Chapter 1 Functions and Graphs

Transcription

1 Section. Modeling and Equation Solving 7 Chapter Functions and Graphs Section. Modeling and Equation Solving Eploration. k = d 00 - = = 7 m = 0.7. t=6.%+0.%=7% or m = d s = d + td k, s = pm p = s m = d + td d k = d + td k + t = = = Yes, because $6.99 *0.80=$ $ =$4.6 Eploration. Because the linear model maintains a constant positive slope, it will eventuall reach the point where 00% of the prisoners are female. It will then continue to rise, giving percentages above 00%, which are impossible.. Yes, because 009 is still close to the data we are modeling. We would have much less confidence in the linear model for predicting the percentage ears from One possible answer: Males are heavil dominant in violent crime statistics, while female crimes tend to be propert crimes like burglar or shoplifting. Since propert crime rates are senstive to economic conditions, a statistician might look for adverse economic factors in 990, especiall those that would affect people near or below the povert level. 4. Yes. Table. shows that the minimum wage worker had less purchasing power in 990 than in an other ear since 9, which gives some evidence of adverse economic conditions among lower-income Americans that ear. Nonetheless, a careful sociologist would certainl want to look at other data before claiming a connection between this statistic and the female crime rate. Quick Review.. (+4)(-4). (9+)(9-). (4h +9) 4h - 9=(4h +9)(h+)(h-) 7. (+4)(-) 9. (-)(-) # k d = d + t # k d Section. Eercises. (d) (q). (a) (p). (e) (l) 7. (g) (t) 9. (i) (m). (a) The percentage increased from 94 to 999 and then decreased slightl from 999 to 004. (b) The greatest increase occurred between 974 and Women ( ), Men + [, ] b [, 9]. To find the equation, first find the slope. change in Women: Slope= = change in = 6. 4 = 0.8. The -intercept is., so the equation of the line is = Men: Slope = The = -9. = intercept is 8., so the equation of the line is = In both cases, represents the number of ears after For the percentages to be the same, we need to set the two equations equal to each other = =. L 64.6 So, approimatel 6 ears after 94 (08), the models predict that the percentages will be about the same. To check: Males: = L 69.9% Females: = L 69.9% 9.. Because all stepping stones have the same thickness, what matters is area. The area of a square stepping stone is A in. = # = 44 Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

2 8 Chapter Functions and Graphs The area of a round stepping stone is A = pa.4(6.) =.66 in. b L The square stones give a greater amount of rock for the same price.. A scatter plot of the data suggests a parabola with its verte at the origin. [, 6] b [, ] The model =.t fits the data.. The lower line shows the minimum salaries, since the are lower than the average salaries. 7. The 99 points are third from the right, Year, on both graphs. There is a clear drop in the average salar right after the 994 strike. 9. Adding v + to both sides gives v =. Divide both sides b to get v = so v=., A v = is equivalent to v -=0. The graph of =v - is zero for v.08 and for v.08. [ 0, 0] b [, ]. +7-4=0,so a=, b=7, and c= 4: = = 7 = _ 0 The graph of = +7-4 is zero for 8.6 and for.6. [ 0, 0] b [ 0, 0] 7. Change to --=0 (see below); this factors to (+)(-)=0, so = or =. Substituting the first of these shows that it is etraneous. += + 4 (+) = =4+6 --=0 The graph of = is zero for =. [, ] b [, ]. -+= =0 (-4)(+)=0-4=0 or +=0 =4 or = -+=(-)(-)+ is equivalent to -8+-(-)(-)=0. The graph of = -8+-(-)(-) is zero for = and for = [ 0, 0] b [ 0, 0] [ 0, 0] b [ 0, 0] [ 0, 0] b [ 0, 0]. Rewrite as --=0; the left side factors to (+)(-4)=0: +=0 or -4=0 = =4 =.. The graph of = -- is zero for =. and for = or =4 [ 0, 0] b [ 0, 0] Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

3 Section. Modeling and Equation Solving [, ] b [ 0, 0] [ 4, 4] b [ 0, 0] 47. Model the situation using C=0.8+, where is the number of miles driven and C is the cost of a da s rental. (a) Elaine s cost is 0.8(8)+=$ (b) If for Ramon C=$69.80, then = =0 miles (a) = 00 >00 = 00>00 = = for all Ú 0. (b) The graph looks like this: [0, ] b [0, ] (c) Yes, this is different from the graph of =. (d) For values of close to 0, 00 is so small that the calculator is unable to distinguish it from zero. It returns a value of 0 /00 =0 rather than.. (a) = or =. or =.. The product of two odd integers is odd. The product of two even integers is a multiple of 4, since each even integer in the product contributes a factor of to the product. Therefore, n +n is either odd or a multiple of 4.. False. A product is zero if an factor is zero. That is, it takes onl one zero factor to make the product zero. 7. This is a line with a negative slope and a -intercept of. The answer is C. (The graph checks.) 9. As increases b ones, the -values get farther and farther apart, which implies an increasing slope and suggests a quadratic equation. The answer is B. (The equation checks.) 6. (a) March (b) $0 (c) June, after three months of poor performance (d) Ahmad paid (00)($0)=$,000 for the stock and sold it for (00)($00)=$0,000. He lost $,000 on the stock. (e) After reaching a low in June, the stock climbed back to a price near $40 b December. LaToa s shares had gained $000 b that point. (f) One possible graph: 6. (a) Jan. Feb. Mar. Apr. Ma June Jul Aug. Sept. Oct. Nov. Dec. 0 Stock Inde Subscribers Monthl Bills [, ] b [ 00, 00] (b) = onl. [7, 8] b [, 80] [7, 8] b [, ] (b) Slope of the line between the points (8, 69.) and (6,.0) is m = = 6.8 = Using the point-slope form equation for the line, we have = 0.47( - 8), so = The linear model for subscribers as a function of time is = (c) The fit is ver good. The line goes through or is close to all the points. Subscribers [, ] b [ 0, 0]. Let n be an integer. n +n=n(n+), which is either the product of two odd integers or the product of two even integers. [7, 8] b [, 80] Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

