Function `f(x)=x^3-3x^2+3x-100` is increasing at `x=?`

Options

A. `-2`
B. `1`
C. `0`
D. None

Explanation

To determine where the function \( f(x) = x^3 - 3x^2 + 3x - 100 \) is increasing, we need to follow these steps:

### Step 1: Find the derivative of the function
The first step is to calculate the derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 3x - 100) \]

Using the power rule for differentiation:

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

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

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

Notice that \( x^2 - 2x + 1 \) can be factored further:

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

### Step 3: Analyze the sign of the derivative
The expression \( (x - 1)^2 \) is a perfect square, which means it is always non-negative. Therefore, we can conclude:

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

### Step 4: Conclusion about the function
Since \( f'(x) \geq 0 \) for all \( x \), this means that the function \( f(x) \) is increasing for all values of \( x \).

### Final Answer
Thus, the function \( f(x) \) is increasing at all values of \( x \).

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