In subpoint (3), we are given the following problem:

$f''(x) = 2x - 6, \quad f(0) = 1, \quad f'(1) = 0.$

We need to determine the function $f(x)$ that satisfies these conditions.

Step-by-Step Solution

1. Integrate $f''(x) = 2x - 6$ to find $f'(x)$: $f'(x) = \int (2x - 6) \, dx = x^2 - 6x + C_1,$ where $C_1$ is a constant.

2. Use $f'(1) = 0$ to find $C_1$: Substitute $x = 1$ and $f'(1) = 0$: $0 = (1)^2 - 6 \cdot 1 + C_1,$ $0 = 1 - 6 + C_1,$ $C_1 = 5.$ Thus, $f'(x) = x^2 - 6x + 5$.

3. Integrate $f'(x) = x^2 - 6x + 5$ to find $f(x)$: $f(x) = \int (x^2 - 6x + 5) \, dx = \frac{x^3}{3} - 3x^2 + 5x + C_2,$ where $C_2$ is a constant.

4. Use $f(0) = 1$ to find $C_2$: Substitute $x = 0$ and $f(0) = 1$: $1 = \frac{0^3}{3} - 3 \cdot 0^2 + 5 \cdot 0 + C_2,$ $C_2 = 1.$ Thus, $f(x) = \frac{x^3}{3} - 3x^2 + 5x + 1$.

Final Answer

The function $f(x)$ that satisfies all given conditions is: $f(x) = \frac{x^3}{3} - 3x^2 + 5x + 1.$

Would you like a detailed explanation of any step?

Here are some related questions to explore:

1. How do you verify that a solution satisfies initial conditions?
2. What techniques can be used for integrating polynomial functions?
3. How do higher-order derivatives influence the behavior of a function?
4. What role do integration constants play in solving differential equations?
5. How can you generalize this process for different forms of $f''(x)$?

Tip: When solving for a function with given derivative properties, work backwards from the highest derivative given, applying constants as you integrate each step.