The function \( f \) is defined as follows. \( f(x)=\left\{\begin{array}{l}|2 x| \text { if }-3 \leq x<0 \\ x^{3} \quad \text { if } x \geq 0\end{array}\right. \) (a) Find the domain of the function. (b) Locate any intercepts. (c) Graph the function. (d) Based on the graph, find the range. (a) The domain of the function \( f \) is [-3, \( \infty \) ). (Type your answer in interval notation.) (b) Locate any intercepts. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The intercept(s) is/are (Type an ordered pair. Use a comma to separate answers as needed.) B. There are no intercepts.

We are given the function   f(x) = { |2x|  if –3 ≤ x < 0        x³   if x ≥ 0 } Below are the answers to the parts: (a) Domain: Since the first piece is defined for –3 ≤ x < 0 and the second piece for x ≥ 0, the overall domain is all x from –3 to ∞. In interval notation, the domain is:   [–3, ∞) (b) Intercepts: • y-intercept: Set x = 0.   For x = 0, we use the branch for x ≥ 0: f(0) = 0³ = 0.   Thus, the y-intercept is (0, 0). • x-intercept: Solve f(x) = 0.   For x ≥ 0: x³ = 0 ⟹ x = 0.   For –3 ≤ x < 0: |2x| = 0 ⟹ 2x = 0 ⟹ x = 0, but note that x = 0 is not in the interval [–3, 0), so it doesn’t come from here.   Thus, the only x-intercept is again (0, 0). Thus, the correct choice is:  A. The intercept(s) is/are (0, 0). (c) Graph of the function: 1. For –3 ≤ x < 0: f(x) = |2x|.   Since x is negative in this interval, |2x| simplifies to –2x.   At x = –3, f(–3) = –2(–3) = 6. So the point (–3, 6) is on the graph.   As x increases toward 0, f(x) decreases linearly. The endpoint at x = 0 is not included in this branch (open circle if drawn) because the inequality is x < 0. 2. For x ≥ 0: f(x) = x³.   At x = 0, f(0) = 0³ = 0. This gives the point (0, 0). Note that (0, 0) is included (filled dot).   For x > 0, the cubic function increases, so the graph moves upward and to the right. When you sketch the graph:  – Plot the line segment from (–3, 6) to (0, 0). Use a closed dot at (–3, 6) and an open circle at (0, 0) for this piece.  – From (0, 0) onward, draw the cubic curve, which passes through (0, 0) (with a filled dot) and rises for x > 0. (d) Range: 1. For –3 ≤ x < 0 (first branch): f(x) = –2x.   As x increases from –3 to 0, –2x decreases from 6 to values approaching 0 (but these values are positive). Thus, this piece covers values from just above 0 up to 6 (with 6 included). 2. For x ≥ 0 (second branch): f(x) = x³ covers all values from 0 to ∞ (since x³ is 0 at x = 0 and increases without bound as x increases). Therefore, combining both pieces, the range is:   [0, ∞) Summary of Answers: (a) Domain: [–3, ∞) (b) Intercept: (0, 0) (c) See description above for the detailed sketch construction. (d) Range: [0, ∞)