1. The function given is $$h(x) = 4\sqrt{-5x^2 + 2x + 1}$$. We need to find the domain where the expression inside the square root is non-negative because the square root of a negative number is undefined in the real numbers. 2. Set the inside expression $$-5x^2 + 2x + 1 \geq 0$$. 3. Multiply both sides by -1 to make it easier to factor and flip the inequality sign: $$5x^2 - 2x - 1 \leq 0$$. 4. Factor the quadratic: $$5x^2 - 2x -1 = (5x + 1)(x - 1)$$. 5. The roots are $$x = -\frac{1}{5}$$ and $$x = 1$$. Since the parabola opens upwards (leading coefficient positive), the quadratic is less than or equal to zero between the roots. 6. Therefore, the domain of $$h$$ is $$\left[-\frac{1}{5}, 1\right]$$. 7. In interval notation, $$D_h = \left[-\frac{1}{5}, 1\right]$$ (no union needed since it's one continuous interval). 8. Next, since $$h(x) = g(f(x))$$, the inside function is the expression inside the square root, and the outside function is the remaining operation: - Inside function: $$f(x) = -5x^2 + 2x + 1$$ - Outside function: $$g(x) = 4\sqrt{x}$$ Summary: - Domain: $$D_h = \left[-\frac{1}{5}, 1\right]$$ - Inside function: $$f(x) = -5x^2 + 2x + 1$$ - Outside function: $$g(x) = 4\sqrt{x}$$