use newtons method to approximate a root of the equation (4x^{4}-2x^{2}-3 = 0). let (x_{0}=1) and compute…

Explanation:

Step1: Define function and its derivative

Let the given equation be $f(x) = 4x^4 - 2x^2 - 3$. The derivative of $f(x)$ is $f'(x) = 16x^3 - 4x$. Newton's Method formula is $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$.

Step2: Calculate $x_1$ using $x_0 = 1$

Substitute $x_0 = 1$ into $f(x)$ and $f'(x)$: $f(1) = 4(1)^4 - 2(1)^2 - 3 = 4 - 2 - 3 = -1$ $f'(1) = 16(1)^3 - 4(1) = 16 - 4 = 12$ Now, calculate $x_1$: $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{-1}{12} = 1 + \frac{1}{12} = \frac{13}{12}$ $x_1 \approx 1.0833333333...$

Step3: Calculate $x_2$ using $x_1 = \frac{13}{12}$

Substitute $x_1 = \frac{13}{12}$ into $f(x)$ and $f'(x)$: $f\left(\frac{13}{12}\right) = 4\left(\frac{13}{12}\right)^4 - 2\left(\frac{13}{12}\right)^2 - 3 = 4\left(\frac{28561}{20736}\right) - 2\left(\frac{169}{144}\right) - 3 = \frac{28561}{5184} - \frac{169}{72} - 3 = \frac{28561 - 12168 - 15552}{5184} = \frac{841}{5184}$ $f'\left(\frac{13}{12}\right) = 16\left(\frac{13}{12}\right)^3 - 4\left(\frac{13}{12}\right) = 16\left(\frac{2197}{1728}\right) - \frac{52}{12} = \frac{2197}{108} - \frac{13}{3} = \frac{2197 - 468}{108} = \frac{1729}{108}$ Now, calculate $x_2$: $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = \frac{13}{12} - \frac{\frac{841}{5184}}{\frac{1729}{108}} = \frac{13}{12} - \frac{841}{5184} \cdot \frac{108}{1729} = \frac{13}{12} - \frac{841}{48 \cdot 1729} = \frac{13}{12} - \frac{841}{82992}$ $x_2 = \frac{13 \cdot 6916 - 841}{82992} = \frac{89908 - 841}{82992} = \frac{89067}{82992}$ $x_2 \approx 1.073190866...$

Step4: Round the results to six decimal places

$x_1 \approx 1.083333$ $x_2 \approx 1.073191$

Answer:

$x_1 \approx 1.083333$ $x_2 \approx 1.073191$