Selected Answers and Solutions Go to Hotmath.com for step-by-step solutions of most odd-numbered exercises free of charge.

Transcription

1 Go to Hotmath.com for step-b-step solutions of most odd-numbered eercises free of charge. CHAPTER Functions from a Calculus Perspective Chapter Get Read = + 9. = ± - 7. = D = n ;. dozen donuts n - n + 77 Lesson -. { >, }; (,. { -, }; (-, -]. { < < 99, }; (, { < -9 or >, }; (-, -9 (, 9. {.n =, n -, n }. { - or >, }; (-, -] (,. { = n, n }. function 7. function 9. function. function. function. function 7. not a function 9a. {(, 7, (, 7, (, 7, (,, (, } 9b. Yes; there is eactl one estimated high temperature each da. Yes; there is eactl one low temperature each da. a. -7 b c. -7 b - 7 b - b - 9 (- a. g(- = ( - + (- - (- = + (- - = - - = ( b. g( = ( + ( - ( = = + - ( - b c. g( - b = ( - b + ( - b - = ( - 7b + b - b - b + b + - b - = -9 b + b - b + b - b + = - b + b - 7b + b - 7b + 7 a. -. b. - - c. - b + b - 7b + b - 7b + 7 7a. 7b. 7c. 7 + n 9. (-, - (-, - (-,. (-,. (.,. (-, - (-, (, 7. Yes; sample answer: Because length must be positive, the domain of the function is (,. 9. ; 7. ; a.. million; 9. million b. integral values inside the interval [, ]. No; a vertical line would pass through infinitel man points. 7. No; the -ais is a vertical line that passes through two points on the graph, (, and (, -. 9 Sample answer: Because presidential elections are held ever ears, and do not have a finite end, it is impractical to displa the set in interval notation. If set-builder notation is used, the interval can be taken into account and a finite interval is not necessar. Therefore, the set of presidential election ears beginning in 79 can be described in setbuilder notation as { = n + 79, n }.. D = { 7, }. -; -;. a + ; a + h + ; - 7. a - a + ; a + ah + a + h + a + ah + h - a - h + ; a + h 9. - a ; - a - a h - a h - a h - ah - h ; - a - a h - a h - ah - h 7. 7a - ; 7a + 7h - ; 7 7. a ; a + a h + ah + h ; a + ah + h 7a. A(l = l ; [,.]. 7b. A(h =. h ; [.,.] 7c..9 in 77. No; sample answer: Most nonnegative -values are paired with two -values because it is necessar to take both the positive and negative values of the absolute value of when solving the equation for. 79a. f ( = [-, ] scl: b [-, ] scl: f( = [-, ] scl: b [-, ] scl: f( = [-, ] scl: b [-, ] scl: f(= [-, ] scl: b [-, ] scl: f( = [-, ] scl: b [-, ] scl: f( = [-, ] scl: b [-, ] scl: connected.mcgraw-hill.com R

2 79b. n Range (-, [, (-, [, (-, [, 79c. Sample answer: When n is even in f ( = n, the range is [,. 79d. Sample answer: When n is odd in f ( = n, the range is (-,. When the denominator of is zero, the ( + ( + ( - epression is undefined. Therefore, f( is undefined for = -, = -, and =. In interval notation the domain can be described as (-, - (-, - (-, (,, and in set-builder notation the domain can be described as { -, -, -, }. I prefer set-builder notation because instead of listing the four intervals in which can be an element, it lists the three real numbers that cannot be equal to.. true. False; sample answer: Two or more elements in X ma be matched with the same element in Y. 7. Sample answer: If each of the possible inputs is assigned to eactl one output in the verbal description, the relation is a function. 9. Sample answer: If each input value in the table is paired with a unique output value, the relation is a function. 9. Sample answer: If each -value can be paired with eactl one -value after the equation is solved for, then the relation is a function , 97. (-., 99. (9, ; consistent and independent. (, ; consistent and independent. {,,,,, 9,, }. 7. B 9. A Lesson - a. 7.7 b.. c.. The function value at = - appears to be about. Find f (-. f ( = + f (- = - + = + or The function value at = - appears to be about. Find f(-. f (- = - + = + or The function value at = appears to be about. Find f (. f ( = + = a. b. - c. undefined 7a., tons,, tons,, tons;,7 tons,, tons,, tons 7b. about or = ; f ( = -.( +. ( -. ( +.( +.7 9; f ( = -.( +. ( -. ( +.( D = (-,, R = [,. D = (-, ], R = [-, ]. D = [-,, R = [-, a. Sample answer: copper: D = [-, ], R = [.7]; aluminum: D = [-, ], R = [.,.]; zinc: D = [-, ], R = [.,.]; steel: D = [-, ], R = [.,.7] b. Sample answer: copper.7 J, aluminum. J, zinc. J, steel. J 7 From the graph, it appears that f ( intersects the -ais at approimatel (,, so the -intercept is. Find f (. f ( = ( - ( - ( f ( = Therefore, the -intercept is. From the graph, the -intercepts appear to be at about - and. Let f ( = and solve for. - - = ( - ( + = = or - = or + = = = - Therefore, the zeros of f are,, and intercept: ; no zeros; + = -. -intercept: -; zeros: -, ; - - = ( + ( - = + = or - = = - =. -intercept: ; zeros: -, -; + + = ( + ( + = + = or + = = - = -. -ais; (, (, - (, - (, - (, - (, - (, ais, -ais, and origin; (, (, - (, - (, - (, - (, - (, - Because 9 - (- = is equivalent to 9 - =,the graph is smmetric with respect to the -ais. (Continued on the net page Because = (- - is equivalent to = -, the graph is smmetric with respect to the -ais. (, - (-, - (-, - (-, (, (, (, Because 9 (- - = is equivalent to 9 - =, the graph is smmetric with respect to the -ais. R Selected Answers

