The expression you provided is \(\arccos\left(\frac{x}{2}\right)\). This represents the inverse cosine function, which finds the angle whose cosine is \(\frac{x}{2}\).

### Key Points:
- **Domain**: For the expression \(\arccos\left(\frac{x}{2}\right)\) to be valid, \(\frac{x}{2}\) must lie within the interval \([-1, 1]\). This translates to:
  \[
  -1 \leq \frac{x}{2} \leq 1
  \]
  Simplifying this, we get:
  \[
  -2 \leq x \leq 2
  \]
  So, \(x\) must be within the interval \([-2, 2]\).

- **Range**: The range of the \(\arccos\) function is \([0, \pi]\). This means that the output of \(\arccos\left(\frac{x}{2}\right)\) will always be an angle in radians within this interval.

### Expression Interpretation:
The function \(\arccos\left(\frac{x}{2}\right)\) gives the angle \(\theta\) such that:
\[
\cos(\theta) = \frac{x}{2}
\]
where \(\theta\) lies in the range \([0, \pi]\).

Would you like further details on how this function behaves or any specific examples?

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Here are 5 related questions:
1. How would you differentiate \(\arccos\left(\frac{x}{2}\right)\) with respect to \(x\)?
2. What is the value of \(\arccos\left(\frac{1}{2}\right)\)?
3. How can \(\arccos\left(\frac{x}{2}\right)\) be expressed in terms of sine?
4. Can you find the Taylor series expansion for \(\arccos\left(\frac{x}{2}\right)\) around \(x = 0\)?
5. What is the graph of \(\arccos\left(\frac{x}{2}\right)\) over its domain?

**Tip**: Always check the domain of the argument when working with inverse trigonometric functions to ensure the expression is valid.

Explore the definition and properties of the arccos(x/2) function, which represents the inverse cosine function. Learn about its domain, range, and interpretation with examples. Suitable for high school students studying trigonometry and inverse functions.

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