4 0 Chapter Functions and Graphs (d) The monthl bill scatter plot has a curved shape that could be modeled more effectivel b a function with a curved graph. Some possibilities include a quadratic function (parabola), a logarithmic function, a power function (e.g., square root), a logistic function, or a sine function. (e) Subscribers Monthl Bills [4, 8] b [0, 80] [4, 8] b [, ] (f) In 99, cellular phone technolog was still emerging, so the growth rate was not as fast as it was in more recent ears. Thus, the slope from 99 (t=) to 998 (t=8) is lower than the slope from 998 to 004. Cellular technolog was more epensive before competition brought prices down. This eplains the anomal on the monthl bill scatter plot. 7. We need We need - 0 and +<0 < 6 - and Ú. Section. Eercises. Yes, = - 4 is a function of, because when a number is substituted for, there is at most one value produced for No, = does not determine as a function of, because when a positive number is substituted for, can be either or. A A. Yes 7. No 9. We need + 4 Ú 0; this is true for all real. Domain: ( q, q). Section. Functions and Their Properties Eploration. From left to right, the tables are (c) constant, (b) decreasing, and (a) increasing.. X X X moves X Y moves X Y moves X Y from from from to 0 to to to 0 0 to 0 to 0 0 to 0 0 to 0 to to 0 to 4 to to to to For an increasing function, Y/ X is positive. For a decreasing function, Y/ X is negative. For a constant function, Y/ X is For lines, Y/ X is the slope. Lines with positive slope are increasing, lines with negative slope are decreasing, and lines with 0 slope are constant, so this supports our answers to part. Quick Review = 0 = 6 = ; As we have seen, the denominator of a function cannot be zero. We need - 6 = 0 = 6. [, ] b [, ]. We need + Z 0 and - Z 0. Domain: ( q, ) ª (, ) ª (, q). [ 0, 0] b [ 0, 0]. We notice that g = - = -. As a result, - Z 0 and Z 0. Domain: ( q,0) ª (0, ) ª (, q). [ 0, 0] b [, ]. We need + Z 0, + Z 0, and 4 - Ú 0. The first requirement means Z, the second is true for all, and the last means 4. The domain is therefore ( q, ) ª (, 4]. [, ] b [, ] Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

5 Section. Functions and Their Properties 7. f()=0- can take on an negative value. Because is nonnegative, f() cannot be greater than 0. The range is ( q, 0]. 9. The range of a function is most simpl found b graphing it. As our graph shows, the range of f() is ( q, ) ª [0, q).. Increasing on ( q, ]; decreasing on [, q). [ 0, 0] b [ 0, 0]. Yes, non-removable. [ 0, 0] b [ 0, 0]. Yes, non-removable. [ 4, 6] b [, ]. Constant functions are alwas bounded. 7. >0 for all, so is bounded below b =0. 9. Since = - is alwas positive, we know that Ú 0 for all. We must also check for an upper bound: Thus, is bounded. 4. f has a local minimum when =0., where =.7. It has no maimum. [ 0, 0] b [, ]. Local maima at (, 4) and (, ), local minimum at (, ). The function increases on ( q, ], decreases on [, ], increases on [, ], and decreases on [, q). 7. (, ) and (, ) are neither. (, ) is a local maimum, and (, ) is a local minimum. The function increases on ( q,], decreases on [, ], and increases on [, q). 9. Decreasing on ( q, ]; increasing on [, q). [, ] b [0, 6] 4. Local minimum: 4.09 at 0.8. Local maimum:.9 at 0.8. [, ] b [ 0, 0] 4. Local maimum: 9.6 at.0. Local minima: =0 at =0 and =0 at = 4. [ 0, 0] b [, 8]. Decreasing on ( q, ]; constant on [, ]; increasing on [, q). [ 0, 0] b [0, 0] [, ] b [0, 80] 47. Even: f- = - 4 = 4 = f 49. Even: f- = - + = + = f. Neither: f( )= ( ) +0.0( )+= -0.0+, which is neither f() nor f().. Odd: g- = = += g() Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