3 (, - - (-, (-, (-, - (, (, (, Because 9(- - (- = is equivalent to 9 - =, the graph is smmetric with respect to the origin. 9.. [-, ] scl: b [-, 7] scl: Even; the graph of f ( is smmetric with respect to the -ais. f (- = (- = - = f( neither; g(- = (- - + = origin;. -ais (, - (-, - (-, - (-, - (, - - (, - - (, - Because - = - (- is equivalent to = -, the graph is smmetric with respect to the origin.. -ais; (, - (-, - (-, - + (-, + (, + (, + (, - 9 (-, (-, (-, -7-7 (, -7 - (, 9 (, 9 Because = (- - (- is equivalent to = -, the graph is smmetric with respect to the -ais. Because ( - + (- = is equivalent to ( - + =, the graph is smmetric with respect to the -ais. [-, ] scl: b [-, ] scl: a. undefined b. c.. a. Because represents ears since and the model is onl valid from to, the relevant domain of the function is {, }. The graph does not appear to etend below f( =. or above f (.. Therefore, the range of the function is approimatel R = {. million. million, }. b. Because is ears after, =. Sample answer: From the graph, it appears that the function value when = is about.. Find h(. h( =.( +.( +. = + +. or. c. Sample answer: From the graph, it appears that the -intercept is about.. Find h(. h( =.( +.( +. =. The -intercept represents the number of millions of households with onl wireless service in. d. There are no zeros because there were more than zero households with onl wireless phone service for all of the ears in the domain. 7a. (, [-, ] scl: b [-, ] scl:. 7. [-, ] scl: b [-9, ] scl: [-, ] scl: b [-, ] scl: neither; f (- = -(- + (- - = Even; the graph of h( is smmetric with respect to the -ais. h(- = (- - 9 = - 9 = h( 7b. [, ]; The relevant domain represents the interval of time beginning when the dose was first taken and ending when the pain reliever left the bloodstream. Because time cannot be negative,. The amount of pain reliever in the bloodstream is zero when the dose is first taken at =, and is zero again at =. Therefore,. Combining these restrictions, the relevant domain is {, } or [, ]. 7c. about milligrams 9. no zeros [-, ] scl: b [-, ] scl: For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R

4 .. [-, ] scl: b [-, ] scl: [-, ] scl: b [-, ] scl: + - = + = + = + = + = = - = - + = =. D = (-, -] (-,, R = (-, 7. D = (-, -] (, [7,, R = [-, 9a. D = [, 7], R [-9.9,.] 9b. Sample answers:,.; The -intercept represents the percent population change in 9. 9c..; The zeros represent the times at which the percent population change was and will be. 9d..%; Sample answer: This value does not seem realistic because it would mean that the population increased b about %, which does not seem likel in comparison to the population trends from 9 to. a f ( - - undefi ned b. Sample answer: As approaches from the left, the value of f ( becomes greater and greater. As approaches from the right, the value of f ( becomes more and more negative. c. [-, ] scl: b [-, ] scl: Sample answer: As approaches from the left, the graph decreases without bound. As approaches from the right, the graph increases without bound. d. Sample answer: As the value of becomes increasingl large, or when >, the denominator of the fraction will also become increasingl large. This creates smaller fractions that will continue to decrease but will never reach zero or cross the -ais. This also applies as decreases at an increasing rate, or when < -. However, since is negative, the value will also be negative but it too will never reach. As the value of approaches, the difference of and becomes smaller and smaller. When the difference d is - < d <, the denominator is smaller than the numerator, thus producing a larger number. If the difference is positive, the fraction will approach infinit. If the difference is negative, the fraction will approach negative infinit. Onl at will the fraction fail to eist because the difference in the denominator cannot be [-, ] scl: b [-, ] scl: [-, ] scl: b [-, ] scl: [-, ] scl: b [-, ] scl: 9. Sample answer: 7 The graph of a relation is smmetric with respect to the origin if, for ever point (, on the graph, the point (-, - is also on the graph. Therefore, the points (,, (, 9, and (, are also on the graph. Even; the graph of g(n is smmetric with respect to the -ais. g(-n = n - 7 = (-n - 7 = n - 7 = g(n Odd; the graph of f (g is smmetric with respect to the origin. f (-g = (-g 9 = - g 9 = -f ( g Odd; the graph of h( is smmetric with respect to the origin. h(- = (- - 7(- + (- = = -h( 7. Sample answer: For a relation to be a function, each -value must map to eactl one -value. A relation with two -intercepts located at (, and (, cannot be a function because two different -values cannot map to the same -value. A relation with two -intercepts located at (, and (, can be a function because two different -values can map to the same -value. 7. False; sample answer: If n is, the range is { = }. If n is negative, the range is {, }. The range will onl be {, } if n is positive. 77. False; sample answer: Consider the odd function = and the point (, on its graph. Reflecting the point (, in the line = - produces the point (-, - which is not on the graph of =. 79. odd; Sample answer: Proof: Given: a is an odd function and b( = a(- Prove: b is an odd function. b( = a(- (Given. a is an odd function (Given. a( = -a( (Def. odd function. b( = -a( (Trans. Prop. of Equalit using and. b( = a( (Div. Prop. of Equalit. a( = -b( (Refleive Prop. of Equalit 7. b (- = a(-(- (Substitute for in R Selected Answers

