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Let's solve for the inverses of the given functions.

### Problem (a)
**Given:** \( f : \mathbb{R} \to \mathbb{R}, \ f(x) = 2 - 3x \)

**Solution:**

1. Let \( y = f(x) \). Thus, we have:
   \[
   y = 2 - 3x
   \]

2. Solve for \( x \) in terms of \( y \):
   \[
   y - 2 = -3x
   \]
   \[
   x = \frac{2 - y}{3}
   \]

3. Now, replace \( y \) with \( x \) to write the inverse function:
   \[
   f^{-1}(x) = \frac{2 - x}{3}
   \]

### Answer for (a):
\[
f^{-1}(x) = \frac{2 - x}{3}
\]

---

### Problem (b)
**Given:** \( f : (-\infty, 0) \to (-1, \infty), \ f(x) = x^2 - 1 \)

**Solution:**

1. Let \( y = f(x) \). Thus, we have:
   \[
   y = x^2 - 1
   \]

2. Solve for \( x \) in terms of \( y \):
   \[
   y + 1 = x^2
   \]
   \[
   x = \pm \sqrt{y + 1}
   \]

3. Since the domain of \( f(x) \) is \( (-\infty, 0) \), we are only interested in the negative square root branch (as \( x \) is negative in this domain). Therefore:
   \[
   x = -\sqrt{y + 1}
   \]

4. Replace \( y \) with \( x \) to write the inverse function:
   \[
   f^{-1}(x) = -\sqrt{x + 1}
   \]

### Answer for (b):
\[
f^{-1}(x) = -\sqrt{x + 1}
\]

---

Would you like additional details on any part of the solution?

Here are five related questions to expand your understanding:

1. What properties must a function have for an inverse to exist?
2. How would the solution change if the domain in part (b) was positive values only?
3. What is the graphical relationship between a function and its inverse?
4. How can you verify that these inverses are correct?
5. What are real-world applications of finding inverses of functions?

**Tip:** When finding inverses, remember to consider the domain restrictions carefully, as they determine which branch or sign to use when solving for \( x \) in terms of \( y \).

Find the inverse of the following functions on the given domains and codomains:

(a) \( f : \mathbb{R} \to \mathbb{R}, \ f(x) = 2 - 3x \)
(b) \( f : (-\infty, 0) \to (-1, \infty), \ f(x) = x^2 - 1 \)

Learn how to find the inverse of functions with specified domains and codomains. This solution involves calculating the inverse for a linear function \( f(x) = 2 - 3x \) and a quadratic function \( f(x) = x^2 - 1 \) within restricted intervals. Suitable for students in Grades 10-12.

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