The second approximation \(x_2\) is -0.112
Newton iteration formula is:
\(x_{n+1}=x_n - \frac{f(x_n)}{f'(x_{n+1})}\)
We have the initial root is \(x_1 = 1\) function:
\(\cos(x^2 + 3) = x^3 \\ \cos(x^2 + 3) - x^3 =0\\f(x)=\cos(x^2 + 3) - x^3\\f'(x)=-2x \ sinx(x^2 + 3) - 3x^2\\f(1)=\cos(1^2 + 3) - 1^3\\f(1)=\cos(4) - 1\\f(1)=-1.6534\\f'(1)=-2.1 \ sinx(1^2 + 3) - 3.1^2\\f'(1)=-1.486\)
Now, substitute the values in the newton iteration formula, we get
\(x_{2}=x_1- \frac{f(x_1)}{f'(x_{1})}\\x_{2}=1-\frac{-1.6534}{-1.486} \\x_{2}=-0.112\)
Therefore, the second root is -0.112
Learn more:https://brainly.com/question/17031314