5 . b (- = a( (Substitute for -(-. 9. b (- = -b( (Trans. Prop. of Equalit using and. b is an odd function. (Def. odd function. even; Sample answer: Proof: Given: a is an odd function and b( = [a(] Prove: b is an even function. b( = [a(] (Given. a is an odd function (Given. a( = -a( (Def. odd function. b( = [a(-] (Substitute - for in. b( = [-a(] (Trans. Prop. of Equalit using and. b(- = [a(] (Mult. Prop. 7. b(- = b( (Trans. Prop. of Equalit using and. b is an even function. (Def. even function. odd; Sample answer: Proof: Given: a is an odd function and b( = [a(] Prove: b is an odd function. b( = [a(] (Given. a is an odd function (Given. a(- = -a( (Def. odd function. b(- = [a(-] (Substitute - for in. b(- = [-a(] (Trans. Prop. of Equalit using and. b(- = [a(] (Mult. Prop. 7. b(- = [a(] (Div. Prop. of Equalit. b(- = b( (Trans. Prop. of Equalit using 7 and 9. b(- = -b( (Div. Prop. of Equalit. b is an odd function. (Def. odd function. Sometimes; sample answer: The relation described b + ( - = is smmetric with respect to the line =, but is not a function. A relation whose points all lie on the line = is smmetric with respect to the line = and is a function. 7. Sometimes; sample answer: The relation described b + = is smmetric with respect to the - and -aes but is not a function. The relation {(, -, } is smmetric with respect to the - and -aes and is a function. 9a. - 9b c. 9 n - n - 9a. 9b c a., 9b. 9c no solution R + RW + W. - + i 7. B 9. B Lesson -. Continuous; f (- = or about., lim., and - lim = f (-.. Discontinuous at = -; h (- is undefined - and lim = -, so h( has a removable discontinuit at = -. - Continuous at =. h ( =, lim h ( =, and lim h ( = h(.. Discontinuous; g( is undefined and g( approaches - as approaches from the left and as approaches from the right, so g( has an infinite discontinuit at =. 7. Discontinuous at = ; h ( is undefined and h( approaches - as approaches from the left and as approaches from the right, so h( has an infinite discontinuit at =. Discontinuous at = ; h ( is undefined and lim h( =, so h( has a removable discontinuit at =. 9. Discontinuous at = -; f ( approaches - as approaches - from the left and as approaches - from the right, so f ( has a jump discontinuit at = -. a. Find f (.. f (w = 7. w 7. f (. = (. =. The function is defined at w =.. Find lim f (w. Construct a table that shows values of f (w w. for w-values approaching. from the left and from the right. w f (w lim f (w =.. w. Because lim = f (., f (w is continuous as w =.. w. b. The function is discontinuous at w =. Construct a table that shows values of f (w for w-values approaching from the right. The function does not have to be eamined for negative values of w since a wall cannot have a negative thickness. w... f (w Because f ( is undefined and f (w approaches as w approaches from the right, f (w is discontinuous at w = and has an infinite discontinuit at w =. c. Evaluate the function for several w-values in its domain. Use these points to construct a graph. f(w = 7. w w. and. - and, and 7. - and, and, and 9. no zeros on the interval. and. From the graph, it appears that f ( as - and f( - as. -, -, f ( From the graph, it appears that f ( - as - and f ( - as. -, -, f ( From the graph, it appears that f ( as - and f ( as. -, -, f ( For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R

6 9. From the graph, it appears that f ( - as - and f ( - as. - - f ( discontinuit: infinite at = and = -; end behavior: lim h( =, lim h( = ; zeros: = - and - 9. a. [, ] scl: b [,.] scl:. b. Sample answer: The end behavior of the graph indicates that as the concentration of the catalst solution is increased, the chemical reaction rate approaches.., R( Sample answer: As, the fraction will decrease, and q( will approach.. Sample answer: As, the fraction will approach, so p( will approach. 7. Sample answer: As, the fraction will decrease, and c( will approach. 9. Sample answer: As, h( will increase without bound and approach.. Sample answer: As the mass of the railwa car continues to increase, the railwa car s kinetic energ will approach. h ( has an infinite discontinuit at = because h ( is undefined and h ( approaches as approaches from the left and the right. w h (.,..,. [-, ] scl: b [-, ] scl: discontinuit: infinite at = - and = ; end behavior: lim h( = -, lim h( = ; zeros: = -,, and -.. [-, ] scl: b [-, ] scl: -, - -, f ( [-, ] scl: b [-, ] scl: end behavior: f ( = -, lim - lim end behavior: f ( =, lim - lim f ( = - f ( = -, - -, f ( From the graph, it appears that as - and, h (. h ( -, undefi ned. -,. - The table supports this conjecture.. n + a. f (n = n b. [, ] scl: b [, ] scl: c. $; Sample answer: As n approaches, f (n approaches. Therefore, the average cost approaches $ as the number of shirts sold increases. 7a. Sample answer: If n is even, f ( approaches as approaches - and. f( = + + [-, ] scl: b [-, ] scl: discontinuit: infinite at = -, =, and = ; end behavior: lim f( =, lim f ( = ; zeros: = b. Sample answer: If n is odd, f ( approaches - as approaches - and as approaches. [-, ] scl: b [-, ] scl: f( = + + R Selected Answers

