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chapter 1 55

chapter 1 55 - CHAPTEM REVIEW B 55 14 y = 2 ﬁ ﬁ Start...

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Unformatted text preview: CHAPTEM REVIEW B 55 14. y = 2 ﬁ ﬁ: Start with the graph of y = (/5 reﬂect about the T—axis. and shift 2 units upward. 15. f(.’L‘) = \$ + 2: Start with the graph of f(.r) : l/T and shift 2 units to the left. —:r if .17 < 0 16' HI): { 6171 ifTZO On (730. 0). graph 3/ : ~T (the line with slope —1 and yiintercept 0) with open endpoint (0, 0). On [0. oo). graph y : e“: — 1 (the graph ofy = e“E shifted 1 unit downward) with closed endpoint (0., 0). 17. (a) The terms of f are a mixture of odd and even powers of T so f is neither even nor odd, (b) The terms of f are all odd powers of .17. so f is odd. (C) f(~.1:) : 6*”): : 6‘9”2 : ﬂat) so f is even. (d) f(~.1:) : 1+ Sin(~;c):17 sin 1 Now f(~:1:) 7E f(\$) and f(~T) gé aﬂr). so f is neither even not odd, 18. For the line segment from (—2. 2) to (~1, 0). the slope is 0‘2 ~- 2 d t" 0» 2~1 _1+2ﬂ .an aanLlZlIOITISJ/i —~ (1+) or. equivalently. y : —2.z? — 2. The circle has equation 352 + y2 : 1: the top half has equation y : \/1 ~ 1:2 —2T—2 if —2§T§—1 \/1~T2if—1<:L‘§1 (we have solved for positive y.) Thus. ﬂat) : { 19. f(;L) : 1111'. D : (0. 00): 9(35) : J72 — 9. D : R. (f 09%)?) : f(9(17)) : f(;1:2 - 9) :1n(:1?2 e 9) Domain: 12 — 9 > 0 i T2 > 9 :> [Tl > 3 :> :1: E (—30. ~3) U (3.90) (go f)(x) : g(f(.r)) : 9(11127) : (111T)2 — 9. Domain: .T > 0. or (0. 00) (f o f)(’lt) : f(f(;1:)) : f(ln:r) :ln(ln:1:). Domain: 111:1: > 0 :> .r > 80 : 1. or (1. 30) (g og)(T) : g(g(;r)) : g(.T2 ~ 9) : (x2 ~ (3)2 e 9, Domain: .I (E R. or (*00. 30) 20. Let M17) : .1: + ﬂ. 9(1‘) : ﬂ. and f(.27) : 1/1'. Then (f Cg 0 MM?) : ff : F(.T) :1: a: ...
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