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Let's solve each part of the problem step by step.

### Part 1: \( f(x) \geq 0 \)

The function is \( f(x) = \frac{x(x+3)}{x-1} \).

To find where \( f(x) \geq 0 \), we need to find the intervals where the function is non-negative. This requires analyzing the sign of the numerator and denominator.

1. **Numerator**: \( x(x + 3) \)
   - \( x(x + 3) = 0 \) at \( x = 0 \) or \( x = -3 \)
   - For \( x < -3 \), \( x(x + 3) > 0 \) (positive).
   - For \( -3 < x < 0 \), \( x(x + 3) < 0 \) (negative).
   - For \( x > 0 \), \( x(x + 3) > 0 \) (positive).

2. **Denominator**: \( x - 1 \)
   - \( x - 1 = 0 \) at \( x = 1 \)
   - For \( x < 1 \), \( x - 1 < 0 \) (negative).
   - For \( x > 1 \), \( x - 1 > 0 \) (positive).

3. **Combined Sign**:
   - For \( x \in (-\infty, -3) \cup (0, 1) \), the function is positive.
   - For \( x \in (-3, 0) \cup (1, \infty) \), the function is negative.
   - At \( x = 0 \) and \( x = -3 \), the function equals zero.

Therefore, \( f(x) \geq 0 \) on the intervals \( (-\infty, -3] \cup [0, 1] \).

So the correct answer is: \(\textbf{]-\infty, -3] \cup [0, 1[}\).

### Part 2: Horizontal/Oblique Asymptotes

To find the horizontal or oblique asymptotes of \( f(x) = \frac{x(x+3)}{x-1} \), we consider the degrees of the polynomials:

- The degree of the numerator is 2.
- The degree of the denominator is 1.

Since the degree of the numerator is greater than the denominator, the function has an oblique asymptote found by polynomial long division.

\[
\frac{x^2 + 3x}{x - 1} = x + 4 + \frac{4}{x - 1}
\]

As \( x \to \infty \), \( \frac{4}{x - 1} \to 0 \). Thus, the oblique asymptote is \( y = x + 4 \).

So the correct answer is: \(\textbf{y = x + 4}\).

### Part 3: Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero (where the function is undefined):

\[
x - 1 = 0 \implies x = 1
\]

Therefore, the vertical asymptote is at \( x = 1 \).

So the correct answer is: \(\textbf{only x = 1}\).

### Part 4: Extreme Points

To find extreme points, we take the derivative and set it to zero:

\[
f(x) = \frac{x(x+3)}{x-1}
\]

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

Setting the numerator to zero:

\[
x^2 - 3 = 

Learn how to analyze the function f(x) = x(x + 3) / (x - 1) step-by-step with detailed solutions to find where the function is non-negative, its asymptotes, and extreme points. Suitable for advanced high school students.

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