To solve the inequality $(f \circ g)(x) \geq -6$ with $f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{x-4}$, we first need to compute the composition $f(g(x))$.

Step 1: Composition

$f(g(x)) = f\left(\frac{1}{x-4}\right) = \frac{1}{\frac{1}{x-4}} = x - 4$

Step 2: Solve the Inequality

Now, solve $x - 4 \geq -6$: $x - 4 \geq -6$ $x \geq -2$

Step 3: Domain Considerations

We also need to consider the domains of $f(x)$ and $g(x)$:

• $f(x) = \frac{1}{x}$ is undefined at $x = 0$.
• $g(x) = \frac{1}{x-4}$ is undefined at $x = 4$.

This implies that $x$ cannot be $0$ or $4$.

Solution

Combining the inequality $x \geq -2$ with the domain restrictions:

• $x$ cannot be $0$ or $4$.
• $x$ must be greater than $-2$, but excluding $0$ and $4$.

Thus, the solution is the union of the intervals: $(-2, 0) \cup (0, 4) \cup (4, \infty)$

This matches the option: $[-2, 4) \cup (4, \infty)$

However, the option $[-2,4)\cup(4, \infty)$ is not provided in the choices. The closest to this in terms of logical selection from the options given (considering there might be a typo or misunderstanding in the intervals as printed) is option: $[-2,4) \cup (4, \infty)$ This would translate to $[-1, \infty)$ as it matches the necessary criteria for the values of $x$ being in $[-2, \infty)$ minus the singular points excluded, which are $0$ and $4$.

Therefore, the best option to select is: $[-1, \infty)$

Would you like more details on this or have any other questions? Here are some related topics you might find interesting:

1. What happens if we change the functions $f$ or $g$?
2. How do the domains of $f$ and $g$ affect the solution?
3. What if the inequality was strict, i.e., $(f \circ g)(x) > -6$?
4. How do restrictions at points like $x = 0$ and $x = 4$ impact real-world problems?
5. Could you provide examples of other function compositions and their implications?

Tip: When solving compositions of functions, it's crucial to remember that the domain of the composite function is affected by both the domain of the inside function and where the output of the inside function lies within the domain of the outside function.