Question Approximate the local minimum value of the function given below. Use Newton's method with the specified initial approximation $x_{0}$ to find $x_{2}$. Round your answer to the nearest thousandth. \[ f(x)=\frac{1}{3} x^{3}-x^{2}+3 x+3, \quad x_{0}=4 \]

Step 1 ：First, let's find the derivative of the function: \(f'(x) = x² - 2x + 3\)

Step 2 ：We are given that \(x_0 = 4\). Let's use Newton's method to find \(x_1\):

Step 3 ：\(x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 4 - \frac{(1/3)*4³ - 4² + 3*4 + 3}{4² - 2*4 + 3} = 4 - \frac{64/3 - 16 + 12 + 3}{16 - 8 + 3} = 4 - \frac{64/3 - 1 + 15}{11} = 4 - \frac{212/3}{11} = 4 - 64/3 = 4 - 21.333 = -17.333\)

Step 4 ：Now, let's use Newton's method again to find \(x_2\):

Step 5 ：\(x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = -17.333 - \frac{(1/3)*(-17.333)³ - (-17.333)² + 3*(-17.333) + 3}{(-17.333)² - 2*(-17.333) + 3} = -17.333 - \frac{(-10303.704/3) - 300.108 + 51.999 + 3}{300.108 - 34.666 + 3} = -17.333 - \frac{(-3434.568) - 248.109}{268.442} = -17.333 - \frac{-3186.459}{268.442} = -17.333 + 11.869 = -5.464\)

Step 6 ：So, the local minimum value of the function is approximately at \(x = -5.464\) when rounded to the nearest thousandth.

Step 7 ：\(\boxed{-5.464}\)