Let f(x) =`x^(3)-3(7-a)X^(2)-3(9-a^(2))x+2`
The values of parameter a if f(x) has a negative point of local minimum are

Options

A. `pi`
B. (-3,3)
C. `(-oo,(58)/(14))`
D. none of these

Explanation

To solve the problem, we need to determine the values of the parameter \( a \) such that the function \( f(x) = x^3 - 3(7-a)x^2 - 3(9-a^2)x + 2 \) has a negative point of local minimum. We will follow these steps:

### Step 1: Find the first derivative \( f'(x) \)
The first derivative \( f'(x) \) is calculated to find the critical points where the local minima or maxima occur.

\[ f'(x) = 3x^2 - 3(7-a) \cdot 2x - 3(9-a^2) \]
\[ = 3x^2 - 6(7-a)x - 3(9-a^2) \]

### Step 2: Set the first derivative to zero
To find the critical points, we set \( f'(x) = 0 \):

\[ 3x^2 - 6(7-a)x - 3(9-a^2) = 0 \]

Dividing the entire equation by 3 simplifies it:

\[ x^2 - 2(7-a)x - (9-a^2) = 0 \]

### Step 3: Use the discriminant condition
For the quadratic equation to have real roots, the discriminant must be non-negative:

\[ D = b^2 - 4ac \geq 0 \]

Here, \( a = 1 \), \( b = -2(7-a) \), and \( c = -(9-a^2) \):

\[ D = [-2(7-a)]^2 - 4 \cdot 1 \cdot (-(9-a^2)) \geq 0 \]
\[ = 4(7-a)^2 + 4(9-a^2) \geq 0 \]
\[ = 4[(7-a)^2 + (9-a^2)] \geq 0 \]

Since \( 4 \) is always positive, we can simplify to:

\[ (7-a)^2 + (9-a^2) \geq 0 \]

### Step 4: Analyze the second derivative \( f''(x) \)
Next, we find the second derivative \( f''(x) \) to determine the nature of the critical points:

\[ f''(x) = 6x - 6(7-a) \]

### Step 5: Set the second derivative condition
For a local minimum, we need \( f''(x) > 0 \):

\[ 6x - 6(7-a) > 0 \]
\[ x > 7-a \]

### Step 6: Establish conditions for negative local minimum
Since we are looking for a negative point of local minimum, we require:

\[ 7-a < 0 \implies a > 7 \]

### Step 7: Combine conditions
From the discriminant condition, we have:

\[ (7-a)^2 + (9-a^2) \geq 0 \]

This condition is always satisfied for real values of \( a \). The critical condition we derived is \( a > 7 \).

### Conclusion
The values of the parameter \( a \) such that \( f(x) \) has a negative point of local minimum are:

\[ \boxed{a > 7} \]