1. **State the problem:** We need to analyze and understand the function $$f(x) = \sin^3\left(\cos\sqrt{x^3 + 2x^2}\right)$$. 2. **Understand the function:** The function is a composition of several functions: - Inside the square root: $$x^3 + 2x^2$$ - Then take the square root: $$\sqrt{x^3 + 2x^2}$$ - Then take the cosine of that: $$\cos\sqrt{x^3 + 2x^2}$$ - Finally, cube the sine of that: $$\sin^3(\cdots) = \left(\sin(\cdots)\right)^3$$ 3. **Domain considerations:** The expression inside the square root must be non-negative: $$x^3 + 2x^2 \geq 0$$ Factor: $$x^2(x + 2) \geq 0$$ Since $$x^2 \geq 0$$ always, the inequality depends on $$x + 2 \geq 0$$, so: $$x \geq -2$$ 4. **Summary of domain:** $$x \geq -2$$ 5. **Behavior and range:** - The cosine function outputs values in $$[-1,1]$$. - The sine function also outputs values in $$[-1,1]$$. - Cubing the sine preserves the sign and compresses values between -1 and 1. 6. **Final function:** $$f(x) = \left(\sin\left(\cos\sqrt{x^3 + 2x^2}\right)\right)^3$$ This function is continuous and defined for $$x \geq -2$$. **Final answer:** The function is defined for $$x \geq -2$$ and is given by $$f(x) = \sin^3\left(\cos\sqrt{x^3 + 2x^2}\right)$$.