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

Options

A. `0 lt x lt 2`
B. `x lt 2`
C. `x gt 2` or `x lt 0`
D. all x

Answer

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

### Step 1: Find the derivative of the function
We start by differentiating the function \( f(x) \) with respect to \( x \).

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

### Step 2: Set the derivative greater than zero
For the function to be increasing, the derivative must be greater than zero.

\[ f'(x) > 0 \implies 3x^2 - 6x > 0 \]

### Step 3: Factor the derivative
We can factor the expression on the left-hand side.

\[ 3x(x - 2) > 0 \]

### Step 4: Determine the critical points
To find where the expression is greater than zero, we first find the critical points by setting the derivative equal to zero.

\[ 3x(x - 2) = 0 \]

This gives us the critical points:

\[ x = 0 \quad \text{and} \quad x = 2 \]

### Step 5: Test intervals around the critical points
We will test the sign of \( f'(x) \) in the intervals determined by the critical points: \( (-\infty, 0) \), \( (0, 2) \), and \( (2, \infty) \).

1. **Interval \( (-\infty, 0) \)**: Choose \( x = -1 \)
\[ f'(-1) = 3(-1)(-1 - 2) = 3(-1)(-3) = 9 > 0 \]
(Increasing)

2. **Interval \( (0, 2) \)**: Choose \( x = 1 \)
\[ f'(1) = 3(1)(1 - 2) = 3(1)(-1) = -3 < 0 \]
(Decreasing)

3. **Interval \( (2, \infty) \)**: Choose \( x = 3 \)
\[ f'(3) = 3(3)(3 - 2) = 3(3)(1) = 9 > 0 \]
(Increasing)

### Step 6: Combine the intervals
From our tests, we find that \( f(x) \) is increasing in the intervals \( (-\infty, 0) \) and \( (2, \infty) \).

### Conclusion
Thus, the function \( f(x) = x^3 - 3x^2 + 6 \) is increasing for \( x < 0 \) and \( x > 2 \).

### Final Answer
The function is increasing for \( x < 0 \) and \( x > 2 \).

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