`f(x)=-(1)/(3)x^(3)+5x^(2)+12`

Options

A. `I_(1)=(0,10),I_(2)=(-infty, 0) uu(10, infty)`
B. `I_(1)=(-infty, 0), I_(2)=(0, 10) uu(10, infty)`
C. `I_(1)=(-infty, 10), I_(2)=(10,infty) uu(0, 10)`
D. none of these

Answer

To determine where the function \( f(x) = -\frac{1}{3}x^3 + 5x^2 + 12 \) is increasing or decreasing, we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function \( f(x) \).

\[ f'(x) = \frac{d}{dx} \left(-\frac{1}{3}x^3 + 5x^2 + 12\right) \]

Using the power rule:

\[ f'(x) = -\frac{1}{3} \cdot 3x^2 + 5 \cdot 2x + 0 \]

Simplifying this gives:

\[ f'(x) = -x^2 + 10x \]

### Step 2: Set the derivative equal to zero to find critical points
Next, we set the derivative equal to zero to find the critical points:

\[ -x^2 + 10x = 0 \]

Factoring out \( x \):

\[ x(-x + 10) = 0 \]

This gives us the critical points:

\[ x = 0 \quad \text{and} \quad x = 10 \]

### Step 3: Test intervals around the critical points
We will test the sign of \( f'(x) \) in the intervals determined by the critical points \( (-\infty, 0) \), \( (0, 10) \), and \( (10, \infty) \).

1. **Interval \( (-\infty, 0) \)**: Choose \( x = -1 \)
\[ f'(-1) = -(-1)^2 + 10(-1) = -1 - 10 = -11 \quad (\text{negative}) \]

2. **Interval \( (0, 10) \)**: Choose \( x = 5 \)
\[ f'(5) = -(5)^2 + 10(5) = -25 + 50 = 25 \quad (\text{positive}) \]

3. **Interval \( (10, \infty) \)**: Choose \( x = 11 \)
\[ f'(11) = -(11)^2 + 10(11) = -121 + 110 = -11 \quad (\text{negative}) \]

### Step 4: Determine where the function is increasing or decreasing
From our tests:
- \( f'(x) < 0 \) in the intervals \( (-\infty, 0) \) and \( (10, \infty) \) (decreasing).
- \( f'(x) > 0 \) in the interval \( (0, 10) \) (increasing).

### Conclusion
Thus, we can summarize the intervals:
- The function is **increasing** on \( (0, 10) \).
- The function is **decreasing** on \( (-\infty, 0) \) and \( (10, \infty) \).

### Final Answer
- Increasing interval \( i_1: (0, 10) \)
- Decreasing interval \( i_2: (-\infty, 0) \cup (10, \infty) \)

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