IF `f(x)=2x^3+3x^2-12x+5` , then the interval in which `(l_1)` increases and `(l_2)` decreases is

Options

A. `l_1=(-infty,-2)cup(1,infty),l_2=(-2,1)`
B. `l_1=(-infty,1),l_2=(-1,2)cup(2,infty)`
C. `l_1=(-infty,-1)cup(2,infty),l_2=(-1,2)`
D. `l_1=(-infty,2)cup(-1,infty),l_2=(5,2)`

Answer

To solve the problem, we need to find the intervals in which the function \( f(x) = 2x^3 + 3x^2 - 12x + 5 \) is increasing and decreasing. We will do this by finding the derivative of the function and analyzing its sign.

### Step-by-Step Solution:

1. **Find the derivative of the function**:
\[ f'(x) = \frac{d}{dx}(2x^3 + 3x^2 - 12x + 5) \]
Using the power rule, we differentiate each term:
\[ f'(x) = 6x^2 + 6x - 12 \]

2. **Set the derivative equal to zero to find critical points**:
\[ 6x^2 + 6x - 12 = 0 \]
Dividing the entire equation by 6 simplifies it:
\[ x^2 + x - 2 = 0 \]

3. **Factor the quadratic equation**:
\[ (x + 2)(x - 1) = 0 \]
This gives us the critical points:
\[ x = -2 \quad \text{and} \quad x = 1 \]

4. **Determine the intervals**:
The critical points divide the number line into three intervals:
- \( (-\infty, -2) \)
- \( (-2, 1) \)
- \( (1, \infty) \)

5. **Test the sign of the derivative in each interval**:
- For \( x < -2 \) (e.g., \( x = -3 \)):
\[ f'(-3) = 6(-3)^2 + 6(-3) - 12 = 54 - 18 - 12 = 24 \quad (\text{positive}) \]
- For \( -2 < x < 1 \) (e.g., \( x = 0 \)):
\[ f'(0) = 6(0)^2 + 6(0) - 12 = -12 \quad (\text{negative}) \]
- For \( x > 1 \) (e.g., \( x = 2 \)):
\[ f'(2) = 6(2)^2 + 6(2) - 12 = 24 + 12 - 12 = 24 \quad (\text{positive}) \]

6. **Conclusion about increasing and decreasing intervals**:
- The function is **increasing** on the intervals where \( f'(x) > 0 \):
\[ L_1: (-\infty, -2) \cup (1, \infty) \]
- The function is **decreasing** on the interval where \( f'(x) < 0 \):
\[ L_2: (-2, 1) \]

### Final Answer:
- The interval in which \( L_1 \) (the function increases) is \( (-\infty, -2) \cup (1, \infty) \).
- The interval in which \( L_2 \) (the function decreases) is \( (-2, 1) \).