The function \( f \) is defined as follows. \( f(x)=\left\{\begin{array}{ll}-2 x+3 & \text { if } x<1 \\ 2 x-1 & \text { if } x \geq 1\end{array}\right. \) (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 \( (-\infty, \infty) \). (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 piecewise function   f(x) = { –2x + 3 if x < 1        2x – 1 if x ≥ 1 } Let’s work through each part. ───────────────────────────── (a) Domain Both pieces of the function have formulas defined for all x-values in their respective regions. Since every real number falls into one of these intervals, the domain of f is   (–∞, ∞). ───────────────────────────── (b) Intercepts Intercepts include any x-intercepts (where f(x) = 0) and the y-intercept (where x = 0). 1. Y-intercept:  Set x = 0. Since 0 < 1, we use the first piece.   f(0) = –2(0) + 3 = 3.  Thus, the y-intercept is (0, 3). 2. X-intercepts:  We need to solve f(x) = 0 in each region.  • For x < 1: Solve –2x + 3 = 0   –2x = –3 ⇒ x = 3/2 = 1.5.   But 1.5 is not less than 1, so this solution does not belong to the first piece.  • For x ≥ 1: Solve 2x – 1 = 0   2x = 1 ⇒ x = 1/2 = 0.5.   However, 0.5 is not in the domain x ≥ 1.  Since neither branch provides an acceptable x-value, there are no x-intercepts. Thus, the only intercept is the y-intercept (0, 3). ───────────────────────────── (c) Graph of the Function 1. For x < 1, the function f(x) = –2x + 3 is a straight line with a slope of –2 and a y-intercept of 3.  • When x is very small (large negative), f(x) is large positive.  • As x approaches 1 from the left, f(1⁻) = –2(1) + 3 = 1.  • Note: The point (1, 1) is not included in this piece (open circle at (1, 1)). 2. For x ≥ 1, the function f(x) = 2x – 1 is also a straight line, with a slope of 2.  • At x = 1, f(1) = 2(1) – 1 = 1, so the graph includes the point (1, 1) (closed dot).  • As x increases, f(x) increases linearly. The overall graph consists of two ray segments that meet at (1, 1); the left piece approaches (1, 1) with an open circle, while the right piece starts at (1, 1) with a closed circle. ───────────────────────────── (d) Range To find the range, notice the following:  • For x ≥ 1: f(x) = 2x – 1 is an increasing line. Its minimum value is at x = 1, giving f(1) = 1, and f(x) increases without bound as x increases.  • For x < 1: f(x) = –2x + 3 is a decreasing function as x increases toward 1. As x approaches 1 from the left, f(x) approaches 1 from above. For x very negative, f(x) becomes very large.  Thus, the smallest value that f actually takes is f(1) = 1, and all values greater than or equal to 1 are attained. Therefore, the range is   [1, ∞). ───────────────────────────── Summary Answers (a) Domain: (–∞, ∞) (b) Intercepts:  • Y-intercept: (0, 3)  • No x-intercepts (c) Graph: Two line segments—one for x < 1 given by y = –2x + 3 (with an open circle at (1, 1)) and one for x ≥ 1 given by y = 2x – 1 (starting with a closed dot at (1, 1)). (d) Range: [1, ∞)