Section 1.3 Functions and Their Graphs 19

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1 y y Section 1.3 Functions and Their Graphs 19 3, Ÿ 1, y œ 26. y œ œ 2, 1 œ, 0 Ÿ " 27. (a) Line through a!ß! band a"ß " b: y œ Line through a"ß " band aß! b: y œ 2, 0 Ÿ Ÿ 1 f() œ œ 2, 1 Ÿ 2 Ú Ý 2,!Ÿ "!ß " Ÿ (b) f() œ Û Ý 2ß Ÿ $ Ü!ß $ Ÿ Ÿ % 28. (a) Line through a!ß 2 band aß! b: y œ 2 Line through a2 ß " band a&ß! b: m œ œ œ, so y œ a 2b " œ f() œ, 0 Ÿ, Ÿ & œ " & $ $! " " " " " & & $ $ $ $ $ $!! Ð "Ñ " $ %! (b) Line through a "ß! band a!ß $ b: m œ œ $, so y œ $ $ Line through a!ß $ band aß " b: m œ œ œ, so y œ $ $ $, " Ÿ! f() œ œ $,! Ÿ 29. (a) Line through a "ß " band a!ß! b: y œ Line through a!ß " band a"ß " b: y œ " Line through a"ß" band a$ß! b: m œ œ œ, so y œ a " b "œ Ú " Ÿ! f() œ Û "! Ÿ " Ü " $ (b) Line through a ß " band a!ß! b: y œ " Line through a!ß band a"ß! b: y œ Line through a"ß " band a$ß " b: y œ "! " " " " " $ $ "

2 20 Chapter 1 Preliminaries Ú " Ÿ Ÿ! f() œ Û! Ÿ " Ü " " Ÿ $ 30. (a) Line through ˆ T "! and at b: m, so y ˆ T ß! ß" œ œ œ 0 œ "!, 0 Ÿ Ÿ f() œ T T ", ŸT Ú T A,!Ÿ Ý T Aß Ÿ T (b) f() œ Û $ T Aß T Ÿ Ý $ T Ü Aß Ÿ Ÿ T T atî b T T T ( 4)( 2) (a) From the graph, 1 Ê ( 2ß0) (%ß_) (b) : 1 0 Ê 0 Ê 0 Ê 4 since is positive; ( 4)( 2) 0: Ê 2 0 Ê 0 Ê 2 since is negative; sign of ( 4)( 2) ïïïïïðïïïïïðïïïïî 2 % Solution interval: ( ß 0) (%ß _) T ( 1) (a) From the graph, Ê ( _ß 5) ( 1ß1) (b) Case 1: Ê 2 Ê Ê 5. Thus, ( _ß 5) solves the inequality ( 1) Case 1 1: Ê 2 Ê Ê 5 which is true if 1. Thus, ( 1ß1) solves the inequality. 3 2 Case 1 : Ê Ê which is never true if 1, so no solution here. In conclusion, ( _ß 5) ( 1ß1). 33. (a) ÚÛœ 0 for [0ß1) (b) ÜÝœ 0 for ß ( 1 0] 34. ÚÛœÜÝ only when is an integer. 35. For any real number, n Ÿ Ÿn ", where n is an integer. Now: n Ÿ Ÿn "Ê Ðn "ÑŸ Ÿ n. By definition: Ü Ýœ n and ÚÛœnÊ ÚÛœ n. So Ü Ýœ Ú Ûfor all d.

3 Section 1.3 Functions and Their Graphs To find f() you delete the decimal or fractional portion of, leaving only the integer part. $ 37. v œ f() œ Ð"% 2ÑÐ22 2Ñ œ % 72 $!) ;! 7Þ È 38. (a) Let h œ height of the triangle. Since the triangle is isosceles, AB AB œ 2 Ê AB œ 2ÞSo, h " œš È2 Êh œ"ê B is at a!ß" b Êslope of AB œ "ÊThe equation of AB is y œ f() œ B "; Ò!ß "Ó. (b) AÐÑœ2 y œ2ð "Ñœ 2 ; Ò!ß"Ó. 39. (a) Because the circumference of the original circle was )1 and a piece of length was removed. ) 1 (b) r œ œ % "' 1 È "' % 1 1 % 1 % 1 % 1 1 È "' 1 a ) 1 È "' 1 $ $ 1 1 % 1 (c) h œ È"' r œ É"' ˆ % œ É"' ˆ 16 œ É œ É œ (d) V œ " r h œ " ) 1 1 1ˆ œ 40. (a) Note that 2 mi = 10,560 ft, so there are È)!! feet of river cable at $180 per foot and a"!ß &'! bfeet of land cable at $100 per foot. The cost is Cab œ ")! È)!! "!! a"!ß &'! b. (b) C a! b œ $ "ß!!ß!!! C a&!! b $ "ß "(&ß )" C a"!!! b $ "ß ")'ß &" C a"&!! b $ "ß "ß!!! C a!!! b $ "ß %$ß ($ C a&!! b $ "ß ()ß %(* C a$!!! b $ "ß $"%ß )(! Values beyond this are all larger. It would appear that the least epensive location is less than 2000 feet from the point P. 41. A curve symmetric about the -ais will not pass the vertical line test because the points a, y band a, y blie on the same vertical line. The graph of the function y œ fa b œ! is the -ais, a horizontal line for which there is a single y-value,!, for any. 42. Pick 11, for eample: "" &œ"'ä "'œ$ä$ 'œ'ä œ"$ä"$ œ"", the original number. fab œ a & b ' œ, the number you started with. '