6 Chapter Functions and Graphs. The quotient is undefined at =, indicating that - = is a vertical asmptote. Similarl, lim : q - =, indicating a horizontal asmptote at =. The graph confirms these. [ 0, 0] b [ 0, 0] + 7. The quotient is undefined at =, indicating - a possible vertical asmptote at =. Similarl, + lim indicating a possible horizontal asmptote at =. The graph confirms these : q - = -, asmptotes. + behaves much like so there is a horizontal = +, asmptote at = The graph matching this description. is (b). 6. The denominator cannot equal zero, so there is no vertical + asmptote. When is ver large, behaves much + like which for large is close to zero. So there =, is a horizontal asmptote at =0. The graph matching this description is (a). 67. (a) Since, lim we epect a horizontal : q - = 0, asmptote at =0. To find where our function crosses =0, we solve the equation - = 0 = 0# - = 0. The graph confirms that f() crosses the horizontal asmptote at (0, 0). [ 8, ] b [ 0, 0] + 9. The quotient is undefined at = and = -. - So we epect two vertical asmptotes. Similarl, the + lim so we epect a horizontal asmptote : q - =, at =. The graph confirms these asmptotes. [ 0, 0] b [ 0, 0] The quotient does not eist at =, - 8 so we epect a vertical asmptote there. Similarl, 4-4 lim so we epect a horizontal asmptote : q + 8 = 0, at =0. The graph confirms these asmptotes. [ 4, 6] b [, ] 6. The denominator is zero when = -, so there is a vertical asmptote at = - When is ver large,. [ 0, 0] b [ 0, 0] (b) Since lim we epect a horizontal : q + = 0, asmptote at =0. To find where our function crosses =0, we solve the equation: + = 0 = 0# + = 0. The graph confirms that g() crosses the horizontal asmptote at (0, 0). [ 0, 0] b [, ] lim : q (c) Since we epect a horizontal + = 0, asmptote at =0. To find where h() crosses =0, we solve the equation + = 0 = 0# + = 0 = 0. Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

7 Section. Functions and Their Properties The graph confirms that h() intersects the horizontal asmptote at (0, 0). 79. One possible graph: [, ] b [, ] 69. (a) The vertical asmptote is =0, and this function is undefined at =0 (because a denominator can t be zero). (b) 8. One possible graph: [ 0, 0] b [ 0, 0] Add the point (0, 0). (c) Yes. It passes the vertical line test. 7. True. This is what it means for a set of points to be the graph of a function. 7. Temperature is a continuous variable, whereas the other quantities all var in steps. The answer is B. 7. Air pressure drops with increasing height. All the other functions either steadil increase or else go both up and down. The answer is C. 77. (a) [, ] b [, ] k= (b) But the discriminant of -+ is negative ( ), so the graph never crosses the -ais on the interval (0, q). (c) k= (d) Ú But the discriminant of ++ is negative ( ), so the graph never crosses the -ais on the interval ( q, 0). 8. (a) f() is bounded above b =. To determine if = is in the range, we must solve the equation for : = = = 0 =. Since f() eists at =0, = is in the range. (b) lim Thus, g() is : q + = lim : q = lim =. : q bounded b =. However, when we solve for, we get = + + = 9 + = 9 = 0. Since 9 Z 0, = is not in the range of g(). (c) h() is not bounded above. (d) For all values of, we know that sin() is bounded above b =. Similarl, sin() is bounded above b =# =. It is in the range. 4 (e) lim : q + + = lim 4 : q + = lim 4a : + ba + b = lim 4 : q + (since + for ver large )=0. c 4 Similarl, lim As a result, we : -q + + = 0 d know that g() is bounded b =0 as goes to q and q. However, g()>0 for all >0 (since (+) >0 Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

8 4 Chapter Functions and Graphs alwas and 4>0 when >0), so we must check points near =0 to determine where the function is at its maimum. [Since g()<0 for all <0 (since (+) >0 alwas and 4<0 when <0) we can ignore those values of since we are concerned onl with the upper bound of g().] Eamining our graph, we see that g() has an upper bound at =, which occurs when =. The least upper bound of g()=, and it is in the range of g(). 4. The graph of g()= +e (shown below) does have a horizontal asmptote at =0. [, ] b [, ]. 8. Since f is odd, f( )= f() for all. In particular, f( 0)= f(0). This is equivalent to saing that f(0)= f(0), and the onl number which equals its opposite is 0. Therefore, f(0)=0, which means the graph must pass through the origin. 87. (a) f is continuous on [, 4]; the maimum value is, which occurs at = 4, and the minimum value is, which occurs at = 0. (b) f is continuous on [, ]; the maimum value is, which occurs at =, and the minimum value is 0., which occurs at =. (c) f is continuous on [ 4, ]; the maimum value is, which occurs at = -4, and the minimum value is, which occurs at = -. (d) f is continuous on [ 4, 4]; the maimum value is, which occurs at = -4, and the minimum value is, which occurs at = 0. Section. Twelve Basic Functions Eploration. The graphs of f()= and f()=ln have vertical asmptotes at =0.. The graph of g()= +ln (shown below) does have a vertical asmptote at =0. [.7, 6.7] b [.,.]. The graphs of f()=, f()=e, and f()= + e - have horizontal asmptotes at =0. [ 4.7, 4.7] b [.,.] Both f()= and g()= have - = - vertical asmptotes at =0, but h()=f()+g() does not; it has a removable discontinuit. Quick Review p Section. Eercises. = +; (e). = - ; (j). = ; (i) 7. =int(+); (k) 9. =(+) ; (d) 4. - (l) + e - ;. Eercise 8. Eercises 7, 8 7. Eercises, 4, 6, 0,, 9. =, =, = =sin,. =, = = ƒƒ,. = =e, =, + e -. = =sin, =cos, =, + e - 7. =, =, = =sin, Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

