The sum of all the local minimum values of the twice differentiable function `f : R to R` defined by `f (x) = x ^(3) - 3x ^(2) - (3f''(2))/2 x + f ''(1)` is :

Options

A. `0`
B. `5`
C. `-22`
D. `-27`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write down the function
The function given is:
\[ f(x) = x^3 - 3x^2 - \frac{3f''(2)}{2}x + f''(1) \]

### Step 2: Differentiate the function
We differentiate \( f(x) \) to find \( f'(x) \):
\[ f'(x) = 3x^2 - 6x - \frac{3f''(2)}{2} \]

### Step 3: Set the first derivative to zero
To find the local minima, we set \( f'(x) = 0 \):
\[ 3x^2 - 6x - \frac{3f''(2)}{2} = 0 \]

### Step 4: Solve for \( x \)
Dividing the entire equation by 3 simplifies it:
\[ x^2 - 2x - \frac{f''(2)}{2} = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot \left(-\frac{f''(2)}{2}\right)}}{2 \cdot 1} \]
\[ x = \frac{2 \pm \sqrt{4 + 2f''(2)}}{2} \]
\[ x = 1 \pm \sqrt{1 + \frac{f''(2)}{2}} \]

### Step 5: Find the second derivative
Now we differentiate \( f'(x) \) to find \( f''(x) \):
\[ f''(x) = 6x - 6 \]

### Step 6: Evaluate \( f''(1) \) and \( f''(2) \)
Calculating \( f''(1) \):
\[ f''(1) = 6(1) - 6 = 0 \]
Calculating \( f''(2) \):
\[ f''(2) = 6(2) - 6 = 6 \]

### Step 7: Substitute back into the equation
Now substituting \( f''(2) = 6 \) into our expression for \( x \):
\[ x = 1 \pm \sqrt{1 + \frac{6}{2}} = 1 \pm \sqrt{4} = 1 \pm 2 \]
Thus, we have two critical points:
\[ x = 3 \quad \text{and} \quad x = -1 \]

### Step 8: Determine local minimum values
Now we evaluate \( f(-1) \) and \( f(3) \):
1. For \( x = 3 \):
\[ f(3) = 3^3 - 3(3^2) - \frac{3 \cdot 6}{2} \cdot 3 + 0 \]
\[ = 27 - 27 - 27 = -27 \]

2. For \( x = -1 \):
\[ f(-1) = (-1)^3 - 3(-1)^2 - \frac{3 \cdot 6}{2} \cdot (-1) + 0 \]
\[ = -1 - 3 + 9 = 5 \]

### Step 9: Sum of local minimum values
Now, we sum the local minimum values:
\[ \text{Sum} = f(3) + f(-1) = -27 + 5 = -22 \]

### Final Answer
The sum of all the local minimum values is:
\[ \boxed{-22} \]