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

Options

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

Answer

To solve the problem step by step, we will first clarify the function and then find the intervals where the function is increasing and decreasing.

### Step 1: Define the function
The given function is:
\[ f(x) = 2x^3 + 3x^2 - 12x + 5 \]

### Step 2: Find the derivative of the function
To determine where the function is increasing or decreasing, we need to find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(2x^3) + \frac{d}{dx}(3x^2) - \frac{d}{dx}(12x) + \frac{d}{dx}(5) \]
Calculating each term:
- The derivative of \( 2x^3 \) is \( 6x^2 \)
- The derivative of \( 3x^2 \) is \( 6x \)
- The derivative of \( -12x \) is \( -12 \)
- The derivative of \( 5 \) is \( 0 \)

Thus, we have:
\[ f'(x) = 6x^2 + 6x - 12 \]

### Step 3: Set the derivative to zero to find critical points
To find the intervals of increase and decrease, we set the derivative equal to zero:
\[ 6x^2 + 6x - 12 = 0 \]
Dividing the entire equation by 6:
\[ x^2 + x - 2 = 0 \]

### Step 4: Factor the quadratic equation
We can factor the quadratic:
\[ (x + 2)(x - 1) = 0 \]
Setting each factor to zero gives us the critical points:
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

### Step 5: Determine the sign of \( f'(x) \) in the intervals
The critical points divide the number line into three intervals: \( (-\infty, -2) \), \( (-2, 1) \), and \( (1, \infty) \). We will test the sign of \( f'(x) \) in each interval.

1. **Interval \( (-\infty, -2) \)**: Choose \( x = -3 \)
\[ f'(-3) = 6(-3)^2 + 6(-3) - 12 = 54 - 18 - 12 = 24 \quad (\text{positive}) \]

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

3. **Interval \( (1, \infty) \)**: Choose \( x = 2 \)
\[ f'(2) = 6(2)^2 + 6(2) - 12 = 24 + 12 - 12 = 24 \quad (\text{positive}) \]

### Step 6: Conclusion about intervals
- The function is **increasing** in the intervals where \( f'(x) > 0 \):
- \( (-\infty, -2) \) and \( (1, \infty) \)

- The function is **decreasing** in the interval where \( f'(x) < 0 \):
- \( (-2, 1) \)

### Final Result
- **Increasing Intervals (I1)**: \( (-\infty, -2) \cup (1, \infty) \)
- **Decreasing Interval (I2)**: \( (-2, 1) \)