We are given the function:

$f(x) = \sqrt{x - 2} + 7$

We are asked to determine $f(x-1) \cdot f(x-1)$, which is essentially $(f(x-1))^2$.

Step 1: Substitute $x-1$ into the function

To find $f(x-1)$, substitute $x-1$ for $x$ in the original function:

$f(x-1) = \sqrt{(x-1) - 2} + 7 = \sqrt{x - 3} + 7$

Step 2: Square the expression

Now, to compute $f(x-1) \cdot f(x-1)$, we square the expression for $f(x-1)$:

$(f(x-1))^2 = \left( \sqrt{x - 3} + 7 \right)^2$

Step 3: Expand the square

Use the identity $(a + b)^2 = a^2 + 2ab + b^2$ to expand the expression:

$(f(x-1))^2 = \left( \sqrt{x - 3} \right)^2 + 2 \cdot \sqrt{x - 3} \cdot 7 + 7^2$

$(f(x-1))^2 = (x - 3) + 14 \sqrt{x - 3} + 49$

Step 4: Simplify

Combine like terms:

$(f(x-1))^2 = x - 3 + 49 + 14 \sqrt{x - 3}$

$(f(x-1))^2 = x + 46 + 14 \sqrt{x - 3}$

Thus, the expression for $f(x-1) \cdot f(x-1)$ is:

$f(x-1) \cdot f(x-1) = x + 46 + 14 \sqrt{x - 3}$

Do you need more details or have any questions?

Here are 5 related questions for deeper understanding:

1. What is the domain of the function $f(x) = \sqrt{x - 2} + 7$?
2. How would $f(x-2) \cdot f(x-2)$ differ from $f(x-1) \cdot f(x-1)$?
3. Can you find the derivative of $f(x) = \sqrt{x - 2} + 7$?
4. How would you determine $f^{-1}(x)$, the inverse of $f(x)$?
5. What happens to $f(x-1) \cdot f(x-1)$ as $x \to 3$?

Tip: When squaring expressions that include square roots, remember to apply the expansion carefully to avoid common mistakes!