7 9. Infinite; f ( is undefined, and f ( approaches - as approaches from the left and as approaches from the right. The graph is continuous at =, then b substitution: ( + a = (b + a = 9 + a = b + a Simplif. 9 = b Subtract. b = Divide. The graph is continuous at = -, then b substitution: (-b + a = -b + = - (- + a = - + b = -9 + a = Simplif. a = 9 Subtract.. lim f ( = ; because f is odd, f (- = -f(.. lim - - f ( = ; with -ais smmetr, f ( = f (-. 7. [-, ] scl: b [-, ] scl: 9. no zeros [-, ] scl: b [-, ] scl: Even; the graph of h( is smmetric with respect to the -ais. h(- = (- - = - = h( 7. D = (-, - ( -, + ( +, 7a., 7b. 79 7c. 999,9, infinite solutions 79. (,, - -. J. D Lesson -. f is increasing on (-, -., decreasing on (-.,, and increasing on (,.. f is decreasing on (-,. and increasing on (.,.. f is increasing on (-, and increasing on (,. 7. f is increasing on (-, -, decreasing on (,, and increasing on (,. 9. f is constant on (-, -, increasing on (-, -., and decreasing on (-.,. a. ( = b. about.9 ft f ( Because f (-. > f (- and f (-. > f (-, there is an absolute maimum in the interval (-, -. The approimate value of the absolute maimum is f ( Because f ( < f (-. and f ( < f (., there is a relative maimum in the interval (-.,.. The approimate value of this relative minimum is.. f ( Because f (. > f ( and f (. > f (, there is also an absolute maimum in the interval (,. The approimate value of the absolute maimum is.9.. abs. min: (-.7, -.; rel. ma: (.,.; rel. min: (.9, rel. min: (-, -; rel. ma: (.,. 9. rel. min: (., -.; rel. ma: (,. rel. ma: (-., -.; rel. min: (., -.. rel. ma: (.,.; rel. min: (-., -.. abs. min: (-., rel. ma: (.,.; rel. min: (-.7, rel. ma: (-.,.; rel. min: (, -. rel. ma: (-.,.; rel. min: (.9, -.. r =. in.; h =.7 in a. f ( = and f ( =.. - m = Rate of change - =. or. million pounds more per ear b. f ( =.9 and f ( = m = Rate of change - =.7 or.7 million pounds more per ear 9a. [, ]: ; [, ]: ; [, ]:. 9b. The object is increasing in speed, or accelerating, along all three intervals. It is accelerating at the fastest rate for the interval [, ]. While it is ver slowl accelerating for the interval [, ], it is still increasing its speed. 9c. Steep graph = high magnitude rate of change = rapidl increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing... t. Your etrema values should be close to the actual etrema values given. It appears that f ( has an absolute maimum at = -., a relative minimum at =, and an absolute maimum at =.. It also appears that lim f ( = - and lim f ( - =-, so we conjecture that this function has no absolute minimum. a.,,,, O I( = For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R7

8 b. about $.9; Sample answer: This value represents the average change in net personal income in the U.S. from to 7. c. The rate of change was lowest from to at about $. and highest from to 7 at about $7.... in.. in.. in.; Sample answer: The function for the surface area of the bo in terms of the length of one side of the base is f(w = w +,9 w. [, ] scl: b [, ] scl: Graph the function. The absolute minimum occurred when w =. in.. Sample answer: 7. Sample answer: f( 9. Sample answer:. (-, -; maimum. (., ; minimum. (, -; minimum f ( = 7. Sample answer: One reason for the variation in average rate of change ma be that Simeon s famil encountered traffic later in their trip, thus slowing them down. Another ma be that Simeon s famil traveled on residential roads for the first hour before entering a highwa for the net three hours. The two instances on the graph where Simeon s famil appeared to not travel an distance ma be as a result of the famil needing to stop to eat or take a break. The ma have also encountered an accident where traffic was completel stopped. 9. Sample answer: ; Sample answer: When a function is constant on an interval, the -coordinates of each point in the interval are the same. Because f ( is constant on the interval [a, b], f (a = f (b. f (a - f (b m = a - b = a - b or So, the slope of the secant line is zero. f( 7. There are an infinite number of relative maima and minima. The relative maimum is and occurs at { = 9 + n, n }. The relative minimum is - and occurs at [-, ] scl: 9 b [-, ] scl:. { = 7 + n, n }. 7. Sample answer: When a function is increasing on an interval, the average rate of change is positive. When a function is decreasing on an interval, the average rate of change is negative. When a function is constant on an interval, the average rate of change is zero. 77. Continuous; f ( =, lim f ( =, and lim f ( = f (. 79. Even; the graph of f ( is smmetric with respect to the -ais. f (- = (- = = f (. [-, ] scl: b [-, ] scl: [-, ] scl: b [-, ] scl: neither; g(- = (- - + = - +. (-, -] [,. (,, 7. -, 9. 7, i 9. or amps 9. G 97. G Lesson -. D = { }, R = { }. The graph has a -intercept at (, and -intercepts for {, }. The graph has no smmetr. The graph has a jump discontinuit for { }. lim - f ( = - and lim f ( =. The graph is constant for { }. The graph increases for { }.. D = { }, R = { }. The graph has an intercept at (,. The graph is smmetric with respect to the origin. The graph is continuous. lim f ( = - and lim f ( =. The graph is - increasing on (-,.. D = { }, R = { = c, c }. If c =, all real numbers are -intercepts. If c, there are no -intercepts. The graph has a -intercept at (, c. If c, the graph is smmetric with respect to the -ais. If c =, the graph is smmetric with respect to the -ais, -ais, and origin. The graph is continuous. lim - f ( = c and lim f ( = c. The graph is constant on (-,. 7. f( g( 9. f( g( R Selected Answers

