Calculate one iterations of Newton's Method to approximate a zero for f(x) = x ^ 2 - 3 where x 1 =2 Write your answer in decimal form.

Question

Calculate one iterations of Newton's Method to approximate a zero for f(x) = x ^ 2 - 3 where x 1 =2 Write your answer in decimal form.

Answer 1

Recall that

x = x - f/f'
So, starting with x1=2
x2 = x1 - f(x1)/f'(x1) = 2 - f(2)/f'(2) = 2 - (2^2-3)/(2*2) = 2 - 1/4 = 7/4
f(x2) = 1/16 ... getting closer, eh?

Answer 2

f(x) = x^2 - 3

f ' (x) = 2x

x_new = x - f(x)/f ' (x)
= x - (x^2 - 3)/2x
= (2x^2 - x^2 + 3)/2x
= (x^2 + 3)/(2x)

x_new x
2 1.75 <===== one iteration
1.75 1.7321
1.7321 1.73205
1.73205 1.73205 ---- repeat to 5 decimals

(2 more iterations gave the answer correct to 9 decimals on my calculator)

so x^2 - 3 = 0 has a root at x = 1.73205

actual answer:
x^2 - 3 = 0
x^2 = 3
x = ± √3 = appr ± 1.73201..

Answer 3

To calculate one iteration of Newton's Method, we need to find the tangent line at the point (x_1, f(x_1)) and then find its x-intercept. The x-intercept of the tangent line will be the new approximation, denoted as x_2.

Given that f(x) = x^2 - 3 and x_1 = 2, we need to find the slope of the tangent line at x_1.

Step 1: Find the derivative of f(x).
f'(x) = 2x

Step 2: Evaluate f'(x_1).
f'(2) = 2 * 2 = 4

Step 3: Find the equation of the tangent line.
Using the point-slope form, the equation of the tangent line is:
y - f(x_1) = m(x - x_1)
y - (f(2)) = 4(x - 2)
y - (2^2 - 3) = 4(x - 2)
y - (4 - 3) = 4(x - 2)
y - 1 = 4x - 8
y = 4x - 7

Step 4: Find the x-intercept of the tangent line.
Setting y to zero, we get:
0 = 4x - 7
4x = 7
x = 7/4 = 1.75

Therefore, one iteration of Newton's Method using x_1 = 2 gives us an approximation of the zero for f(x) = x^2 - 3 as x_2 = 1.75 in decimal form.