9 Section. Twelve Basic Functions 9. Domain: All reals Range: [, q) 9. [, ] b [ 0, 0] [ 0, 0] b [ 0, 0]. Domain: ( 6, q) Range: All reals (a) h() is increasing on [0, q) and decreasing on ( q,0]. (b) h() is even, because it is smmetric about the -ais. (c) The one etremum is a minimum value of 0 at =0. (d) h()= ƒ ƒ - 0 is the absolute value function, ƒ ƒ, shifted 0 units down. 4. [ 0, 0] b [ 0, 0]. Domain: All reals Range: All integers [.7, 6.7] b [.,.]. [ 0, 0] b [ 0, 0] (a) s() is increasing on [, q) and decreasing on ( q, ]. (b) s() is neither odd nor even. (c) The one etremum is a minimum value of 0 at =. (d) s()= ƒ - ƒ is the absolute value function, ƒ ƒ, shifted units to the right. 4. The end behavior approaches the horizontal asmptotes = and =. 4. [0, 0] b [, ] (a) r() is increasing on [0, q). (b) r() is neither odd nor even. (c) The one etreme is a minimum value of 0 at =0. (d) r()= - 0 is the square root function, shifted 0 units right. 7. There are no points of discontinuit. 47. [, ] b [, 4] (a) f() is increasing on ( q, q). (b) f() is neither odd nor even. (c) There are no etrema. (d) f = is the logistic function, + e - + e -, stretched verticall b a factor of. There are no points of discontinuit. Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

10 6 Chapter Functions and Graphs 49. There are no points of discontinuit.. There is a point of discontinuit at =0.. (a) This is g()= ƒ ƒ. (b) Squaring and taking the (positive) square root has the same effect as the absolute value function.. (a) [, ] b [, ] f = = ƒƒ = ƒ ƒ = g [, ] b [, ] This is the function f()=. (b) The fact that lne = shows that the natural logarithm function takes on arbitraril large values. In particular, it takes on the value L when =e L. 7. The Greatest Integer Function f()=int () [ 4.7, 4.7] b [.,.] Domain: all real numbers Range: all integers Continuit: There is a discontinuit at each integer value of. Increasing/decreasing behavior: constant on intervals of the form [k, k+), where k is an integer Smmetr: none Boundedness: not bounded Local etrema: Ever non-integer is both a local minimum and local maimum. Horizontal asmptotes: none Vertical asmptotes: none End behavior: int() : q as : q and int() : q as : q. 9. True. The asmptotes are =0 and =. 6. <+ <4. The answer is D. + e - 6. The answer is E. The others all have either a restricted domain or intervals where the function is decreasing or constant. 6. (a) A product of two odd functions is even. (b) A product of two even functions is even. (c) A product of an odd function and an even function is odd. 67. (a) Pepperoni count ought to be proportional to the area of the pizza, which is proportional to the square of the radius. (b) = k4 k = 6 = 4 = 0.7 (c) Yes, ver well. (d) The fact that the pepperoni count fits the epected quadratic model so perfectl suggests that the pizzeria uses such a chart. If repeated observations produced the same results, there would be little doubt. 69. (a) At =0, does not eist, e =, ln is not defined, cos =, and - =. + e (b) for f()=, f(+)=+=f()+f() (c) for f()=e, f()=e =e e =f() # f() (d) for f()=ln, f(+)=ln()=ln()+ln() =f()+f() (e) The odd functions:,, sin, Section.4 Building Functions from Functions Eploration If f=- and g = +, then f g= a + b - = + - =. - + If f = ƒ + 4ƒ and g =, - + then f g= ` a b + 4 ` Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

11 Section.4 Building Functions from Functions 7 = ƒ ƒ = ƒ ƒ = ƒ ƒ =. 7. ( q, q) If f = and g =, then f g= = Note, we use 9. (, ) ƒƒ. the absolute value of because g is defined for - q q, while f is defined onl for positive values of. The absolute value function is alwas positive. If f = and g = 0.6, then f g= 0.6 =. If f = - and g=ln(e ), then f g=ln(e )-= ln(e )+ln -= ln e+ln -=+ln -= ln. If f= sin cos and g = then f g= sin, cos = sinaa bb = sin. This is the double angle formula (see Section.4). You can see this graphicall. If f = - and g = sina b, [0, ] b [, ] then f g=- asin a bb = cosaa bb = cos. (The double angle formula for cos is cos =cos - sin =(-sin )-sin =- sin. See Section..) This can be seen graphicall: Section.4 Eercises. (f+g)()=-+ ; (f-g)()=-- ; (fg)()=(-)( )= -. There are no restrictions on an of the domains, so all three domains are ( q, q).. (f+g)()= +sin ; (f-g)()= -sin ; (fg)()= sin. Domain in each case is [0, q). For, 0. For sin, q<<q. +. (f/g)()= + Ú 0 and Z 0, ; so the domain is [, 0) ª (0, q). (g/f)()= + Ú 0, + ; so the domain is (, q). 7. f>g =. The denominator cannot be zero - and the term under the square root must be positive, so Therefore, 6, which means that The domain is -,. - g>f = The term under the square root. must be nonnegative, so - Ú 0 (or ). The denominator cannot be zero, so Z 0. Therefore, <0 or 0 6. The domain is -, 0 h 0, [0, ] b [, ] f g f g 4 ( ) ( ) 0.6 ln(e ) ln sin cos sin sin cos Quick Review.4. ( q, ) ª (, q). ( q, ]. [, q) [0, ] b [0, ]. (f g)()=f(g())=f(4)=; (g f)( )=g(f( ))=g( 7)= 6. (f g)()=f(g())= f + =f()= +4=8; (g f)( )=g(f( ))=g(( ) +4) =g(8)= 8 + =. f(g())=(-)+=-+=-. Because both f and g have domain ( q, q), the domain of f(g()) is ( q, q). g(f())=(+)-=+; again, the domain is ( q, q). 7. f(g())= + -=+-=-.The domain of g is Ú -, while the domain of f is ( q, q), so the domain of f(g()) is Ú -, or [, q). g(f())= - + = -. The domain of f is ( q, q), while the domain of g is [, q), so g(f()) requires that f Ú -. This means - Ú -, or Ú, which means - or Ú. Therefore the domain of g(f()) is ( q, ] ª [, q). Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

