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Let's solve problem #13, starting with part (a).

The function provided is:

\[
f(x) = 1 + \sqrt{9 + 8x}
\]

To find the inverse of \( f(x) \), we follow these steps:

### Step 1: Replace \( f(x) \) with \( y \)
\[
y = 1 + \sqrt{9 + 8x}
\]

### Step 2: Solve for \( x \) in terms of \( y \)

First, subtract 1 from both sides:
\[
y - 1 = \sqrt{9 + 8x}
\]

Now, square both sides to eliminate the square root:
\[
(y - 1)^2 = 9 + 8x
\]

Next, subtract 9 from both sides:
\[
(y - 1)^2 - 9 = 8x
\]

Finally, divide by 8 to solve for \( x \):
\[
x = \frac{(y - 1)^2 - 9}{8}
\]

### Step 3: Swap \( x \) and \( y \) to express the inverse function

\[
f^{-1}(x) = \frac{(x - 1)^2 - 9}{8}
\]

So, the inverse function is:

\[
f^{-1}(x) = \frac{(x - 1)^2 - 9}{8}
\]

---

For part (b), we need to determine for what values of \( x \) the inverse function is defined. Let's first look at the domain of the original function:

\[
f(x) = 1 + \sqrt{9 + 8x}
\]

The expression inside the square root must be non-negative:
\[
9 + 8x \geq 0
\]

Solving for \( x \):
\[
x \geq -\frac{9}{8}
\]

Thus, the domain of \( f(x) \) is \( x \geq -\frac{9}{8} \). Since the domain of \( f(x) \) becomes the range of the inverse function \( f^{-1}(x) \), the inverse is defined for values of \( x \geq -\frac{9}{8} \).

---

Would you like further clarification on any of these steps or details?

### Related questions:
1. How do we verify if a function and its inverse are correct?
2. What are the common properties of inverse functions?
3. Can every function have an inverse? Why or why not?
4. How does restricting the domain affect the existence of an inverse?
5. What is the graphical interpretation of an inverse function?

**Tip:** When finding an inverse function, always ensure the original function is one-to-one by checking if it passes the horizontal line test.

Find a formula for the inverse of f(x) = 1 + √(9 + 8x).

Learn how to find the inverse of the function f(x) = 1 + √(9 + 8x) in this detailed algebra tutorial. The solution involves solving for x, applying square roots, and understanding the domain and range of the function. Suitable for students in Grades 10-12.

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