) - $a
-4*pow($x0,-5)
round($x0-$fx0/$fpx0,6)
pow($x1,-4)-$a
-4*pow($x1,-5)
round($x1-$fx1/$fpx1,6)
pow($x2,-4)-$a
-4*pow($x2,-5)
round($x2-$fx2/$fpx2,6)
round($x3,3)

Approximate to three decimal places using Newton's Method and compare with the value from a calculator.

$a-1/4

1.2 Step 1

Newton's Method is used to approximate a root of f(x) = 0.

Question Sequence

Question 1.1

In order to use this method, we must find a function f(x) that has x = $a-1/4 as a root. If x = $a-1/4, then x-4 = nc1ItEz0kR4=.

Correct.
Incorrect.

Question 1.2

Therefore, we choose f(x) = x-4 - nc1ItEz0kR4=.

Correct.
Incorrect.

1.3 Step 2

In order to use Newton's method to find a root of f(x) = x-4 - $a, we need an initial guess, x0, which is close to the root, and the formula for Newton's method to generate successive approximations.

Question Sequence

Question 1.3

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
Correct.
Incorrect.

Question 1.4

We see that the method requires f'(x).

Find f'(x).

f'(x) = a·xb

where a = sWNiRN4ZHUU= and b = /euMlW48Z7w=

Correct.
Incorrect.

Question 1.5

Now choose an initial guess x0 which is close to the root. If we find two x-values a and b where f(a) < 0 < f(b) or f(a) > 0 > f(b), then we can pick our initial guess x0 to be a number between a and bbecause the root must be between those numbers.

Let's select a = 0.5 and b = 1.

f(0.5) = nh/UkZmBhHI=i4H1MqFIfHIXjgQa3GgUug== 0.

f(1) = wXYIfDe/F4g=8SjSn/KdQtEH7LgXNHs+Vw== 0.

So we can make an initial guess between these two points. We guess x0 = $x0.

Correct.
Incorrect.

1.4 Step 3

To approximate accurately to three decimal places, perform Newton's method several times until the successive approximations agree up to three decimal places.

Question Sequence

Question 1.6

Find x1.

= +QaHWdBXbCc=

(Round your answer to six decimal places.)

Correct.
Incorrect.

Question 1.7

Continue finding successive approximations for x2 and x3 up to six decimal places.

x2 = 4MLAwx+65zo=

x3 = PU3rm14CqBw=

Correct.
Incorrect.

Question 1.8

Since x2 and x3 agree up to three decimal places, the approximation of

$a-1/4 ≈ fpzogoeOkbmsBux4.

Correct.
Incorrect.

1.5 Step 4

Question 1.9

Using a calculator, find the value of $a−1/4 to six decimal places to show that our approximation using Newton's Method is very close to the calculator value.

$a−1/4 ≈ qjqZz1N+poo=

Correct.
Incorrect.