use newtons method to approximate the given number correct to eight decimal places.\n$$\\sqrt4{73}$$

Explanation:

Step1: Define the function and its derivative

To find $\sqrt[4]{73}$, let $x = \sqrt[4]{73}$, so $x^4 = 73$. Define $f(x) = x^4 - 73$ and $f'(x) = 4x^3$.

Step2: Set up the Newton's method formula

The iteration formula is $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{x_n^4 - 73}{4x_n^3}$.

Step3: Choose an initial guess

Since $2^4 = 16$ and $3^4 = 81$, the value is close to 3. Let $x_0 = 3$.

Step4: Perform the first iteration

$$x_1 = 3 - \frac{3^4 - 73}{4(3^3)} = 3 - \frac{81 - 73}{108} = 3 - \frac{8}{108} \approx 2.92592593$$

Step5: Perform the second iteration

$$x_2 = 2.92592593 - \frac{(2.92592593)^4 - 73}{4(2.92592593)^3} \approx 2.92302111$$

Step6: Perform the third iteration

$$x_3 = 2.92302111 - \frac{(2.92302111)^4 - 73}{4(2.92302111)^3} \approx 2.92301773$$

Step7: Verify convergence to eight decimal places

The next iteration $x_4$ remains $2.92301773$ when rounded to eight decimal places.

Answer:

2.92301773