12 8 Chapter Functions and Graphs 9. fg = f - = - = - ; the domain is [, ]. gf = g = - = - 4 ; the domain is [, ].. fg = fa b = > = > = ; the domain is - q, 0 h 0, q. gf = ga ; b = > = > = the domain is - q, 0 h 0, q.. One possibilit: f = and g = -.. One possibilit: f = ƒƒ and g = One possibilit: f = + and g = One possibilit: f = cos and g =.. r = t in., so V = 4 pr = 4 p t ; when t=00, V = 4 p = 46,94p L 77,74.6 in.. The initial area is ()(7)= km. The new length and width are l=+t and w=7+t, so A=lw= (+t)(7+t). Solve (7+t)(+t)=7 ( times its original size), either graphicall or algebraicall: the positive solution is t.6 sec.. ()+4()=+4=7 (4)+4( )=-8=4 ()+4( )=9-4= The answer is (, ). 7. =-, = - and = = -, = - and = ƒƒ = Q ƒƒ = - + Q = - + or = = - and = - 4. = Q = and = - or = ƒƒ and = -ƒƒ fg = fa + b = =. The domain is Z 0. If f = - + and g = +, then fg = f + = = + - =. The domain is- q, q. If f = a + b and g = then -, fg = fa - b = ± - + = - ± =. The domain is Z. -. (a) (f+g)()=(g+f)()=f() if g()=0. (b) (fg)()=(gf)()=f() if g()=. (c) (f g)()=(g f)()=f() if g()=.. + -=0. Using the quadratic formula, so, and a + - b = = - ; - 4- = - ; = = a + = - b 4. False. If g()=0, then a f () is not defined and 0 is g b not in the domain of a f even though 0 ma be in g b, the domains of both f() and g(). 47. Composition of functions isn t necessaril commutative. The answer is C. 49. (f f)()=f( +)=( +) += ( 4 + +)+= The answer is E.. If f = e and g = ln, then fg = f ln = e ln = e ln =. The domain is 0, q. If f = + and g = -, then fg = f - = - + = - + =. The domain is, q. If f = - and g = -, then fg = f - = - - = - - =. The domain is - q, 4. If f = and g = +, then - [ 9.4, 9.4] b [ 6., 6.] f g D e ln (0, ) ( ) [, ) ( ) (, ] ( ) 0 (, ) Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

13 Section. Parametric Relations and Inverses 9 Section. Parametric Relations and Inverses Eploration. T starts at 4, at the point (8, ). It stops at T=, at the point (8, ). 6 points are computed.. The graph is smoother because the plotted points are closer together.. The graph is less smooth because the plotted points are further apart. In CONNECT mode, the are connected b straight lines. 4. The smaller the Tstep, the slower the graphing proceeds. This is because the calculator has to compute more X and Y values.. The grapher skips directl from the point (0, ) to the point (0, ), corresponding to the T-values T= and T=0. The two points are connected b a straight line, hidden b the Y-ais. 6. With the Tmin set at, the grapher begins at the point (, 0), missing the bottom of the curve entirel. 7. Leave everthing else the same, but change Tmin back to 4 and Tma to. Quick Review =+6,so= = +. =-4, so = ; - 4. (+) =- + =- - = - (-) = (+) + = - = (-4) =+ -4 =+ - =4+ (-) = = - 9. = +, Ú - and Ú 04 = +, Ú -, and Ú 0 = -, Ú -, and Ú 0 Section. Eercises. = = 6, = + = 9. The answer is (6, 9).. = - 4 =, = + =. The answer is (, ).. (a) t (, )=(t, t-) ( 6, 0) ( 4, 7) (, 4) 0 (0, ) (, ) (4, ) (6, 8) (b) t= =a -=.-. This is a function., b (c) [, ] b [, ] 7. (a) t (, )=(t, t-) (9, ) (4, 4) (, ) 0 (0, ) (, ) (4, 0) (9, ) (b) t=, =(+). This is not a function. (c) 9. (a) B the vertical line test, the relation is not a function. (b) B the horizontal line test, the relation s inverse is a function.. (a) B the vertical line test, the relation is a function. (b) B the horizontal line test, the relation s inverse is a function.. = - 6 Q. = - + Q [, ] b [, ] = ;( q, q) + + = - + = - - = = - + f - = = - + ; - = + - ( q, ) ª (, q) 7. = -, Ú, Ú 0 Q = -, Ú 0, Ú = -, Ú 0, Ú f - = = +, Ú 0 9. = Q = f - = = ; ( q, q). = + Q = - 6 = + 6 f - = = + 6 = - + = + = + f - = = - ; ( q, q) Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

