To analyze the problem:

We are tasked with determining which expressions are not defined at $x = 1$ given the following:

• $f(x) = \sqrt{x + 2}$
• $g(x) = \frac{1}{x - 1}$

Step 1: Examine each function at $x = 1$

• $f(x) = \sqrt{x + 2}$: At $x = 1$, $f(1) = \sqrt{1 + 2} = \sqrt{3}$, which is defined.

• $g(x) = \frac{1}{x - 1}$: At $x = 1$, $g(1) = \frac{1}{1 - 1} = \frac{1}{0}$, which is undefined.

Step 2: Analyze each option

1. $f(x) + g(x)$: $f(1) + g(1)$ involves adding $\sqrt{3}$ (defined) and $\frac{1}{0}$ (undefined), so this expression is not defined at $x = 1$.

2. $g(x) - f(x)$: $g(1) - f(1)$ involves subtracting $\sqrt{3}$ (defined) from $\frac{1}{0}$ (undefined), so this expression is not defined at $x = 1$.

3. $(fg)(x)$ (product of $f(x)$ and $g(x)$): $f(1) \cdot g(1) = \sqrt{3} \cdot \frac{1}{0}$, which involves multiplication with an undefined term, so this expression is not defined at $x = 1$.

4. $g(f(x))$ (composition of $g(x)$ with $f(x)$): $f(1) = \sqrt{3}$, so $g(f(1)) = g(\sqrt{3}) = \frac{1}{\sqrt{3} - 1}$. This is defined because $\sqrt{3} - 1 \neq 0$.

Conclusion

The expressions not defined at $x = 1$ are:

• $f(x) + g(x)$
• $g(x) - f(x)$
• $(fg)(x)$

Would you like further clarification or detailed steps? Here are related questions to expand your understanding:

1. What happens if $x = 2$ for these functions?
2. How does $g(f(x))$ behave for other values of $x$?
3. Why does division by zero make a function undefined?
4. Could $f(x)$ ever cause undefined behavior for $g(f(x))$?
5. How can these concepts apply to composite functions in calculus?

Tip: Always evaluate domain restrictions carefully when dealing with square roots or divisions.