9 . g( f(. g( f(... The graph of g( is the graph of f ( translated units to the right and g( = -, or translated units down and g( = The graph of g( is the graph of f ( reflected in the -ais and translated units right when g( = -, or reflected in the -ais and translated units up when g( = Because there is a two-month dela, d( is the graph of p( translated two units to the right. Therefore, d( = ( ( - + ( - +. The graph of g( is the graph of f ( translated units down; g( = -.. The graph of g( is the graph of f ( translated unit to the right and units down; g( = f ( = ; the graph of g( graph of f ( translated units to the left and epanded verticall. is the f( g( 7. 9c. [, 7] scl: b [, ] scl:. f( 9a. The graph of c ( is the graph of f ( compressed verticall and translated.99 units up. 9b. c ( = d. Yes; the plans will equal each other at minutes. g( 7. f ( = ; the graph of g( graph of f ( translated f( units to the right and epanded verticall. is the g( 9. f ( = ; g( is the graph translated units to the left, f( epanded verticall, and reflected in the -ais.. f ( = ; g( is the graph of translated units to the left and compressed verticall. g( g( f( of f ( f ( h(. f( g( h( For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R9

10 . f( h( g(. translated one unit to the left; translated one unit down. translated units to the left; epanded verticall; translated 7 units down. g( = + 7. g( = a. g( =.f ( 9b. There is no affect on f (. While the opening is delaed, the number of das after the opening, which determines the domain of the function, is unaffected. 9c. g( = f ( -.. f ( = ; Sample answer: The graph of g( is the graph of f ( translated units to the right, epanded verticall, reflected in the -ais, and translated units up.. f( = ; Sample answer: The graph of g( is the graph of f ( translated units to the right, reflected in the -ais, and translated units up a. f ( =. +. if =.+ +. if < 7b. Tai Cost Price (dollars c. The graph of f ( is translated. unit up. 7 9 Fare Units 9. A vertical shift of is f ( + or +. A horizontal shift of -7 is f ( - (-7 or + 7. A vertical epansion b a factor of is f ( or ( = Combining the attributes, g( = f ( - (-7 + or Price (dollars g( = Tai Cost 7 9 Fare Units g( = g( = g( = g( = Sample answer: Both; for the greatest integer function, a shift of a units left is identical to a shift of a units up. 7. Sample answer: Order is important because different graphs can be obtained depending on the order the transformations are performed. For eample, if (a, b is on the original graph and there is a translation units up and then a reflection in the -ais, the resulting point on the transformation will be (a, -b -. However, if (a, b is reflected in the -ais first and then translated units up, the resulting point will be (a, -b Sometimes; sample answer: f ( = is an odd function and f (- - f ( when = -. However, f ( = is an odd function and f (- = - f ( for all. 79 Sample answer: There are an infinite number of functions that meet these criteria. An values of a, b, and c for g( = a + b + c are valid such that f (- = -. To simplif this process, let a =. f ( = a + b - c - = - + b - c c - = - + b (c - = - + b (c - + = b An values of b and c that satisf this equation will be valid. If c =, b = and g( = ; Sample answer: As, the denominator of the fraction will increase and the value of the fraction will approach, so g( will approach. 7. ; Sample answer: As, the fraction will get closer and closer to, so p( will approach. R Selected Answers

