The function `f(x)=x^(3)-6x^(2)+12x-16, x in R` is

Options

A. [1,2]
B. [1,2)
C. (1,2]
D. (1,2)

Answer

To analyze the function \( f(x) = x^3 - 6x^2 + 12x - 16 \) in terms of its increasing and decreasing behavior, we will follow these steps:

### Step 1: Find the derivative of the function
To determine where the function is increasing or decreasing, we first need to find the derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 12x - 16) \]

Using the power rule, we differentiate each term:

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

### Step 2: Factor the derivative
Next, we will factor the derivative to find critical points.

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

### Step 3: Find critical points
To find the critical points, we set the derivative equal to zero:

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

This gives us:

\[ (x - 2)^2 = 0 \implies x - 2 = 0 \implies x = 2 \]

### Step 4: Analyze the sign of the derivative
Now, we need to analyze the sign of \( f'(x) \) around the critical point \( x = 2 \).

1. For \( x < 2 \):
\[ f'(x) = 3(x - 2)^2 > 0 \quad (\text{since the square of any real number is non-negative}) \]

2. For \( x > 2 \):
\[ f'(x) = 3(x - 2)^2 > 0 \quad (\text{for the same reason}) \]

### Step 5: Conclusion about increasing/decreasing behavior
Since \( f'(x) > 0 \) for all \( x \neq 2 \), we conclude that:

- \( f(x) \) is increasing for all \( x \in \mathbb{R} \) except at \( x = 2 \), where the derivative is zero but does not change sign.

### Final Result
Thus, the function \( f(x) = x^3 - 6x^2 + 12x - 16 \) is increasing for all \( x \in \mathbb{R} \) except at \( x = 2 \).

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