` y=[ x(x - 3)^2 ]` increases for all values of x lying in the interval

Options

A. `0 lt x lt 1`
B. ` 0 lt x lt oo`
C. ` - oo lt x lt 0`
D. ` 1 lt x lt 3`

Answer

To determine the intervals where the function \( y = x(x - 3)^2 \) is increasing, we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function with respect to \( x \).

\[ y = x(x - 3)^2 \]

Using the product rule, where \( u = x \) and \( v = (x - 3)^2 \):

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

Calculating \( \frac{dv}{dx} \):

\[ \frac{dv}{dx} = 2(x - 3) \]

Calculating \( \frac{du}{dx} \):

\[ \frac{du}{dx} = 1 \]

Now substituting back into 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 \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = 2x(x - 3) + (x - 3)^2 \]

Factoring out \( (x - 3) \):

\[ \frac{dy}{dx} = (x - 3)(2x + (x - 3)) \]

This simplifies to:

\[ \frac{dy}{dx} = (x - 3)(3x - 3) \]

Factoring further gives:

\[ \frac{dy}{dx} = 3(x - 1)(x - 3) \]

### Step 3: Set the derivative greater than or equal to zero
To find the intervals where the function is increasing, we set the derivative greater than or equal to zero:

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

### Step 4: Find the critical points
The critical points occur when \( \frac{dy}{dx} = 0 \):

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

This gives us the critical points \( x = 1 \) and \( x = 3 \).

### Step 5: Test intervals around the critical points
We will test the intervals determined by the critical points \( (-\infty, 1) \), \( (1, 3) \), and \( (3, \infty) \):

1. **Interval \( (-\infty, 1) \)**: Choose \( x = 0 \)
\[ 3(0 - 1)(0 - 3) = 3(-1)(-3) = 9 \geq 0 \quad \text{(Increasing)} \]

2. **Interval \( (1, 3) \)**: Choose \( x = 2 \)
\[ 3(2 - 1)(2 - 3) = 3(1)(-1) = -3 < 0 \quad \text{(Decreasing)} \]

3. **Interval \( (3, \infty) \)**: Choose \( x = 4 \)
\[ 3(4 - 1)(4 - 3) = 3(3)(1) = 9 \geq 0 \quad \text{(Increasing)} \]

### Step 6: Conclusion
The function \( y = x(x - 3)^2 \) is increasing in the intervals:

\[ (-\infty, 1] \quad \text{and} \quad [3, \infty) \]