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The problem asks to identify which of the given functions are odd functions. Let's go through the functions step by step to check whether each one is odd.

### Recall:
A function \( f(x) \) is **odd** if:
\[ f(-x) = -f(x) \]
This means that if you replace \( x \) with \( -x \), the result must be the negative of the original function.

Now, let's analyze the functions:

### 1. \( y = x - 2 \)
Substitute \( -x \) for \( x \):
\[ f(-x) = -x - 2 \]

Clearly, \( f(-x) \neq -(x - 2) = -x + 2 \), so this function is **not odd**.

### 2. \( y = x^3 + 5x \)
Substitute \( -x \) for \( x \):
\[ f(-x) = (-x)^3 + 5(-x) = -x^3 - 5x \]

Now, check if this is the negative of the original function:
\[ -(x^3 + 5x) = -x^3 - 5x \]

Since \( f(-x) = -f(x) \), this function **is odd**.

### 3. \( y = \frac{x}{4} \)
Substitute \( -x \) for \( x \):
\[ f(-x) = \frac{-x}{4} = -\frac{x}{4} \]

Check if this is the negative of the original function:
\[ -\frac{x}{4} = -f(x) \]

Thus, this function **is odd**.

### 4. \( y = 4x^3 - x + 2 \)
Substitute \( -x \) for \( x \):
\[ f(-x) = 4(-x)^3 - (-x) + 2 = -4x^3 + x + 2 \]

Clearly, \( f(-x) \neq -(4x^3 - x + 2) = -4x^3 + x - 2 \), so this function is **not odd**.

### Conclusion:
Among the given functions, the **odd functions** are:
- \( y = x^3 + 5x \)
- \( y = \frac{x}{4} \)

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Would you like further details or clarification on any step?

Here are 5 related questions to explore further:
1. What are the properties of even functions?
2. How can we graphically identify odd functions?
3. Can a function be both even and odd?
4. How does symmetry relate to odd functions?
5. Can a polynomial of even degree be an odd function?

**Tip:** Odd functions have rotational symmetry about the origin, which can help when sketching or analyzing graphs!

Given the functions, identify the odd functions among them.

Learn to identify odd functions from a given set by checking the symmetry properties. This guide provides a step-by-step approach using the definition of odd functions, suitable for high school students.

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