Let f(x) `=x^(3) - 6x^2 +9x +18`, then f(x) is strictly decreasing in …………

Options

A. `(-infty ,1)`
B. ` (3,infty)`
C. ` (-infty,1)`
D. ` (1,3)`

Answer

To determine the intervals where the function \( f(x) = x^3 - 6x^2 + 9x + 18 \) is strictly decreasing, we need to follow these steps:

### Step 1: Find the derivative of \( f(x) \)
We start by differentiating the function to find \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x + 18) \]
Using the power rule, we get:
\[ f'(x) = 3x^2 - 12x + 9 \]

### Step 2: Set the derivative equal to zero
To find the critical points, we set the derivative equal to zero:
\[ 3x^2 - 12x + 9 = 0 \]
Dividing the entire equation by 3 simplifies it:
\[ x^2 - 4x + 3 = 0 \]

### Step 3: Factor the quadratic equation
Next, we factor the quadratic:
\[ (x - 1)(x - 3) = 0 \]
This gives us the critical points:
\[ x = 1 \quad \text{and} \quad x = 3 \]

### Step 4: Test intervals around the critical points
We will test the intervals determined by the critical points \( (-\infty, 1) \), \( (1, 3) \), and \( (3, \infty) \) to see where \( f'(x) < 0 \).

1. **Interval \( (-\infty, 1) \)**:
Choose \( x = 0 \):
\[ f'(0) = 3(0)^2 - 12(0) + 9 = 9 \quad (\text{positive}) \]

2. **Interval \( (1, 3) \)**:
Choose \( x = 2 \):
\[ f'(2) = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3 \quad (\text{negative}) \]

3. **Interval \( (3, \infty) \)**:
Choose \( x = 4 \):
\[ f'(4) = 3(4)^2 - 12(4) + 9 = 48 - 48 + 9 = 9 \quad (\text{positive}) \]

### Step 5: Conclusion
From our tests, we find that:
- \( f'(x) > 0 \) in \( (-\infty, 1) \) and \( (3, \infty) \)
- \( f'(x) < 0 \) in \( (1, 3) \)

Thus, the function \( f(x) \) is strictly decreasing in the interval \( (1, 3) \).

### Final Answer
The function \( f(x) \) is strictly decreasing in \( (1, 3) \).

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