Test whether the following function `f(x)=2-3x+3x^(2)-x^(3), x in R` is increasing or decreasing.

Answer

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

### Step 1: Find the derivative of the function
To analyze the behavior of the function, we first need to compute its derivative \( f'(x) \).

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

Calculating the derivative term by term:

- The derivative of \( 2 \) is \( 0 \).
- The derivative of \( -3x \) is \( -3 \).
- The derivative of \( 3x^2 \) is \( 6x \).
- The derivative of \( -x^3 \) is \( -3x^2 \).

Combining these results, we get:

\[ f'(x) = -3 + 6x - 3x^2 \]

### Step 2: Simplify the derivative
Now, we can rearrange the derivative:

\[ f'(x) = -3x^2 + 6x - 3 \]

Factoring out \( -3 \):

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

Recognizing that \( x^2 - 2x + 1 \) is a perfect square:

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

### Step 3: Analyze the sign of the derivative
The expression \( (x - 1)^2 \) is always non-negative (i.e., \( (x - 1)^2 \geq 0 \)) for all \( x \in \mathbb{R} \). Therefore, \( -3(x - 1)^2 \) is always less than or equal to zero:

\[ f'(x) \leq 0 \]

This indicates that the derivative is either negative or zero, meaning the function is either decreasing or constant.

### Step 4: Conclusion
Since \( f'(x) \) is less than or equal to zero for all \( x \in \mathbb{R} \), we conclude that the function \( f(x) \) is decreasing for all real numbers \( x \).