14 0 Chapter Functions and Graphs. One-to-one. One-to-one 7. fg = c + d - = + - = ; gf = = = 9. fg = - > 4 + = - + gf = > = > = =. fg = = - + = ; = + - = ; = - a - + b gf = = # P - Q = = + - =. (a) =(.08)(00)=08 euros (b) = =. This converts euros () to dollars () (c) =(0.99)(48)=$ =e and =ln are inverses. If we restrict the domain of the function = to the interval 0, q, then the restricted function and = are inverses. 7. = ƒƒ 9. True. All the ordered pairs swap domain and range values. 4. The inverse of the relation given b +=9 is the relation given b +=9. + = + 0 = Z = = - Z = -4 - = -9 Z = + 0 = Z = 4 + = 9 The answer is E. 4. f()=- =- The inverse relation is =- += + = + f ()=. The answer is C. 4. (Answers ma var.) (a) If the graph of f is unbroken, its reflection in the line = will be also. (b) Both f and its inverse must be one-to-one in order to be inverse functions. (c) Since f is odd, (, ) is on the graph whenever (, ) is. This implies that (, ) is on the graph of f whenever (, ) is. That implies that f is odd. (d) Let =f(). Since the ratio of to is positive, the ratio of to is positive. An ratio of to on the graph of f is the same as some ratio of to on the graph of f, hence positive. This implies that f is increasing (a) = which gives us the slope 88 - = 7 6 = 0.7, of the equation. To find the rest of the equation, we use one of the initial points - 70 = = = (b) To find the inverse, we substitute for and for, and then solve for : = = 0.7 = 4 -. The inverse function converts scaled scores to raw scores. 49. (a) It does not clear the fence. [0, 0] b [0, 00] Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

15 Section.6 Graphical Transformations (b) It still does not clear the fence. [0, 0] b [0, 00] Yes [, ] b [, ] [.7,.7] b [.,.] (c) Optimal angle is 4. It clears the fence. Eploration. [0, 0] b [0, 00]. When k=, the scaling function is linear. Opinions will var as to which is the best value of k. Section.6 Graphical Transformations Graph C. Points with positive -coordinates remain unchanged, while points with negative -coordinates are reflected across the -ais.. Eploration. [, ] b [, ] The raise or lower the parabola along the -ais.. Graph A. Points with positive -coordinates remain unchanged. Since the new function is even, the graph for negative -values will be a reflection of the graph for positive -values.. [, ] b [, ] The move the parabola left or right along the -ais.. Graph F. The graph will be a reflection across the -ais of graph C. 4. Graph D. The points with negative -coordinates in graph A are reflected across the -ais. [, ] b [, ] [, ] b [, ] Eploration. [, ] b [, ] [, ] b [, ] [ 4.7, 4.7] b [.,.] The. and the stretch the graph verticall; the 0. and the 0. shrink the graph verticall. Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

16 Chapter Functions and Graphs.. h [ 4.7, 4.7] b [.,.] The. and the shrink the graph horizontall; the 0. and the 0. stretch the graph horizontall. 6 f g 6 6 Quick Review.6. (+). (+6). (-/) = ( - +-)+( -+)-+ = = -6+ Section.6 Eercises. Vertical translation down units.. Horizontal translation left 4 units.. Horizontal translation to the right 00 units. 7. Horizontal translation to the right unit, and vertical translation up units. 9. Reflection across -ais.. Reflection across -ais. For # 0, recognize =c # (c>0) as a vertical stretch (if c>) or shrink (if 0<c<) of factor c, and =(c # ) as a horizontal shrink (if c>) or stretch (if 0<c<) of factor /c. Note also that =(c # ) =c, so that for this function, an horizontal stretch/shrink can be interpreted as an equivalent vertical shrink/stretch (and vice versa).. Verticall stretch b.. Horizontall stretch b >0. =, or verticall shrink b 0. = g()= = f - 6; starting with f, translate right 6 units to get g. 9. g()= (+4-) = f(+4); starting with f, translate left 4 units, and reflect across the -ais to get g.. 0. Since the graph is translated left units, f = The graph is reflected across the -ais, translated left units, and translated up units. = - would be reflected across the -ais, = - + adds the horizontal translation, and finall, the vertical translation gives f = = (a) = f()= ( - -+) = + +- (b) =f( )=( ) -( ) -( )+ = (a) = -f() = -( 8) = - (b) = f( ) = 8(-) = -8 = -. Let f be an odd function; that is, f( )= f() for all in the domain of f. To reflect the graph of =f() across the -ais, we make the transformation =f( ). But f( ) = f() for all in the domain of f, so this transformation results in = f(). That is eactl the translation that reflects the graph of f across the -ais, so the two reflections ield the same graph.. 7. f g h 6 9. (a) ==( -4)= -8 (b) = f =f()=() -4()=7-4. (a) ==( +-)= +-4 (b) =f()=() +-= Starting with =, translate right units, verticall stretch b, and translate down 4 units. 4. Starting with =, horizontall shrink b and translate down 4 units. Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

