The function `f(x) = x^(3) - 6x^(2) + 15x - 12` is

Options

A. strictly decreasing on R
B. strictly increasing on R
C. increasing in `(-oo, 2]` and decreasing in `(2, oo)`
D. none of these

Explanation

To determine the nature of the function \( f(x) = x^3 - 6x^2 + 15x - 12 \), we need to analyze its first derivative. Here’s the step-by-step solution:

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

\[ f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 15x - 12) \]

Using the power rule of differentiation:

\[ f'(x) = 3x^2 - 12x + 15 \]

### Step 2: Factor the first derivative
Next, we can factor the first derivative to analyze its sign.

\[ f'(x) = 3(x^2 - 4x + 5) \]

### Step 3: Complete the square
Now, we will complete the square for the quadratic expression \( x^2 - 4x + 5 \).

\[ x^2 - 4x + 4 + 1 = (x - 2)^2 + 1 \]

Thus, we can rewrite the first derivative as:

\[ f'(x) = 3((x - 2)^2 + 1) \]

### Step 4: Analyze the sign of the first derivative
The term \( (x - 2)^2 \) is always non-negative (i.e., \( \geq 0 \)) for all real numbers \( x \). Therefore, \( (x - 2)^2 + 1 \) is always positive since it is the sum of a non-negative number and 1.

Thus, we conclude:

\[ f'(x) > 0 \quad \text{for all } x \in \mathbb{R} \]

### Step 5: Conclusion about the function
Since the first derivative \( f'(x) \) is positive for all \( x \), it indicates that the function \( f(x) \) is strictly increasing on the entire real line \( \mathbb{R} \).

Therefore, the correct answer is that the function \( f(x) \) is strictly increasing.

### Final Answer:
The function \( f(x) = x^3 - 6x^2 + 15x - 12 \) is strictly increasing on \( \mathbb{R} \).

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