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Published bySamuel Scott Modified over 3 years ago

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Piecewise-defined Functions ½ x – 2, x > 2 f(x) =f(x) =3, x = 1 –2x + 3, –2 x < 1 Ex. 1: x 2 4 6... y –1 0 1... h/d h d d... x –2 –1 0 1 y 7 5 3 1 h/d d d d h y x O x – 1, x < –3 f(x) =f(x) = x – 3, x = 4 –2 / 3 x + 2, –3 x < 3 Ex. 2: x –3 –4 –5... y –4 –5 –6... h/d h d d... x –3 0 3 y 4 2 0 h/d d d h y x O Domain: (–, 3) {4}Range: (–, –4) (0, 4] Evaluate: a) f(–1) = 2 2 / 3 b) f(3) = undefinedc) f(4) = 1 Domain: [–2, 1] (2, ) Range: (–1, )Evaluate: a) f(0) = 3b) f(1) = 3c) f(2) = undefined (1, 3) (4, 1)

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Piecewise-defined Functions (contd) –3 / 2 x + 4, x > 2 f(x) =f(x) = 2x – 1, –3 < x 2 Ex. 3: x 2 4 6... y 1 –2 –5... h/d h d d... x –3 –2 0 1 2 y –7 –5 –1 1 3 h/d h d d d d y x O |x|, –3 x < 1 f(x) =f(x) = x – 5, x = 3 3, 1 x < 2 Ex. 4: x –3 –2 –1 0 1 y 3 2 1 0 1 h/d d d d d h x 1 2 y 3 3 h/d d h y x O Evaluate: a) f(–2) = –5 b) f(2) = 3c) f(4) = –2Domain: (–3, )Range: (–, 3] Domain: [–3, 2) {3} Range: _{–2} [0, 3]_ Evaluate: a) f(–3) = 3 b) f(1.5) = 3c) f(3) = –2 (3, –2)

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Piecewise-defined Functions (contd) x, x > 1 f(x) =f(x) = –2x – 7, x –2 Ex. 5: x 1 4 9 … y 1 2 3 … h/d h d d … x –2 –1 0 1 y 4 1 0 1 h/d h d d d Domain: (–, )Range: [–3, )Evaluate: a) f(4) = 2 b) f(1) = 1 c) f(–3) = –1 f(x) =f(x) = Ex. 6: y x O x 2, –2 < x 1 x –2 –3 –4 … y –3 –1 1 … h/d d d d … Domain: (–, ) Range: (–, –1) {1} [2, )Increasing: (–, –3) (1, ) Decreasing: None ½x + 3/2,x 1 1,–3 x < 1 x + 2, x < –3 y x O

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Piecewise-defined Functions (contd) f(x) =f(x) = Ex. 7: y x O f(x) =f(x) = Ex. 8: y x O Domain: (–, )Range: (–, ) Increasing: (–, –2) (0, 3) Decreasing: (–2, 0) (3, ) Domain: (–, –1) (–1, )Range: (–, –1) (0, ) Increasing: (0, 2) Decreasing: (–, –1) (–1, 0) (2, ) –½x, x (2, ) x 2, x (–1, 2] –2x – 1, x (–, –1) is in 2x – 6,x [3, ) |x|, x [–2, 3) x + 4, x (–, –2) OR f(x) = 2x – 6,x [3, ) |x|, x (–2, 3) x + 4, x (–, –2]

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