11 9. -intercept = ; zeros: -,, ; - - = ( - - = ( - ( + = = or - = or + = = = - 9. about. 9. B 9. G Lesson -. ( f + g( = + + ; D = [, ; ( f - g( = - + ; D = [, ; ( f g( = + ; D = [, ; ( f g ( = + ; D = (,. ( f + g( = + + ; D = (-, ; ( f - g( = + + ; D = (-, ; ( f g( = ; D = (-, ; ( f g ( = + ; D (-,- (-,. ( f + g( = + ; D = (-, ; ( f - g( = - ; D = (-, ; ( f g( = ; D = (-, ; ( f g ( = + 9 ; D = (-, (, 7. ( f + g( = + + ; D = (-, (, ; ( f - g( = ; D = (-, (, ; ( f g( = + ; D = (-, (, ; ( f g ( = + ; D = (-, (, 9. ( f + g( = - ; D = (, ; ( f - g( = - ; D = (, ; ( f g( = ; D = (, ; ( f g ( = ; D = (,. ( f + g( = ; D = [-, ; (f - g( = ; D = [-, ; ( f g( = ; D = [-, ; ( f g ( = ; D = [-, (, a. ( f + g( = - + ; b. ( f + g( represents all factors that influence the budget. c. $7; The budget for weeks.. [ f g]( = - 9; [ g f ]( = - ; [ f g]( = 9 7. [ f g]( = ; [ g f ]( = ; [ f g]( = - 9. [ f g]( = ; [ g f ]( = ; [ f g]( = -7,. [ f g]( = for ±. [ f g]( = -. [ f g]( = - for < 7. [ f g]( = a. Since the denominator cannot be zero, - c. Thus, v v c. Because velocit cannot be negative, D = {v v < c, v }. b. m( = = - ( m(, = =., - ( m(,, = =. - ( ( v c. lim m(v =, since as m approaches c, approaches, v c c making - v ver small. divided b a ver small c number results in a ver large quotient. d. Sample answer: Let g(v = - v and f (v =. c v m(v = f [g(v] = - v c. Sample answer: f ( = - ; g( = +. Sample answer: f ( = -; g( = - 9. Sample answer: f ( = ; g( = + 7. Sample answer: f ( = ; g( = - 9. Sample answer: f ( = + ; g( = - a. h[ f (]; The bonus is found after subtracting, from the total sales. b. $. Sample answer: f ( = - + ; g( = -. Sample answer: f ( = + + ; g( = 7. Sample answer: f ( = ; g( = + 9. f (. = -.7, f (- =, f ( + = + +. f (. = -., f (- = -, f ( + = The domain of h( = + is {, } and of g( = - is all real numbers. Therefore the domain of g h is {, }. [g h]( = g[(h(] = g( + = [( + - ] = [ ] = + + Since the domain of f ( = + is all real numbers, the domain of f g h is {, }. [ f g h]( = f ( g[(h(] = f ( + + = ( = + + Therefore, [f g h]( = + + for.. [ f g h]( = + for 7a. g( = + 7b. g( = + 9a. g( = 9b. g( =... 7a. {m m >, m }; The molar mass of the gas cannot be negative or zero. 7b. about 7. m/s 7c. The velocit will decrease. 7d. Sample answer: v(m = f [ g(]; f (m = m ; g(m = (.9( m 9. Sample answer: f ( = ; g( = + ; h( = - 7. Sample answer: f ( = ; g( = + ; h( = + 7. [ f g]( = for - 9; [ g f ]( = [ f g]( = 9 - for - ; [g f ]( = 9 - for [ f g]( = for and ; [g f ]( = for - and 79.. h( f( O (h - f( g( h( (h + g( For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R

12 . { - - or, }. {, } 7. odd 9. even 9. Sample answer: f ( =, 9. Sample answer: This can occur when the range of g( is a subset of the domain of f (. For eample, when f ( = and g( = +, the range of g( is all real numbers greater than or equal to. This range is a subset of the domain of f (, which is all real numbers greater than or equal to. Thus, f [ g(] eists for all real numbers. 9 The domain of h( = is {, }. To evaluate g h, ou must be able to evaluate g( = + for each of these h(-values, which is onl possible when h( -. That is, when -. Since we do not know if is positive or negative, we cannot multipl both sides of this inequalit b to solve for. Instead, rewrite the inequalit as -. When = -, + = and when =, + is undefined. Test intervals to determine the solutions of + are therefore -, - <, and >. Test points in these intervals to determine the solutions of the inequalit. Since + is true for = - and =, but not for = -, the solution is - or >. Therefore the domain of g h is { - or, }. To find f g h, ou must be able to evaluate f( =, for each of these g[h(]-values, - which can onl be done for g[h(] -. This means that we must eclude from the domain those values for which + = -. + = - + = = + = = Therefore, the domain of f g h is { - or < < or > }. In interval notation, the domain is (-, -] (, (,. 97. Sample answer: Order is important because there usuall will be different outputs for f [ g(] and g[ f (]. For eample, if f ( = and g( = +, f [g(] = + + and g[ f (] = [-, ] scl: b [-, ] scl: rel. ma: (, ; rel. min: (,. [-, ] scl: b [-, ] scl: abs. min: (-.7, ,.7 a. RBI HR b. D = {,, }, R = {7,,,, } c. No; each of the domain values and is paired with two different range values. 7. G Lesson -7. no. no. es 7. es 9. es. es. no. no 7. no 9. es; f - ( = + ; -. es; f - ( = ; >. es; f - ( = + ;. no -( - 7. f [ g(] = + = = = ; - (- + g[ f (] = = ( = 9. f [ g(] = ( - = + = - + = ; g[ f (]= - (- + = = =. f [ g(] = ( + ( + - = - = + - = ; g[ f (] = - + = = =. f [ g(] = ( = = + - = ; g[ f (] = = = ( + - = + =. f [ g(] = + = + = + = = ; g[ f (] = = = + + = 7 a. If =.m, the inverses comes from = m. = m Original equation. = m = m Divide. Take the square root and reject the negative root. g( = m Replace with g (. g( = velocit in m/s, = kinetic energ in J and m = mass in kg. b. f [ g(] = f ( m = m ( m = m ( m = ; g[f(] = g m ( m = ( ( m = = ; because f [g(] = g [ f (] =, the functions are inverses. However the are not inverse functions until the domain of f( is restricted to D = [,. (Continued on the net page R Selected Answers

