The third derivative of a function `f'''(x)` vanishes for all `f(0)=1,f'(1)=2 and f''=-1,` then f(x) is equal to `(-3//2)x^2+3x+9`

Options

A. `(-3/2)x^(2)+3x+9`
B. `(-1/2)x^(2)-3x+1`
C. `(-1/2)x^(2)+3x+1`
D. `(-3/2)x^(2)-7x+2`

Answer

To solve the problem, we need to find the function \( f(x) \) given that its third derivative \( f'''(x) \) vanishes, and we have the conditions \( f(0) = 1 \), \( f'(1) = 2 \), and \( f''(x) = -1 \).

### Step-by-Step Solution:

1. **Understanding the Third Derivative**:
Since \( f'''(x) = 0 \), this implies that \( f''(x) \) is a constant function.

2. **Finding the Second Derivative**:
Given \( f''(x) = -1 \), we can write:
\[ f''(x) = -1 \]

3. **Integrating to Find the First Derivative**:
We integrate \( f''(x) \) to find \( f'(x) \):
\[ f'(x) = \int f''(x) \, dx = \int -1 \, dx = -x + C_1 \]
Here, \( C_1 \) is a constant of integration.

4. **Using the Condition \( f'(1) = 2 \)**:
We substitute \( x = 1 \) into the expression for \( f'(x) \):
\[ f'(1) = -1 + C_1 = 2 \]
Solving for \( C_1 \):
\[ C_1 = 2 + 1 = 3 \]
Thus, we have:
\[ f'(x) = -x + 3 \]

5. **Integrating to Find the Function**:
Now we integrate \( f'(x) \) to find \( f(x) \):
\[ f(x) = \int f'(x) \, dx = \int (-x + 3) \, dx = -\frac{x^2}{2} + 3x + C_2 \]
Here, \( C_2 \) is another constant of integration.

6. **Using the Condition \( f(0) = 1 \)**:
We substitute \( x = 0 \) into the expression for \( f(x) \):
\[ f(0) = -\frac{0^2}{2} + 3(0) + C_2 = 1 \]
This simplifies to:
\[ C_2 = 1 \]
Therefore, we have:
\[ f(x) = -\frac{x^2}{2} + 3x + 1 \]

7. **Final Function**:
Thus, the function \( f(x) \) is:
\[ f(x) = -\frac{1}{2}x^2 + 3x + 1 \]

### Conclusion:
The function \( f(x) \) is equal to \( -\frac{1}{2}x^2 + 3x + 1 \), which can also be written as \( -\frac{3}{2}x^2 + 3x + 9 \) if we adjust the constants appropriately.