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The function given is \( f(x) = 2 \arccos(x) - 1 \).

Let's break down each part of the problem:

### (a) Domain and Range
- The **domain** of \( \arccos(x) \) is \( x \in [-1, 1] \) because the inverse cosine function only takes values in this range.
- The **range** of \( \arccos(x) \) is \( [0, \pi] \), since \( \arccos(x) \) outputs angles between 0 and \( \pi \).
- The function \( f(x) = 2\arccos(x) - 1 \) will multiply the range of \( \arccos(x) \) by 2 and subtract 1.

So, the range of \( f(x) \) is:
\[
f(x) = 2[0, \pi] - 1 = [-1, 2\pi - 1]
\]
Thus:
- **Domain:** \( x \in [-1, 1] \)
- **Range:** \( f(x) \in [-1, 2\pi - 1] \)

### (b) Exact value of \( f(-0.5) \)
We need to compute \( f(-0.5) \).

\[
f(x) = 2 \arccos(x) - 1
\]
Substitute \( x = -0.5 \):

\[
f(-0.5) = 2 \arccos(-0.5) - 1
\]
The value of \( \arccos(-0.5) \) is \( \frac{2\pi}{3} \) (in radians):

\[
f(-0.5) = 2 \times \frac{2\pi}{3} - 1 = \frac{4\pi}{3} - 1
\]

This is the exact value of \( f(-0.5) \).

### (c) Inverse function \( f^{-1}(x) \)
To find the inverse function, start with \( f(x) = 2 \arccos(x) - 1 \) and solve for \( x \).

\[
y = 2 \arccos(x) - 1
\]
Add 1 to both sides:
\[
y + 1 = 2 \arccos(x)
\]
Divide by 2:
\[
\frac{y + 1}{2} = \arccos(x)
\]
Now apply the cosine to both sides:
\[
x = \cos\left(\frac{y + 1}{2}\right)
\]
Thus, the inverse function is:
\[
f^{-1}(x) = \cos\left(\frac{x + 1}{2}\right)
\]

---

Would you like more details on any of these steps?

Here are 5 related questions to explore further:
1. What are the properties of the arccosine function?
2. How would the domain and range change if \( f(x) \) were \( 3 \arccos(x) - 2 \)?
3. What is the derivative of \( f(x) = 2 \arccos(x) - 1 \)?
4. How does the value of \( f^{-1}(x) \) change for different inputs?
5. Can you compute the value of \( f(x) \) for another input, such as \( f(0.7) \)?

**Tip:** When working with inverse trigonometric functions, always keep track of their restricted domains and ranges to avoid calculation errors!

A function is defined by f(x) = 2 arccos(x) - 1. (a) Write down the maximum domain and range of this function. (b) Find the exact value of f(-0.5). (c) Find the inverse function f^{-1}(x).

Explore the function f(x) = 2 arccos(x) - 1, including how to find its domain, range, the exact value of f(-0.5), and the inverse function f^{-1}(x). This problem covers advanced trigonometric concepts and is suitable for high school students.

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