13 c. [, ] scl: b [, ] scl: 9.. = g - (. = g - ( = g( = g( = g( = g - ( a. Sample answer: The graph of the function is linear, so it passes the horizontal line test. Therefore, it is a one-to-one function and it has an inverse; f ( =.. b. represents the value of the currenc in U.S. dollars, and f ( represents the value of the currenc in Euros. c. ; You cannot echange negative mone. d.. 7. f - eists. 9. f - eists.. f - does not eist.. f - eists. 7 f - ( Since the verte of the parabola is at (,, either or will make the inverse one-to-one. Choose. = ( - Original function. = - Interchange and. + = Add to each side. Thus, f - ( = Sample answer:, f - ( = + 9. f: D = {, }, R = {, }; f - : D = {, }, R = {, }. f: D = {, }, R = {, }; f - : D = {, }, R = {, } a. Sample answer: f ( = b. f ( = 7. ; represents the number of nesting pairs 7. and f ( represents the number of ears after 9. c. 99. f ( = f - ( - if. -. if > f( 7a. v - ( = ; represents the formula for the flow rate of the gas. 7b. ft s 7c.. ft s 9. [ f - g - ]( = + 7. [f g ] - ( = 7. (f g - ( + 9 -, [ f - g - ]( = + for 77. [ f g ] - ( = + for 79. ( f g - ( = + + for a. b. D = { }, C( =... Cost (dollars c. Time (minutes 7 9 Time (minutes Cost (dollars R = { positive multiples of.} d. D = { positive multiples of.}; R = { } e. The inverse gives the number of possible minutes spent using the scanner that costs dollars. The inverse of (, is (,. Thus, if f contains a zero at, (, is a point on the graph. B definition of an inverse, the graph of the inverse must contain (,, which is its -intercept.. Sample answer: False; constant functions are linear, but the do not pass the horizontal line test. Therefore, constant functions are not one-to-one functions and do not have inverse functions. 7. Sample answer: Yes; one function that does this is f ( =. Even though both limits approach, the do it from opposite sides of and no -values ever share a corresponding -value. Therefore, the function passes the horizontal line test. 9. Sample answer: If f ( =, f does not have an inverse because it is not one-to-one. If the domain is restricted to, then the function is now one-to-one and f - eists; f - ( =. 9. [ f g]( = - for { }; [g f ]( = - for { } 9a. epanded horizontall 9b. translated units to the right and units down 9c. epanded verticall, translated units up 9a. compressed horizontall 9b. translated units to the right 9c. compressed horizontall, translated units down 97. (, -, 99. (, -7,. D. A Chapter Stud Guide and Review. true. true. false; end behavior 7. true 9. true. function. function. 7. D = { } 9. D = {a a -, a }. D = [-, ], R = [, ]. -9; 9. ;,, - 7. continuous at = ; The function is defined when =. The function approaches when approaches from both sides, and f ( =. 9. continuous at = ; The function is defined when =. For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R

14 The function approaches when approaches from both sides, and f ( =. continuous at = 7; The function is defined when = 7. The function approaches. when approaches 7 from both sides, and f (7 =... continuous at = ; The function is defined when =. The function approaches when approaches from both sides, and f ( =.. From the graph, it appears that as, f ( ; as -, f (.. f is decreasing on (-, -, increasing on (-, -., decreasing on (-.,., and increasing on (., ; rel. min. at (-,, rel. ma. at (-., and rel. min. at (., f ( = ; g( is the. f ( = ; g( is the graph of f ( reflected in the -ais and translated graph of f ( compressed verticall b a factor units to the right and of and translated units down. units up. + f( = g( =-( - - f( =. The graph is reflected in the -ais and translated units right and unit up; g( = ( f + g( = + - ; D = (-, ; (f - g( = - ; D = (-, ; (f g( = ; D = (-, ; ( f g ( = - - ; D = (-, (, 7. (f + g( = + ; D = (-, (, ; (f - g( = - ; D = (-, (, ; ( f g( = ( ; D = (-, (, ; f g ( = ; D = (-, (, ; + + ;. [f g]( = for 9. no. es f - ( = + 9. h - ( = -, a. f( = b. $9.99; $9.99 c if ( - if >. No; sample answer: At the time of her promotion, her income had a jump discontinuit.. 7a. A( =. cm 7 7b. A - ( =. in v(t = (9.t + 9 Speed (m/s Chapter Time (s Connect to AP Calculus.. Sample answer:. Sample answer: The secant line appears to become a line tangent to the graph of f. 7. Sample answer: The eact rate of change is which is equivalent to what the other values appeared to be approaching. 9. Sample answer: To find the rate of change of a function f ( at the point, find the difference quotient, and simplif the epression. Substitute h = into the simplified epression, and the result is the rate of change of the function at. CHAPTER Polnomial and Rational Functions Chapter Get Read. ( - ( +. ( - ( - 7. ( - ( f( =- + - f( = -.. {, }; (-, ] 7. { - < < 9, }; (-, 9 9. { < - or >, }; (, - (,. { 9.99 < < 9.99, f( = - - }; (9.99, 9.99 Lesson -. f( = D = (-,, R = [, ; intercept: ; lim f( = and lim f( = ; - continuous for all real numbers; decreasing: (-, ; increasing: (, O R Selected Answers

