Examine whether the function given by `f(x)=x^(3)-3x^(2)+3x-5` is increasing in R.

Answer

To determine whether the function \( f(x) = x^3 - 3x^2 + 3x - 5 \) is increasing in \( \mathbb{R} \), we need to analyze its derivative \( f'(x) \). A function is increasing on an interval if its derivative is greater than or equal to zero throughout that interval.

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

\[ f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 3x - 5) \]

Using the power rule for differentiation:

- The derivative of \( x^3 \) is \( 3x^2 \).
- The derivative of \( -3x^2 \) is \( -6x \).
- The derivative of \( 3x \) is \( 3 \).
- The derivative of the constant \( -5 \) is \( 0 \).

Thus, we have:

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

### Step 2: Simplify the derivative
Next, we can factor the derivative:

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

Notice that \( x^2 - 2x + 1 \) can be rewritten as a perfect square:

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

### Step 3: Analyze the sign of the derivative
Now, we analyze \( f'(x) \):

Since \( (x - 1)^2 \) is a square term, it is always non-negative. Therefore:

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

Specifically, \( f'(x) = 0 \) when \( x = 1 \) and \( f'(x) > 0 \) for all \( x \neq 1 \).

### Step 4: Conclusion
Since \( f'(x) \geq 0 \) for all \( x \) in \( \mathbb{R} \), the function \( f(x) \) is increasing on the entire set of real numbers \( \mathbb{R} \).

### Final Answer
The function \( f(x) = x^3 - 3x^2 + 3x - 5 \) is increasing in \( \mathbb{R} \).

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