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Let's solve for the inverse function \( f^{-1}(x) \) for the given function:

\[
f(x) = 7 + \sqrt{6x - 5}
\]

### Step-by-step solution:

1. **Express \( y = f(x) \):**
   \[
   y = 7 + \sqrt{6x - 5}
   \]

2. **Solve for \( x \) in terms of \( y \):**

   First, isolate the square root:
   \[
   y - 7 = \sqrt{6x - 5}
   \]

   Square both sides to eliminate the square root:
   \[
   (y - 7)^2 = 6x - 5
   \]

   Now, solve for \( x \):
   \[
   6x = (y - 7)^2 + 5
   \]
   \[
   x = \frac{(y - 7)^2 + 5}{6}
   \]

3. **Replace \( y \) with \( x \):**

   Since we are solving for the inverse function, replace \( y \) with \( x \):
   \[
   f^{-1}(x) = \frac{(x - 7)^2 + 5}{6}
   \]

Thus, the inverse function is:
\[
f^{-1}(x) = \frac{(x - 7)^2 + 5}{6}
\]

This holds for \( x \geq 7 \).

Do you want more details or have any questions about the process?

Here are 5 related questions to consider:
1. What is the domain and range of \( f(x) \) and \( f^{-1}(x) \)?
2. How does the square root function affect the behavior of the inverse?
3. Can you find the derivative of \( f^{-1}(x) \)?
4. How do you verify that the two functions \( f(x) \) and \( f^{-1}(x) \) are indeed inverses?
5. What are the graphical interpretations of inverse functions?

**Tip:** Always check the domain and range when finding inverse functions to ensure the inverse exists and is correct for the given values.

Let f(x) = 7 + √(6x - 5). Find f⁻¹(x) for x ≥ 7.

This page provides a detailed step-by-step solution to find the inverse of f(x) = 7 + √(6x - 5), suitable for students in Grades 10-12. The problem explores inverse functions and algebraic manipulation of radicals.

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