If `y=x(x-3)^(2)` decreases for the values of `x` given by

Options

A. `1 lt x lt 3`
B. `x lt 0`
C. `x gt 0`
D. `0 lt x lt 3/2`

Explanation

To determine the values of \( x \) for which the function \( y = x(x - 3)^2 \) is decreasing, we will follow these steps:

### Step 1: Differentiate the function
We need to find the derivative of the function \( y \) with respect to \( x \).

Using the product rule:
\[ y = u \cdot v \quad \text{where } u = x \text{ and } v = (x - 3)^2 \]
The product rule states that \( \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \).

First, we differentiate \( v \):
\[ \frac{dv}{dx} = 2(x - 3) \]

Now, differentiate \( u \):
\[ \frac{du}{dx} = 1 \]

Now applying the product rule:
\[ \frac{dy}{dx} = x \cdot 2(x - 3) + (x - 3)^2 \cdot 1 \]
This simplifies to:
\[ \frac{dy}{dx} = 2x(x - 3) + (x - 3)^2 \]

### Step 2: Simplify the derivative
Now, we simplify the expression:
\[ \frac{dy}{dx} = 2x(x - 3) + (x - 3)(x - 3) \]
\[ = 2x(x - 3) + (x^2 - 6x + 9) \]
\[ = 2x^2 - 6x + x^2 - 6x + 9 \]
\[ = 3x^2 - 12x + 9 \]

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

### Step 4: Factor the quadratic equation
We can factor this quadratic:
\[ (x - 3)(x - 1) = 0 \]
Thus, the critical points are:
\[ x = 1 \quad \text{and} \quad x = 3 \]

### Step 5: Determine the intervals
We will check the sign of \( \frac{dy}{dx} \) in the intervals determined by the critical points \( x = 1 \) and \( x = 3 \):
- Interval 1: \( (-\infty, 1) \)
- Interval 2: \( (1, 3) \)
- Interval 3: \( (3, \infty) \)

### Step 6: Test the intervals
1. **For \( x < 1 \)** (e.g., \( x = 0 \)):
\[ \frac{dy}{dx} = 3(0)^2 - 12(0) + 9 = 9 > 0 \quad \text{(increasing)} \]

2. **For \( 1 < x < 3 \)** (e.g., \( x = 2 \)):
\[ \frac{dy}{dx} = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3 < 0 \quad \text{(decreasing)} \]

3. **For \( x > 3 \)** (e.g., \( x = 4 \)):
\[ \frac{dy}{dx} = 3(4)^2 - 12(4) + 9 = 48 - 48 + 9 = 9 > 0 \quad \text{(increasing)} \]

### Conclusion
The function \( y = x(x - 3)^2 \) is decreasing in the interval \( (1, 3) \).