1. **Problem:** Determine the domain and range of the function $$f$$ defined piecewise as: $$f(x) = \begin{cases} x^3, & x < -1 \\ x^3, & -1 \leq x \leq 1 \\ 2x, & x > 1 \end{cases}$$ 2. **Domain:** The domain is all real numbers because each piece is defined for its respective interval and together they cover all real numbers. 3. **Range:** - For $$x < -1$$, $$f(x) = x^3$$. As $$x \to -\infty$$, $$x^3 \to -\infty$$, and at $$x = -1$$, $$f(-1) = (-1)^3 = -1$$. - For $$-1 \leq x \leq 1$$, $$f(x) = x^3$$. Since $$x^3$$ is continuous and increasing, $$f(-1) = -1$$ and $$f(1) = 1$$. - For $$x > 1$$, $$f(x) = 2x$$. At $$x=1$$, $$f(1) = 2$$ (but from the middle piece, $$f(1) = 1$$), so there is a jump discontinuity at $$x=1$$. 4. **Range calculation:** - From $$x < -1$$, range is $$(-\infty, -1)$$. - From $$-1 \leq x \leq 1$$, range is $$[-1, 1]$$. - From $$x > 1$$, range is $$(2, \infty)$$. 5. **Final range:** $$(-\infty, -1) \cup [-1, 1] \cup (2, \infty)$$. --- **Summary:** - Domain: $$(-\infty, \infty)$$ - Range: $$(-\infty, -1) \cup [-1, 1] \cup (2, \infty)$$