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!