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chapter 1 52

# chapter 1 52 - 52 Cl CHAPTER1 FUNCTIONS AND MODELS 1 False...

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Unformatted text preview: 52 Cl CHAPTER1 FUNCTIONS AND MODELS 1. False. 2. False. 3. False. 4. True. 5. True. 6. False. 7. False. 8. True. 9. True. 10. False. 11. False. TRUE-FALSE QUIZ ): 232. s = —1.andt = 1. Then f(s+t) 2 (—1 +1)2 2 02 z 0. but f(s) + f(1t)=(-1)2 +12 = 2 aé 0 = f(3 + t)- ) ) If 331 < x2 and f is a decreasing function, then the y—values get smaller as we move from left to right. Thus. ﬂan) > f(a:2). See the Vertical Line Test. Let f(m) : m2 and 9(36) 2 2x. Then (f o g)(:p) : f(g(:n)) : f(2a:) : (2202 : 4m2 and (9 O f)(\$) = g(f(:r)) : 9(132) = 2:32. 50 f 0 g 75 g o f Let f(;v) : \$3. Then f is one—to-one and f_1(m) : {Vi But 1/f(ac) : 1/5133. which is not equal to f‘1(\$)‘ We can divide by ex since em 75 0 for every 3:. The function 1113: is an increasing function on (0. oo). Letxze. Then (lnm)6 : (lne)6 : 16 : 1.but6lnm:61ne=6-1 267\$ 1 = (lnm)6. 1n :1: ln 62 2 111 e x 62 . Letzc = e2 anda = 6. Then —— = —— = : 2andln— : ln— 2 Inc : 1. so in general the In a ln 6 1n 6 a 6 statement rs false. What rs true. however. Is that In — = 11130 — In a. a _————————’————— EXERCISES __—————————-——— 1. (a) When m : 2. y m 2.7. Thus. f(2) % 2.7. (b) f(:v) : 3 2 :v z 2.3. 5.6 (c) The domain of f is —6 S as g 6. or [—6. 6]. (d) The range of f is —4 g y S 4. or [—4.4]. (e) f is increasing on [—47 4]. that is. on —4 g m g 4. (f) f is not one—to—one since it fails the Horizontal Line Test. (g) f is odd since its graph is symmetric about the origin. 2. (a) When an : 2. y : 3. Thus. g(2) : 3. (e) We reﬂect the graph of g through the line y = x to (b) g is one—to-one because it passes the obtain the graph of g‘l. Horizontal Line Test. (0) When 31 : 2. :1: 5d 0.2. So 94(2) 3 0.2. (d) The range of g is [—1.36]. which is the same as the domain of ng. ...
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