17 Section.7 Modeling with Functions 47. First stretch (multipl right side b ): =, then translate (replace with -4): =(-4). 49. First translate left (replace with +): = +, then stretch (multipl right side b ): = +, then translate down (subtract 4 from the right side): = To make the sketches for # 4, it is useful to appl the described transformations to several selected points on the graph. The original graph here has vertices (, 4), (0, 0), (, ), and (4, 0); in the solutions below, the images of these four points are listed.. Translate left unit, then verticall stretch b, and finall translate up units. The four vertices are transformed to (, 0), (, ), (, 8), and (, ). (b) Change the -value b multipling b the conversion rate from dollars to en, a number that changes according to international market conditions. This results in a vertical stretch b the conversion rate. 67. (a) The original graph is on the left; the graph of = ƒf()ƒ is on the right. [, ] b [ 0, 0] [, ] b [ 0, 0] (b) The original graph is on the left; the graph of = fƒƒ is on the right [, ] b [ 0, 0] [, ] b [ 0, 0] (c). Horizontall shrink b The four vertices are. transformed to (, 4), (0, 0), (, ), (, 0). 4 4 (d). Reflections have more effect on points that are farther awa from the line of reflection. Translations affect the distance of points from the aes, and hence change the effect of the reflections First verticall stretch b then translate up units., 9. False. =f(+) is =f() translated units to the left. 6. To verticall stretch =f() b a factor of, multipl the f() b. The answer is C. 6. To translate =f() units up, add to f(): =f() ±. To reflect the result across the -ais, replace with. The answer is A. 6. (a) Price (dollars) Month Section.7 Modeling with Functions Eploration. n = ; d = 0 n = 4; d = Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

18 4 Chapter Functions and Graphs n = ; d = n = 7; d = 4 n = 6; d = 9 n = 8; d = 0 Section.7 Eercises A= / w=(+)() =(+0.04)= =0.60. Let C be the total cost and n be the number of items produced; C=4,00+.7n.. Let R be the revenue and n be the number of items sold; R=.7n.. The basic formula for the volume of a right circular clinder is V = pr h, where r is the radius and h is height. Since height equals diameter (h=d) and the diameter is two times r (d=r), we know h=r. Then, V = pr r = pr. r.. Linear: r =0.978 Power: r =0.990 Quadratic: R = Cubic: R = Quartic: R = 4. The best-fit curve is quadratic: =0. -..The cubic and quartic regressions give this same curve.. Since the quadratic curve fits the points perfectl, there is nothing to be gained b adding a cubic term or a quartic term. The coefficients of these terms in the regressions are zero. 6. = At =8, =0.(8) -.(8)=8000 Quick Review.7. h=(a/b). h=v/( pr ). r = 7. h = A - pr pr 9. P = n = 9; d = 7 [, ] b [0, 40] V B 4p = A p - r A + r>n nt = Aa + r -nt n b n = 0; d = 7. Let a be the length of the base. Then the other two sides of the triangle have length two times the base, or a. Since the triangle is isoceles, a perpendicular dropped from the top verte to the base is perpendicular. As a result, or h = 4a - a 4 = 6a - a h + a a b = a, 4 a = a, so h = = a. The triangle s area is 4 B 4 A = bh = aaa 9. Let r be the radius of the sphere. Since the sphere is tangent to all si faces of the cube, we know that the height (and width, and depth) of the cube is equal to the sphere s diameter, which is two times r (r). The surface area of the cube is the sum of the area of all si faces, which equals r# r=4r. Thus, A=6# 4r =4r. r a a r h a a h = r r r a r a r b = a. 4 Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

19 Section.7 Modeling with Functions. Solving +4=60 gives =4, so 4=496.The two numbers are 4 and =6,4, so =,00.. 8=t, so t=. hr ()=9.8; 0.7(7)=0.. The $ shirt sells for $9.80. The $7 shirt sells for $0.. The $ shirt is a better bargain, because the sale price is cheaper ( + ) = = There was a 9% increase in cell phone antennas.. (a) (00-)=0.(00). (b) Graph =0.+0.4(00-) and =. Use 7.4 gallons of the 0% solution and about 4.86 gal of the 4% solution.. (a) The height of the bo is, and the base measures 0- b 8-. V()=(0-)(8-) (b) Because one side of the original piece of cardboard measures 0 in, must be greater than 0 but less than 0, so that 0<<. The domain of V() is (0, ). (c) Graphing V() produces a cubic-function curve that between =0 and = has a maimum at approimatel (.06, 68.). The cut-out squares should measure approimatel.06 in b.06 in.. Equation of the parabola, to pass through ( 6, 8) and (6, 8): =k 8=k ( 6) 8 k= 6 = = = [0, 00] b [0, 0] L coordinate of parabola 8 in from center: = (8) = From that point to the top of the dish is 8-=6 in. 7. Original volume of water: V 0 = r h= (9) (4) 0.7 in Volume lost through faucet: V l =time*rate=(0 sec)( in /sec)=600 in Find volume: V f =V 0 -V l = =4.7 Since the final cone-shaped volume of water has radius and height in a 9-to-4 ratio, or r= : V f = a h =4.7 8 hb h = 8 h 64 Solving, we obtain h.6 in. 9. Biccle s speed in feet per second: (* *6 in/rot)( rot/sec)=64 in/sec Unit conversion: (64 in/sec) a ft/inb a mi/ft b(600 sec/hr) 80.4 mi/hr 4. True. The correlation coefficient is close to (or ) if there is a good fit. A correlation coefficient near 0 indicates a ver poor fit. 4. The pattern of points is S-shaped, which suggests a cubic model. The answer is C. 4. The points appear to lie along an upward-opening parabola. The answer is B. 47. (a) C=00,000+0 (b) R=0 (c) 00,000+0=0 00,000=0 =000 pairs of shoes (d) Graph =00,000+0 and =0; these graphs cross when =000 pairs of shoes. The point of intersection corresponds to the break-even point, where C=R. 49. (a) =u()=,000+. (b) =s()=,000++8=,000+. (c) =R u ()=6. (d) 4 =R s ()=79. (e). (a) [0,0000] b [0,00000] (f) You should recommend stringing the rackets; fewer strung rackets need to be sold to begin making a profit (since the intersection of and 4 occurs for smaller than the intersection of and ). [0, ] b [00, 00] (b) List L={., 06., 0., 96.6, 9.0, 87., 8., 79.8, 7.0, 7.7, 68, 64., 6., 8.,.9,.0, 0.8, 47.9, 4., 4.} Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

