271 2.6 Graphs of Basic Functions Piecewise-Defined Functions The absolute value function is a piecewisedefined function. It is defined by different rules over different intervals of its domain. Absolute Value Function ƒ1x2 = ∣ x∣ Domain: 1-∞, ∞2 Range: 30, ∞2 x y -2 2 -1 1 0 0 1 1 2 2 • ƒ1x2 = 0 x 0 decreases on the open interval 1-∞, 02 and increases on the open interval 10, ∞2. • It is continuous on its entire domain, 1-∞, ∞2. x y 0 2 1 2 3 4 f(x) = zxz −10 −10 10 10 f(x) = ) x) Figure 61 EXAMPLE 2 Graphing Piecewise-Defined Functions Graph each function. (a) ƒ1x2 = e -2x + 5 x + 1 if x … 2 if x 72 (b) ƒ1x2 = e 2x + 3 -x2 + 3 if x … 0 if x 70 ALGEBRAIC SOLUTION (a) We graph each interval of the domain separately. If x … 2, the graph of ƒ1x2 = -2x + 5 has an endpoint at x = 2. We find the corresponding y-value by substituting 2 for x in -2x + 5 to obtain y = 1. To find another point on this part of the graph, we choose x = 0, so y = 5. We draw the graph through 12, 12 and 10, 52 as a partial line with endpoint 12, 12. We graph the function for x 72 similarly, using ƒ1x2 = x + 1. This partial line has an open endpoint at 12, 32. We use y = x + 1 to find another point with x-value greater than 2 to complete the graph. See Figure 62. (2, 3) (2, 1) 3 5 1 2 4 –2 6 0 x y x + 1 if x + 2 –2x + 5 if x " 2 f(x) = Figure 62 GRAPHING CALCULATOR SOLUTION (a) We use the TEST feature of the TI-84 Plus to graph the piecewise-defined function. (Press 2ND MATH to display a list containing inequality symbols.) The result of a true statement is 1, and the result of a false statement is 0. We choose x with the appropriate inequality based on how the function is defined. Next we multiply each defining expression by the test condition result. We then add these products to obtain the complete function. The expression for the function in part (a) is shown at the top of the screen in Figure 63. −2 −10 10 10 Figure 63
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