The function `f(x) = x^(3) - 6x^(2) +9x + 3` is decreasing for

Options

A. `1 lt x lt 3`
B. `x gt 1`
C. `x lt 1`
D. `x lt 1 or x gt 3`

Explanation

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

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

\[ f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x + 3) \]

Calculating the derivative term by term:

\[ f'(x) = 3x^2 - 12x + 9 \]

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

\[ 3x^2 - 12x + 9 = 0 \]

### Step 3: Simplify the equation
We can simplify this equation by dividing all terms by 3:

\[ x^2 - 4x + 3 = 0 \]

### Step 4: Factor the quadratic equation
Now, we will factor the quadratic equation:

\[ (x - 3)(x - 1) = 0 \]

### Step 5: Find the critical points
Setting each factor to zero gives us the critical points:

\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

### Step 6: Determine the sign of the derivative
To find where the function is decreasing, we need to test the intervals determined by the critical points \( x = 1 \) and \( x = 3 \). The intervals to test are:

1. \( (-\infty, 1) \)
2. \( (1, 3) \)
3. \( (3, \infty) \)

We can choose test points from each interval:
- For \( x < 1 \) (let's take \( x = 0 \)):
\[ f'(0) = 3(0)^2 - 12(0) + 9 = 9 \quad (\text{positive}) \]
- For \( 1 < x < 3 \) (let's take \( x = 2 \)):
\[ f'(2) = 3(2)^2 - 12(2) + 9 = 3(4) - 24 + 9 = 12 - 24 + 9 = -3 \quad (\text{negative}) \]
- For \( x > 3 \) (let's take \( x = 4 \)):
\[ f'(4) = 3(4)^2 - 12(4) + 9 = 3(16) - 48 + 9 = 48 - 48 + 9 = 9 \quad (\text{positive}) \]

### Step 7: Identify the decreasing interval
From our tests:
- \( f'(x) > 0 \) in \( (-\infty, 1) \)
- \( f'(x) < 0 \) in \( (1, 3) \)
- \( f'(x) > 0 \) in \( (3, \infty) \)

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

### Final Answer
The function \( f(x) = x^3 - 6x^2 + 9x + 3 \) is decreasing for \( x \in (1, 3) \).

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