The problem in the image asks:

Given $g(x) = \sqrt{x}$ and $f(x) = \frac{1}{x - 3}$, for what value of $x$ is $f(g(x))$ undefined?

To solve this, let's break it down:

1. Substitute $g(x)$ into $f(x)$. This gives us: $f(g(x)) = \frac{1}{\sqrt{x} - 3}$

2. $f(g(x))$ will be undefined when the denominator equals zero, i.e., when: $\sqrt{x} - 3 = 0$

3. Solve for $x$: $\sqrt{x} = 3$ Squaring both sides: $x = 9$

Thus, $f(g(x))$ is undefined when $x = 9$.

The correct answer is d. $x = 9$.

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

1. How would the solution change if $g(x)$ were $g(x) = x^2$ instead?
2. What other types of functions make $f(g(x))$ undefined?
3. How do you solve for when a function is defined or undefined in general?
4. Can you describe the domain of the function $f(g(x))$ in this case?
5. How would this problem differ if the function was $f(x) = \frac{1}{x + 3}$?

Tip: When dealing with composite functions like $f(g(x))$, always check the domains of both functions separately.