20 6 Chapter Functions and Graphs (c) The regression equation is =8.07 the data etremel well. * 0.9. It fits (b) Since lim 7 and + 0 = 7 : q lim 7 we epect horizontal + 0 = -7, : -q asmptotes at = 7 and = -7. [0, ] b [00, 00] Chapter Review. (d). (i). (b) 7. (g) 9. (a). (a) All reals (b) All reals. (a) All reals (b) g = + + = +. At = -, g = 0, the function s minimum. The range is 0, q.. (a) All reals (b) ƒƒ Ú 0 for all, so ƒƒ Ú 0 and ƒƒ + 8 Ú 8 for all. The range is 8, q. 7. (a) f = Z 0 and - = ( - ). - Z 0, Z. The domain is all reals ecept 0 and. (b) For 7, f 7 0 and for 6, f 6 0. f does not cross = 0, so the range is all reals ecept f = Continuous [, ] b [ 0, 0]. - q, q [ 4.7, 4.7] b [.,.] 7. As the graph illustrates, is increasing over the intervals - q, -, -,, and, q. [ 4.7, 4.7] b [.,.] 9. - sin, but - q q, so f is not bounded. [ 7, ] b [, 8] [, ] b [, ]. e Ú 0 for all, so - e 0 and - e for all. h is bounded above.. (a) - Z 0, - Z 0, so Z 0 and Z. We epect vertical asmptotes at = 0 and =. (b) =0 [, ] b [ 0, 0]. (a) None (b) -7, at = -. (a) None [ 7, ] b [ 0, 0] [ 6, 4] b [ 0, 0] Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

21 Chapter Review 7. (a) -, at = 0 (b) None. [, ] b [ 0, 0] 7. The function is even since it is smmetrical about the -ais. [ 4.7, 4.7] b [.,.] 9. Since no smmetr is ehibited, the function is neither. [, ] b [, ]. (f g)()= fg = f - 4 = - 4. Since - 4 Ú 0, Ú 4, - or Ú. The domain is - q, -4 h, q.. f# g = f # g = # - 4. Since Ú 0, the domain is 0, q. 7. lim = q. (Large negative values are not in the : q domain.) s 9. r = a s r = B 4 = s b + a s b = s 4,. The area of the circle is ps A = pr = pa s = ps. b = 4 [.,.] b [.,.] 4. = +, = -, = -, so f - = -. r s s s s r = s 4. = so f - =, =, =, d=r, r = d, so the radius of the tank is 0 feet. Volume is. 0 ft V = pr # h = p0 # h = 00ph [, ] b [, ] h [, ] b [, ] 6. Since V = 4000p - t, we know that pr h = 4000p - t. In this case, r = 0, so 4000p - t 00ph = 4000p - t, h = = 40-00p 0 ft t. 0p [, ] b [, ] h h h = 40 ft Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.

22 8 Chapter Functions and Graphs 6. (a) Chapter Project. [, 9] b [60, 970] (b) The regression line is = [, 9] b [60, 970] (c) 79.8() (thousands of barrels) 67. (a) r + a h b =, h 4 = - r, h = - 4r, h = - r h = - 4r, [, 0] b [ 70,,000]. The eponential regression produces L 7.88(.48).. 007: For =0, L 4,84 008: For =, L 49,78 4. The model, which is based on data from the earl, highgrowth period of Starbucks Coffee s compan histor, does not account for the effects of gradual market saturation b Starbucks and its competitors. The actual growth in the number of locations is slowing while the model increases more rapidl.. The logistic regression produces L 6, e h r r h : For=0, L,67 008: For=, L,6 These predictions are less than the actual numbers, but are not off b as much as the numbers derived from the eponential model were. For the ear 00 (=), the logistic model predicts about 6,060 locations, but 00 is too far outside the data set to inspire much confidence in this prediction. h = r (b) V = pr h = pr - r = pr - r (c) Since - r Ú 0, - r Ú 0 Ú r, - r. However, r 0 are invalid values, so the domain is 0, 4. (d) (e).7 in [0, ] b [0, 0] Copright 0 Pearson Education, Inc. Publishing as Addison-Wesle.