Question

write newtons formula used to approximate $sqrt3{7.5}$, a solution of the equation $x^3 - 7.5 = 0$ and find the third iteration value. be sure to verify both parts of the answer are correct when making your selection.$\bigcirc$ $x_{n+1}=2+\frac{x_n^3 - 7.5}{3x_n^2}; x_3approx2.116201451LXB0\bigcirc$ $x_{n+1}=x_n-\frac{x_n^3 - 7.5}{3x_n^2}; x_3approx1.957433821$$\bigcirc$ $x_{n+1}=x_n-\frac{x_n^3 - 7.5}{3x_n^2}; x_3approx2.175010057$

Explanation:

Step1: Recall Newton's Method formula

Newton's method is given by $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

Step2: Define $f(x)$ and find $f'(x)$

Let $f(x) = x^3 - 7.5$. Then $f'(x) = 3x^2$. Substitute into the formula:
$x_{n+1} = x_n - \frac{x_n^3 - 7.5}{3x_n^2}$

Step3: Choose initial guess $x_1$

A reasonable initial guess is $x_1=2$ (since $2^3=8$ which is close to 7.5).

Step4: Calculate second iteration $x_2$

Substitute $x_1=2$ into the formula:
$x_2 = 2 - \frac{2^3 - 7.5}{3(2)^2} = 2 - \frac{8 - 7.5}{12} = 2 - \frac{0.5}{12} = 2 - \frac{1}{24} \approx 1.958333333$

Step5: Calculate third iteration $x_3$

Substitute $x_2\approx1.958333333$ into the formula:
$x_3 = 1.958333333 - \frac{(1.958333333)^3 - 7.5}{3(1.958333333)^2}$
First compute numerator: $(1.958333333)^3 \approx 7.524305556$, so $7.524305556 - 7.5 = 0.024305556$
Denominator: $3(1.958333333)^2 \approx 3(3.835069444) = 11.50520833$
$x_3 \approx 1.958333333 - \frac{0.024305556}{11.50520833} \approx 1.958333333 - 0.002112452 \approx 1.956220881$
(Note: Using more precise calculation of $x_2$: $x_2 = \frac{47}{24} \approx 1.958333333$, $(47/24)^3 = 103823/13824 \approx 7.510375977$, so numerator $=7.510375977-7.5=0.010375977$, denominator $=3*(47/24)^2=3*(2209/576)=6627/576\approx11.50520833$, $x_3=47/24 - (0.010375977/11.50520833)\approx1.958333333-0.0009018\approx1.9574315$, which matches the value in option 3.)

Answer:

$x_{n+1}=x_n-\frac{x_n^3 - 7.5}{3x_n^2}; x_3\approx 1.957433821$