Problem:

We are tasked with the following questions based on the function $f(x) = 1 + \sqrt{9 + 8x}$:

1. Part (a): Find a formula for the inverse of $f(x)$.
2. Part (b): Determine for which values of $x$ the inverse function $f^{-1}(x)$ is defined.

Part (a): Finding the inverse of $f(x)$

The function is given as: $f(x) = 1 + \sqrt{9 + 8x}$

To find the inverse, follow these steps:

1. Replace $f(x)$ with $y$: $y = 1 + \sqrt{9 + 8x}$

2. Solve for $x$ in terms of $y$:

   $y - 1 = \sqrt{9 + 8x}$

   Square both sides to remove the square root: $(y - 1)^2 = 9 + 8x$

3. Isolate $x$: $8x = (y - 1)^2 - 9$ $x = \frac{(y - 1)^2 - 9}{8}$

So, the inverse function $f^{-1}(x)$ is: $f^{-1}(x) = \frac{(x - 1)^2 - 9}{8}$

Part (b): For what values of $x$ is $f^{-1}(x)$ defined?

The original function involves a square root, so we must ensure the expression inside the square root is non-negative. For the function $f(x) = 1 + \sqrt{9 + 8x}$, the expression under the square root, $9 + 8x$, must be greater than or equal to zero.

Solve for $x$: $9 + 8x \geq 0$ $x \geq -\frac{9}{8}$

Therefore, the domain of the original function is $x \geq -\frac{9}{8}$. Since the inverse function $f^{-1}(x)$ must reflect this domain, the inverse function is defined for values of $x \geq 1$, because $f(x) \geq 1$.

Summary of Answers:

• (a): The inverse function is $f^{-1}(x) = \frac{(x - 1)^2 - 9}{8}$.
• (b): The inverse is defined for $x \geq 1$.

Would you like further clarification or additional details on any part? Here are some related questions:

1. How do you verify the correctness of an inverse function?
2. What is the graphical relationship between a function and its inverse?
3. Can you explain the domain and range restrictions for inverses?
4. How do inverse functions apply in real-world contexts?
5. What happens to the inverse when the original function is not one-to-one?

Tip: Remember that a function needs to be one-to-one (each $y$ value is paired with one unique $x$) for its inverse to exist.