4 22 Chapter 1 Preliminaries 1.4 IDENTIFYING FUNCTIONS; MATHEMATICAL MODELS 1. (a) linear, polynomial of degree 1, algebraic. (b) power, algebraic. (c) rational, algebraic. (d) eponential. 2. (a) polynomial of degree 4, algebraic. (b) eponential. (c) algebraic. (d) power, algebraic. 3. (a) rational, algebraic. (b) algebraic. (c) trigonometric. (d) logarithmic. 4. (a) logarithmic. (b) algebraic. (c) eponential. (d) trigonometric. 5. (a) Graph h because it is an even function and rises less rapidly than does Graph g. (b) Graph f because it is an odd function. (c) Graph g because it is an even function and rises more rapidly than does Graph h. 6. (a) Graph f because it is linear. (b) Graph g because it contains a!ß " b. (c) Graph h because it is a nonlinear odd function. 7. Symmetric about the origin 8. Symmetric about the y-ais Dec: _ _ Dec: _! Inc: nowhere Inc:! _ 9. Symmetric about the origin 10. Symmetric about the y-ais Dec: nowhere Dec:! _ Inc: _! Inc: _!! _

5 Section 1.4 Identifying Functions; Mathematical Models Symmetric about the y-ais 12. No symmetry Dec: _ Ÿ! Dec: _ Ÿ! Inc:! _ Inc: nowhere 13. Symmetric about the origin 14. No symmetry Dec: nowhere Dec:!Ÿ _ Inc: _ _ Inc: nowhere 15. No symmetry 16. No symmetry Dec:!Ÿ _ Dec: _ Ÿ! Inc: nowhere Inc: nowhere

6 24 Chapter 1 Preliminaries 17. Symmetric about the y-ais 18. Symmetric about the y-ais Dec: _ Ÿ! Dec:! Ÿ _ Inc:! _ Inc: _! 19. Since a horizontal line not through the origin is symmetric with respect to the y-ais, but not with respect to the origin, the function is even. & " & " 20. fab and fa b a b ˆ " œ œ œ œ œ œ fa b. Thus the function is odd. & a b 21. Since fab œ "œa b "œ fa b. The function is even. & & 22. Since Òfab œ ÓÁÒfa b œa b Óand Òfab œ ÓÁÒ fab œ ab Óthe function is neither even nor odd. $ $ $ 23. Since gab œ, ga b œ œ a b œ ga b. So the function is odd. % % 24. gab œ $ "œa b $ B a b "œga bßthus the function is even. " " " a b " 25. gab œ œ œ ga b. Thus the function is even. " " 26. ga b œ ; ga b œ œ ga b. So the function is odd. " " " t " t " " t 27. hab t œ ; ha t bœ ; hab t œ. Since hab t Á hab t and hab t Á ha t b, the function is neither even nor odd. $ $ 28. Since lt œ la t b, hatb œ ha t band the function is even. 29. hab t œ2t ", ha tb œ 2t ". So hab t Áha t b. hab t œ 2t ", so hab t Á hab t. The function is neither even nor odd. 30. hab t œ2lt l " and ha tb œ2l t l "œ2lt l ". So hab t œha t band the function is even. 31. (a) The graph support the assumption that y is proportional to. = The constant of proportionality is estimated from the slope of the regression line, which is

7 Section 1.4 Identifying Functions; Mathematical Models 25 (b) "Î The graph support = the assumption that y is proportional to. The constant of proportionality is estimated from the slope of the regression line, which is (a) Because of the wide range of values of the data, two graphs are needed to observe all of the points in relation to the regression line. The graphs support the assumption that y is proportional to $. The constant of proportionality is estimated from the slope of the regression line, which is (b) The graph supports the assumption that y is proportional to ln. The constant of proportionality is etimated from the slope of the regression line, which is (a) The scatterplot of y œ reaction distance versus œ speed is Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is approimately 1.1.

8 26 Chapter 1 Preliminaries w (b) Calculate œ speed squared. The scatterplot of versus y œ braking distance is: w Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is approimately $Î * 34. Kepler's 3rd Law is Tadaysb œ!þ%" R, R in millions of miles. "Quaoar" is 4 "! miles from Earth, or about * ' * 4 "! *$ "! % "! miles from the sun. Let R œ 4000 (millions of miles) and 35. (a) T œ a!þ%" ba%!!! b $Î days "!$ß ($ days. The hypothesis is reasonable. )Þ(%"! (b) The constant of proportionality is the slope of the line "!! in./unit mass œ!þ)(% in./unit mass. (c) y(in.) œ a!þ)( in./unit massba"$ unit mass b œ ""Þ$" in. 36. (a) (b) Graph (b) suggests that y œ k $ is the better model. This graph is more linear than is graph (a). 1.5 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS 1. D : _ _, D : 1 Ê D œ D : 1. R : _ y _, R : y 0, R : y 1, R : y 0 f g fbg fg f g fbg fg 2. D f : 1 0 Ê 1, D g: 1 0 Ê 1. Therefore Dfbg œ D fg: 1. R R : y 0, R : y È2, R : y 0 f œ g fbg fg

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