The interval in which the function f defined as `f(x) = 1 - 12x + 9x^(2) - 2x^(3)` is strictly decreasing, is :

Options

A. `(-oo, 1) cup (2, oo)`
B. (1, 2)
C. `(-oo, 1)`
D. `(-oo, 1) cup (1, 2)`

Answer

To determine the interval in which the function \( f(x) = 1 - 12x + 9x^2 - 2x^3 \) is strictly decreasing, we need to follow these steps:

### Step 1: Find the derivative of the function
To find where the function is decreasing, we first compute the derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(1 - 12x + 9x^2 - 2x^3) \]

Calculating the derivative term by term:

- The derivative of \( 1 \) is \( 0 \).
- The derivative of \( -12x \) is \( -12 \).
- The derivative of \( 9x^2 \) is \( 18x \).
- The derivative of \( -2x^3 \) is \( -6x^2 \).

Putting it all together, we have:

\[ f'(x) = -12 + 18x - 6x^2 \]

### Step 2: Rearrange the derivative
We can rearrange the derivative to make it easier to analyze:

\[ f'(x) = -6x^2 + 18x - 12 \]

### Step 3: Factor the derivative
Next, we factor out the common term:

\[ f'(x) = -6(x^2 - 3x + 2) \]

Now, we can factor the quadratic expression:

\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]

Thus, we have:

\[ f'(x) = -6(x - 1)(x - 2) \]

### Step 4: Determine critical points
The critical points occur when \( f'(x) = 0 \):

\[ -6(x - 1)(x - 2) = 0 \]

This gives us:

\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]

### Step 5: Analyze the sign of the derivative
We will now test the intervals determined by the critical points \( x = 1 \) and \( x = 2 \):

1. **Interval \( (-\infty, 1) \)**: Choose \( x = 0 \):
\[ f'(0) = -6(0 - 1)(0 - 2) = -6(-1)(-2) = -12 \quad (\text{negative}) \]

2. **Interval \( (1, 2) \)**: Choose \( x = 1.5 \):
\[ f'(1.5) = -6(1.5 - 1)(1.5 - 2) = -6(0.5)(-0.5) = 1.5 \quad (\text{positive}) \]

3. **Interval \( (2, \infty) \)**: Choose \( x = 3 \):
\[ f'(3) = -6(3 - 1)(3 - 2) = -6(2)(1) = -12 \quad (\text{negative}) \]

### Step 6: Conclusion
The function \( f(x) \) is strictly decreasing in the intervals where \( f'(x) < 0 \):

- From our analysis, \( f'(x) < 0 \) in the intervals \( (-\infty, 1) \) and \( (2, \infty) \).

Thus, the function \( f(x) \) is strictly decreasing in the interval:

\[ (-\infty, 1) \cup (2, \infty) \]

### Final Answer
The interval in which the function \( f \) is strictly decreasing is:

\[ \text{Option 1: } (-\infty, 1) \cup (2, \infty) \]

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