1. We are given the function $$h(x) = -2 \cdot \ln\left(-3\sqrt{x} + 4\right) + 1.$$ We need to determine its domain and identify the inside and outside functions if $$h(x) = g(f(x)).$$ 2. **Domain**: The argument inside the natural logarithm must be positive, so $$-3\sqrt{x} + 4 > 0.$$ 3. Solve for $x$: $$-3\sqrt{x} + 4 > 0 \implies -3\sqrt{x} > -4 \implies \sqrt{x} < \frac{4}{3}$$ because multiplying by -1 reverses the inequality. 4. Since $$\sqrt{x} < \frac{4}{3}$$ and $$\sqrt{x} \ge 0$$ for real $x,$ square both sides: $$x < \left(\frac{4}{3}\right)^2 = \frac{16}{9}.$$ Also, $x \ge 0$ due to the square root. 5. Thus, the domain of $h$ is: $$D_h = \left[0, \frac{16}{9}\right).$$ There is no second interval, so the union with $(\text{none}, \text{none})$ means no additional domain intervals. 6. **Find inside and outside functions:** We can write $h(x)$ as a composite function $g(f(x))$. 7. The inner function is: $$f(x) = -3\sqrt{x} + 4.$$ 8. The outer function is: $$g(t) = -2 \ln(t) + 1,$$ where $$t = f(x).$$ **Final answers:** - Domain: $$\left[0, \frac{16}{9}\right) \cup (\text{none}, \text{none})$$ - Inside function: $$f(x) = -3\sqrt{x} + 4$$ - Outside function: $$g(x) = -2 \ln(x) + 1$$