Question 4 (10 points) Part A) Solve the following initial value problem by finding $f(x)$ which satisfies both of the equations: - $f^{\prime}(x)=3 x^{2}+6 x+5$ - $f(0)=3$ Part B) Calculate $f(2)$ using the function $f(x)$ you found in Part A. Enter the decimal form of $f(2)$ in the box below, using 3 decimal places.

Step 1 ：The given differential equation is \(f^{\prime}(x)=3 x^{2}+6 x+5\).

Step 2 ：To find \(f(x)\), we need to integrate the right-hand side of the equation with respect to \(x\).

Step 3 ：\(\int f^{\prime}(x) dx = \int (3x^2 + 6x + 5) dx\)

Step 4 ：This gives us \(f(x) = \int (3x^2) dx + \int (6x) dx + \int 5 dx\)

Step 5 ：Solving the integrals, we get \(f(x) = x^3 + 3x^2 + 5x + C\)

Step 6 ：We are given that \(f(0) = 3\), so we can substitute \(x = 0\) into the equation to find the constant \(C\).

Step 7 ：Substituting \(x = 0\) gives us \(3 = 0 + 0 + 0 + C\)

Step 8 ：So, \(C = 3\).

Step 9 ：Therefore, the function \(f(x)\) that satisfies the given differential equation and initial condition is \(f(x) = x^3 + 3x^2 + 5x + 3\).

Step 10 ：To find \(f(2)\), we substitute \(x = 2\) into the function \(f(x)\).

Step 11 ：\(f(2) = (2)^3 + 3(2)^2 + 5(2) + 3\)

Step 12 ：Solving the equation gives us \(f(2) = 8 + 12 + 10 + 3\)

Step 13 ：So, \(f(2) = 33\).

Step 14 ：Therefore, \(f(2) = 33.000\) to 3 decimal places.

Step 15 ：So, the final answer is \(\boxed{33.000}\).