Determine the intervals in which the following functions are strictly increasing or strictly decreasing :
`f(x)=2x^(3)-9x^(2)+12x+15`

Answer

To determine the intervals in which the function \( f(x) = 2x^3 - 9x^2 + 12x + 15 \) is strictly increasing or strictly decreasing, we will follow these steps:

### Step 1: Find the derivative of the function
The first step is to differentiate the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(2x^3 - 9x^2 + 12x + 15) \]

Using the power rule of differentiation, we get:

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

### Step 2: Factor the derivative
Next, we will factor the derivative to find the critical points.

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

Now, we can factor the quadratic:

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

### Step 3: Find the critical points
The critical points occur when \( f'(x) = 0 \):

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

This gives us the critical points:

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

### Step 4: Determine the sign of the derivative
Next, we will test the intervals determined by the critical points \( x = 1 \) and \( x = 2 \):

1. **Interval \( (-\infty, 1) \)**:
- Choose a test point, e.g., \( x = 0 \):
\[ f'(0) = 6(0 - 1)(0 - 2) = 6(-1)(-2) = 12 > 0 \]
So, \( f(x) \) is increasing in \( (-\infty, 1) \).

2. **Interval \( (1, 2) \)**:
- Choose a test point, e.g., \( x = 1.5 \):
\[ f'(1.5) = 6(1.5 - 1)(1.5 - 2) = 6(0.5)(-0.5) = -1.5 < 0 \]
So, \( f(x) \) is decreasing in \( (1, 2) \).

3. **Interval \( (2, \infty) \)**:
- Choose a test point, e.g., \( x = 3 \):
\[ f'(3) = 6(3 - 1)(3 - 2) = 6(2)(1) = 12 > 0 \]
So, \( f(x) \) is increasing in \( (2, \infty) \).

### Step 5: Conclusion
From our analysis, we can conclude:

- The function \( f(x) \) is **strictly increasing** on the intervals \( (-\infty, 1) \) and \( (2, \infty) \).
- The function \( f(x) \) is **strictly decreasing** on the interval \( (1, 2) \).

### Summary of Intervals:
- **Increasing**: \( (-\infty, 1) \) and \( (2, \infty) \)
- **Decreasing**: \( (1, 2) \)

---