1. **State the problem:** We have two functions: $$f : [6, \infty) \to \mathbb{R}, \quad f(x) = b - 3\sqrt[3]{x}$$ $$g : (-\infty, 9] \to \mathbb{R}, \quad g(x) = 6x^2 + b$$ We want to find the values of $b$ such that both compositions $f \circ g$ and $g \circ f$ exist. 2. **Recall the domain rules for composition:** - For $f \circ g$ to exist, the output of $g$ must lie in the domain of $f$. - For $g \circ f$ to exist, the output of $f$ must lie in the domain of $g$. 3. **Analyze $f \circ g$:** - Domain of $g$ is $(-\infty, 9]$. - $f$ is defined on $[6, \infty)$. - So, for $f(g(x))$ to exist, we need $g(x) \in [6, \infty)$ for all $x \in (-\infty, 9]$. 4. **Find range condition for $g(x)$:** - $g(x) = 6x^2 + b$. - Since $6x^2 \geq 0$, minimum of $g(x)$ on $(-\infty, 9]$ is at $x=0$: $$g(0) = 6 \cdot 0^2 + b = b$$ - Maximum at $x=9$ is: $$g(9) = 6 \cdot 81 + b = 486 + b$$ - Since $6x^2$ is increasing for $x \geq 0$, the minimum value of $g(x)$ on $(-\infty, 9]$ is $b$. 5. **Condition for $f(g(x))$ to exist:** - We need $g(x) \geq 6$ for all $x \in (-\infty, 9]$. - Minimum $g(x) = b \geq 6$. - So, $b \geq 6$. 6. **Analyze $g \circ f$:** - Domain of $f$ is $[6, \infty)$. - $g$ is defined on $(-\infty, 9]$. - For $g(f(x))$ to exist, we need $f(x) \in (-\infty, 9]$ for all $x \in [6, \infty)$. 7. **Find range condition for $f(x)$:** - $f(x) = b - 3\sqrt[3]{x}$. - Cube root function $\sqrt[3]{x}$ is increasing. - For $x \geq 6$, minimum of $\sqrt[3]{x}$ is at $x=6$: $$\sqrt[3]{6}$$ - As $x \to \infty$, $\sqrt[3]{x} \to \infty$. - So $f(x)$ decreases without bound as $x$ increases. - Maximum value of $f(x)$ is at $x=6$: $$f(6) = b - 3\sqrt[3]{6}$$ 8. **Condition for $g(f(x))$ to exist:** - We need $f(x) \leq 9$ for all $x \geq 6$. - Since $f(x)$ decreases as $x$ increases, the maximum is $f(6)$. - So, $$f(6) = b - 3\sqrt[3]{6} \leq 9$$ - Rearranged: $$b \leq 9 + 3\sqrt[3]{6}$$ 9. **Combine both conditions:** - From $f \circ g$ existence: $b \geq 6$ - From $g \circ f$ existence: $b \leq 9 + 3\sqrt[3]{6}$ 10. **Final answer:** $$\boxed{6 \leq b \leq 9 + 3\sqrt[3]{6}}$$ This is the range of $b$ values for which both compositions exist.