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The provided image contains a mathematical function and a couple of questions related to its analysis. Here's what the image shows:

### Function:
\[
f(x) = \frac{2x}{x^2 + 1}
\]

### Questions:
1. **Positividad y negatividad** (Positivity and negativity)
2. **Crecimiento y decrecimiento. ¿Extremos?** (Growth and decrease. Extrema?)

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Let’s proceed with a detailed step-by-step analysis.

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## Part (a): Positivity and Negativity of \(f(x)\)

- **Positivity**: \(f(x) > 0\)
  \[
  \frac{2x}{x^2 + 1} > 0
  \]
  The numerator is \(2x\), and the denominator \(x^2 + 1\) is always positive for all real \(x\). Thus, the sign of the function depends entirely on the sign of the numerator.
  
  - \(2x > 0 \quad \Longrightarrow \quad x > 0\)

  Hence, the function is positive for:
  \[
  x > 0
  \]

- **Negativity**: \(f(x) < 0\)
  - \(2x < 0 \quad \Longrightarrow \quad x < 0\)

  Hence, the function is negative for:
  \[
  x < 0
  \]

- **At \(x = 0\)**:
  \[
  f(0) = \frac{2(0)}{0^2 + 1} = 0
  \]
  Thus, the function is **zero** at \(x = 0\).

---

## Part (b): Growth, Decrease, and Extrema

To determine intervals of increase and decrease, as well as the extrema, we need the **first derivative** \(f'(x)\).

### First derivative of \(f(x)\):
Using the quotient rule:
\[
f'(x) = \frac{(x^2 + 1)(2) - (2x)(2x)}{(x^2 + 1)^2}
\]
\[
f'(x) = \frac{2x^2 + 2 - 4x^2}{(x^2 + 1)^2} = \frac{-2x^2 + 2}{(x^2 + 1)^2}
\]
\[
f'(x) = \frac{2(1 - x^2)}{(x^2 + 1)^2}
\]

### Critical points:
We set \(f'(x) = 0\):
\[
\frac{2(1 - x^2)}{(x^2 + 1)^2} = 0
\]
This occurs when:
\[
1 - x^2 = 0 \quad \Longrightarrow \quad x^2 = 1 \quad \Longrightarrow \quad x = \pm 1
\]

Thus, the critical points are \(x = 1\) and \(x = -1\).

---

### Increasing and decreasing intervals:
- **Sign of \(f'(x)\)**:
  \[
  f'(x) = \frac{2(1 - x^2)}{(x^2 + 1)^2}
  \]
  - For \(x \in (-\infty, -1) \cup (1, \infty)\), \(1 - x^2 < 0 \Rightarrow f'(x) < 0\): the function is **decreasing**.
  - For \(x \in (-1, 1)\), \(1 - x^2 > 0 \Rightarrow f'(x) > 0\): the function is **increasing**.

---

### Extrema:
- At \(x = -1\) and \(x = 1\), the function switches between increasing and decreasing:
  - **Local minimum** at \(x = -1\).
  - **Local maximum** at \(x = 1\).

---

## Summary:
1. **Positivity**: \(x > 0\)
2. **Negativity**: \(x < 0\)
3. **Zero**: \(x = 0\)
4. **Increasing**: \((-1, 1)\)
5. **Decreasing**: \((-\infty, -1) \cup (1, \infty)\)
6. **Local extrema**:
   - Local maximum at \(x = 1\).
   - Local minimum at \(x = -1\).

---

Would you like to dive deeper into any step or see the graph of the function for more clarity?

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### Related Questions:
1. How do we confirm the nature of extrema using the second derivative?
2. What happens to the function as \(x \to \infty\) or \(x \to -\infty\)?
3. Can the function have any asymptotes? If so, what kind?
4. How does the behavior of this function compare to \(f(x) = \frac{x}{x^2 + 1}\)?
5. How do we find concavity and inflection points for this function?

---

**Tip:** Always remember that the denominator of a rational function can help you determine the overall behavior, especially for signs and limits.

f(x) = 2x / (x^2 + 1). Analyze positivity, negativity, growth, and decrease. Identify extrema.

This problem explores the function f(x) = 2x / (x^2 + 1), analyzing its positivity, negativity, and identifying intervals of growth and decrease. Using the first derivative, we determine critical points and local extrema. This detailed breakdown is ideal for Grades 10-12 students.

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