To find the range of $(f \circ g)(x)$, we first need to evaluate the composition $f(g(x))$. Here are the steps:

Given Functions:

1. $f(x) = x^2 - 9$, where $x < 0$.
2. $g(x) = 10 - \frac{x}{2}$, where $x > 6$.

Composition $f(g(x))$:

$f(g(x)) = f\left(10 - \frac{x}{2}\right)$

Substituting $g(x)$ into $f(x)$: $f\left(10 - \frac{x}{2}\right) = \left(10 - \frac{x}{2}\right)^2 - 9$

Simplify the expression: $f\left(10 - \frac{x}{2}\right) = \left(10 - \frac{x}{2}\right)^2 - 9$ $= \left(100 - 10x + \frac{x^2}{4}\right) - 9$ $= \frac{x^2}{4} - 10x + 91$

Domain of $g(x)$:

Since $g(x) = 10 - \frac{x}{2}$ and $x > 6$, we calculate: $g(x) < 10 - \frac{6}{2} = 10 - 3 = 7$ Thus, $g(x) \in (-\infty, 7)$.

Domain of $f(g(x))$:

From the condition $x < 0$ in $f(x)$, the input $g(x)$ must also satisfy $g(x) < 0$. This imposes: $10 - \frac{x}{2} < 0 \implies \frac{x}{2} > 10 \implies x > 20$

Thus, the domain of $g(x)$ becomes $x \in (20, \infty)$.

Range of $f(g(x))$:

We calculate the range of $f(g(x)) = \frac{x^2}{4} - 10x + 91$ over $x \in (20, \infty)$:

1. As $x \to \infty$, $f(g(x)) \to \infty$ (dominance of $\frac{x^2}{4}$).
2. As $x \to 20^+$, substitute $x = 20$: $f(g(20)) = \frac{20^2}{4} - 10(20) + 91 = 100 - 200 + 91 = -9$

Thus, the range of $f(g(x))$ is: $\boxed{[-9, \infty)}$

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5 Related Questions:

1. How do we determine the domain of composite functions?
2. How does the restriction $x < 0$ for $f(x)$ affect the composition $f(g(x))$?
3. What happens to the range of $f(g(x))$ if $g(x)$ is bounded?
4. Can we find the inverse of $f(x) = x^2 - 9$ if $x < 0$?
5. What changes if the condition $x > 6$ for $g(x)$ is removed?

Tip:

When dealing with composite functions, always consider both the domain of the inner function and how its range aligns with the domain of the outer function.