If `S_(1) and S_(2)` are respectively the sets of local minimum and local maixmum point of the function, `f(x)=9x^(4)+12x^(3)-36x^(2)+2019, x in R`, then

Options

A. `S_1={-2},S_2={0,1}`
B. `S_1={-2,1},S_2={0}`
C. `S_(1)={-1},S_2={0,2}`
D. `S_1={-2,0} ,S_2={1}`

Explanation

To find the sets \( S_1 \) and \( S_2 \) of local minimum and local maximum points of the function \( f(x) = 9x^4 + 12x^3 - 36x^2 + 2019 \), we will follow these steps:

### Step 1: Find the first derivative \( f'(x) \)
To find the critical points, we first need to differentiate the function.

\[ f'(x) = \frac{d}{dx}(9x^4 + 12x^3 - 36x^2 + 2019) \]

Using the power rule for differentiation:

\[ f'(x) = 36x^3 + 36x^2 - 72x \]

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

\[ 36x^3 + 36x^2 - 72x = 0 \]

Factoring out the common term:

\[ 36x(x^2 + x - 2) = 0 \]

### Step 3: Solve for \( x \)
Now, we can solve for \( x \):

1. \( 36x = 0 \) gives \( x = 0 \).
2. For \( x^2 + x - 2 = 0 \), we can factor it:

\[ (x + 2)(x - 1) = 0 \]

This gives us:

\[ x = -2 \quad \text{and} \quad x = 1 \]

So, the critical points are \( x = 0, -2, 1 \).

### Step 4: Find the second derivative \( f''(x) \)
Next, we need to determine whether these critical points are local minima or maxima by finding the second derivative:

\[ f''(x) = \frac{d}{dx}(36x^3 + 36x^2 - 72x) \]

Using the power rule again:

\[ f''(x) = 108x^2 + 72x - 72 \]

### Step 5: Evaluate the second derivative at the critical points
Now we evaluate \( f''(x) \) at each critical point:

1. For \( x = 0 \):

\[ f''(0) = 108(0)^2 + 72(0) - 72 = -72 \quad (\text{less than 0, local maximum}) \]

2. For \( x = -2 \):

\[ f''(-2) = 108(-2)^2 + 72(-2) - 72 = 108(4) - 144 - 72 = 432 - 144 - 72 = 216 \quad (\text{greater than 0, local minimum}) \]

3. For \( x = 1 \):

\[ f''(1) = 108(1)^2 + 72(1) - 72 = 108 + 72 - 72 = 108 \quad (\text{greater than 0, local minimum}) \]

### Step 6: Identify the sets \( S_1 \) and \( S_2 \)
From our evaluations:

- The local maximum point is at \( x = 0 \), so \( S_2 = \{ 0 \} \).
- The local minimum points are at \( x = -2 \) and \( x = 1 \), so \( S_1 = \{ -2, 1 \} \).

### Final Answer
\[ S_1 = \{ -2, 1 \}, \quad S_2 = \{ 0 \} \]