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: Find the first derivative of the function
We start with the function:
\[ y = x(x - 3)^2 \]
We will use the product rule to find the derivative \( \frac{dy}{dx} \).

Using the product rule:
\[ \frac{dy}{dx} = u'v + uv' \]
where \( u = x \) and \( v = (x - 3)^2 \).

Calculating the derivatives:
- \( u' = 1 \)
- \( v = (x - 3)^2 \) and \( v' = 2(x - 3) \)

Now applying the product rule:
\[ \frac{dy}{dx} = 1 \cdot (x - 3)^2 + x \cdot 2(x - 3) \]
\[ = (x - 3)^2 + 2x(x - 3) \]
\[ = (x - 3)^2 + 2x^2 - 6x \]
\[ = (x - 3)^2 + 2x^2 - 6x \]
Now, simplifying further:
\[ = (x - 3)(x - 3) + 2x^2 - 6x \]
\[ = (x - 3)(x - 3) + 2x^2 - 6x \]
\[ = (x - 3)(x - 3) + 2x^2 - 6x \]
\[ = (x - 3)(x - 3 + 2x) \]
\[ = (x - 3)(3x - 3) \]
\[ = 3(x - 1)(x - 3) \]

### Step 2: Set the first derivative to zero to find critical points
Now we set the first derivative equal to zero to find the critical points:
\[ 3(x - 1)(x - 3) = 0 \]
This gives us:
\[ x - 1 = 0 \quad \text{or} \quad x - 3 = 0 \]
Thus, the critical points are:
\[ x = 1 \quad \text{and} \quad x = 3 \]

### Step 3: Determine the intervals of increase and decrease
We will analyze the sign of the first derivative \( \frac{dy}{dx} = 3(x - 1)(x - 3) \) in the intervals determined by the critical points \( x = 1 \) and \( x = 3 \).

- For \( x < 1 \): Choose \( x = 0 \)
\[ \frac{dy}{dx} = 3(0 - 1)(0 - 3) = 3(-1)(-3) = 9 \quad (\text{positive}) \]

- For \( 1 < x < 3 \): Choose \( x = 2 \)
\[ \frac{dy}{dx} = 3(2 - 1)(2 - 3) = 3(1)(-1) = -3 \quad (\text{negative}) \]

- For \( x > 3 \): Choose \( x = 4 \)
\[ \frac{dy}{dx} = 3(4 - 1)(4 - 3) = 3(3)(1) = 9 \quad (\text{positive}) \]

### Step 4: Conclusion
The function \( y = x(x - 3)^2 \) is decreasing in the interval:
\[ (1, 3) \]