Let `f(x)=(x^(2)-2x+1)/(x+3),f i n d x: (i) f(x) gt 0`
` (ii) f(x) lt 0`
`

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{x^2 - 2x + 1}{x + 3} \) under the conditions \( f(x) > 0 \) and \( f(x) < 0 \).

### Step 1: Simplify the function
The numerator \( x^2 - 2x + 1 \) can be factored as \( (x - 1)^2 \). Thus, we can rewrite the function as:
\[ f(x) = \frac{(x - 1)^2}{x + 3} \]

### Step 2: Identify critical points
Next, we need to find the points where the function is equal to zero or undefined. The function is zero when the numerator is zero, and it is undefined when the denominator is zero.

- **Numerator**: \( (x - 1)^2 = 0 \) gives us \( x = 1 \).
- **Denominator**: \( x + 3 = 0 \) gives us \( x = -3 \).

### Step 3: Analyze the sign of \( f(x) \)
We will analyze the sign of \( f(x) \) in the intervals determined by the critical points \( x = -3 \) and \( x = 1 \). The intervals to consider are:
1. \( (-\infty, -3) \)
2. \( (-3, 1) \)
3. \( (1, \infty) \)

### Step 4: Test each interval
1. **Interval \( (-\infty, -3) \)**:
- Choose \( x = -4 \):
\[ f(-4) = \frac{(-4 - 1)^2}{-4 + 3} = \frac{25}{-1} = -25 \quad (\text{Negative}) \]

2. **Interval \( (-3, 1) \)**:
- Choose \( x = 0 \):
\[ f(0) = \frac{(0 - 1)^2}{0 + 3} = \frac{1}{3} \quad (\text{Positive}) \]

3. **Interval \( (1, \infty) \)**:
- Choose \( x = 2 \):
\[ f(2) = \frac{(2 - 1)^2}{2 + 3} = \frac{1}{5} \quad (\text{Positive}) \]

### Step 5: Summarize the results
- \( f(x) > 0 \) in the intervals \( (-3, 1) \) and \( (1, \infty) \).
- \( f(x) < 0 \) in the interval \( (-\infty, -3) \).

### Final Answers
(i) For \( f(x) > 0 \):
\[ x \in (-3, 1) \cup (1, \infty) \]

(ii) For \( f(x) < 0 \):
\[ x \in (-\infty, -3) \]