If the complete set of value(s) of `a` for which the function `f (x) =(ax^(3))/(3)+(a+2) x^(2) +(a-1) x+2` possess a negative point of inflection is `(-oo, alpha)uu(beta,oo)" then " |alpha|+|beta|` is ___________ .

Answer

To solve the problem, we need to find the values of \( a \) for which the function

\[ f(x) = \frac{a x^3}{3} + (a + 2)x^2 + (a - 1)x + 2 \]

has a negative point of inflection. We will follow these steps:

### Step 1: Find the first and second derivatives of \( f(x) \)

1. **First Derivative**:
\[ f'(x) = \frac{d}{dx}\left(\frac{a x^3}{3} + (a + 2)x^2 + (a - 1)x + 2\right) \]
Using the power rule, we get:
\[ f'(x) = ax^2 + 2(a + 2)x + (a - 1) \]

2. **Second Derivative**:
\[ f''(x) = \frac{d}{dx}(ax^2 + 2(a + 2)x + (a - 1)) \]
Again using the power rule, we find:
\[ f''(x) = 2ax + 2(a + 2) \]

### Step 2: Set the second derivative to zero for points of inflection

To find the points of inflection, we set \( f''(x) = 0 \):
\[ 2ax + 2(a + 2) = 0 \]
This simplifies to:
\[ 2ax = -2(a + 2) \quad \Rightarrow \quad ax = -(a + 2) \quad \Rightarrow \quad x = -\frac{a + 2}{a} \quad \text{(for } a \neq 0\text{)} \]

### Step 3: Determine conditions for a negative point of inflection

For the point of inflection to be negative, we require:
\[ -\frac{a + 2}{a} < 0 \]
This inequality will hold true when the numerator and denominator have opposite signs. We analyze the two cases:

1. **Case 1**: \( a + 2 > 0 \) and \( a < 0 \)
- This implies \( a > -2 \) and \( a < 0 \).

2. **Case 2**: \( a + 2 < 0 \) and \( a > 0 \)
- This case is not possible because \( a + 2 < 0 \) implies \( a < -2 \).

### Step 4: Combine the conditions

From Case 1, we have:
\[ -2 < a < 0 \]
Thus, the complete set of values of \( a \) for which the function possesses a negative point of inflection is:
\[ (-\infty, -2) \cup (0, \infty) \]

### Step 5: Identify \( \alpha \) and \( \beta \)

Comparing with the given set \( (-\infty, \alpha) \cup (\beta, \infty) \):
- We identify \( \alpha = -2 \) and \( \beta = 0 \).

### Step 6: Calculate \( | \alpha | + | \beta | \)

Now, we calculate:
\[ | \alpha | + | \beta | = |-2| + |0| = 2 + 0 = 2 \]

### Final Answer

Thus, the final answer is:
\[ \boxed{2} \]