15 . h( =- D = (-,, R = (-, ; intercept: ; lim f ( = and lim f ( = - ; - continuous for all real numbers; decreasing: (-, 7b. V(r V(r = r r. g( = 9 D = (-,, R = (-, ; intercept: ; lim f ( = - and lim f ( = ; - continuous for all real numbers; increasing: (-, 9. f( =- D = (-,, R = (-, ; intercept: ; lim - f ( = and lim f ( = - ; continuous for all real numbers; decreasing: (-, 7. f( = f( = - D = (-,, R = (-, ; intercept: ; lim f ( = and lim f ( = - ; - continuous for all real numbers; decreasing: (-, D = (-, (,, R = (, ; no intercepts; lim f ( = and lim - f( = ; infinite discontinuit at = ; increasing: (-, ; decreasing: (,. f( = -. h( = D = (,, R = (, ; no intercepts; lim f ( = ; continuous on (, ; decreasing: (, D = (-,, R = (-, ; intercept: ; lim f ( = - and - lim f ( = ; continuous for all real numbers; increasing: (-,. f( =- - D = (-, (,, R = (-, (, ; no intercepts; lim f ( = and lim - f ( = ; infinite discontinuit at = ; increasing: (-, and (,. f( = - D = (-, (,, R = (, ; no intercepts; f ( = and lim - lim f ( = ; infinite discontinuit at = ; increasing: (-, ; decreasing: (,. f( = -9 D = (-, (,, R = (-, (, ; no intercepts; lim - f ( = and lim f ( = ; infinite discontinuit at = ; increasing: (-, and (, 7. h( =- 7 D = [,, R = (-, ]; intercept: ; lim f ( = - ; continuous on [, ; decreasing: (,. h( = - D = (-, (,, R = (-, (, ; no intercepts; lim f ( = and lim - f ( = ; infinite discontinuit at = ; decreasing: (-, and (, 7a. D = (,, R = (, For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R

16 9. h( = - D = (-, (,, R = (, ; no intercepts; f ( = and lim - lim f ( = ; infinite discontinuit at = ; increasing: (-, ; decreasing: (, D = 7,, R = [, ; no intercepts; lim f ( = ; continuous on 7, ; increasing: 7, 9. a. b. =.77.7 c. about,9,9 f( =- ( a. [, ] scl: b [, ] scl: [, ] scl: b [, ] scl:. g( =- + b. f ( = c. about 9. F D = (-,, R = (-, ; -intercept: -, -intercept: -; lim f ( = and lim f ( = - ; continuous for all real - numbers; decreasing: (-, 7 Evaluate the function for several -values in its domain. 7 7 f ( Use these points to construct a graph. h( = Since it is an even-degree radical function, the domain is restricted to nonnegative values for the radicand, 7 =. Solve for when the radicand is to find the restriction on the domain. = 7 - = 7 7 = D = (-,, R = (-, -9.]; -intercept: -.; lim f ( = - - and lim f ( = - ; continuous for all real numbers; increasing: (-,.; decreasing: (.,. g( = D = [, ], R = [-, ]; -intercept:.; continuous on [, ]; decreasing: (, a. about.9 Mcal b. about 7. lb. no solution ,.. no solution. 7. Yes; sample n answer: The function follows the form f ( = a, where n is a positive integer. In this case, a = a and n =. 9. Yes; sample n answer: The function follows the form f ( = a, where n is a positive integer. In this case, a = 7 and n = ab.. No; sample answer: The function is not a monomial function because the eponent for is negative a. Volume vs. Pressure Pressure (atmospheres P(v v Volume (liters b. Use the power regression function on the graphing calculator to find values for a and n. P(v =. v - c. Sample answer: Yes; the problem states that the volume and pressure are inversel proportional, and in the power function, the eponent of the volume variable is -.. R Selected Answers

17 d. Graph the regression equation using a graphing calculator. To predict the pressure of the gas if the volume is. liters, use the CALC function on the graphing calculator. [, ] scl:. b [, ] scl:. Let v =.. The pressure of. liters of the gas is about. atmospheres. e. Graph the regression equation using a graphing calculator. To predict the pressure of the gas if the volume is liters, use the CALC function on the graphing calculator. [, ] scl:. b [, ] scl:. Let v =. The pressure of liters of the gas is about. atmospheres. 7. c 7. d 7 The function is undefined at =, therefore; it is a power -n function of the form f ( = a or f ( = a n, where a is a real constant and n is a natural number. Start b assuming n = and solve for a b substituting a set of points (, for and. f ( = a n = a = a = a When n =, f ( = or for the point (,. If this power function is true for the second point (-, -, then f ( can represent the graph. f ( = - = - - = - Since f ( is true for (-, -, f ( = or - is a power function for the graph. 77. f ( = 79a. 7.9 mm 79b. 7. mm 79c. increase. n 7 -n = n ( 7 -n ( = n 7 -n = (n (-n = n + 7 = n + n + = n + n + n + = n +. Sample answer: As n increases, the value of n approaches. This means that the value of n will approach when is positive and - when is negative. Therefore, for positive values of, f ( will approach + or and will resemble the line =. For negative values of, f ( will approach B - + or and will resemble the line =. 7a. r = - + 7b..% 9. ( f + g( = + -, D = (-, - + (-, ; ( f - g( = , D = (-, - + (-, ; ( f g( = - ; D = (-, - (-, ; ( f g ( =, D = (-, - (-, (, O O 9. f( h( O f( O h( g( O g( O 9. f is increasing on (-, - and increasing on (-,. 97. ( - + ( + i i. J. H Lesson -. f( = ( +. f( = (. f( = - For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R7

18 Volume (gallons v(t 9 7 f( = ( - + Draining Water v (t = ( t.... t ( - 9 f( = Time (minutes. The degree is, and the leading coefficient is. Because the degree is even and the leading coefficient is positive, lim f ( = and lim f ( =.. The degree is, and the - leading coefficient is -. Because the degree is even and the leading coefficient is negative, lim f ( = - and lim - f ( = The degree is, and the leading coefficient is -. Because the degree is odd and the leading coefficient is negative, lim f ( = and lim f ( = The degree is, and the leading coefficient is -. Because the degree is even and the leading coefficient is negative, lim f ( = - and lim f ( = -.. The degree is, and the - leading coefficient is. Because the degree is even and the leading coefficient is positive, lim - f ( = and lim f ( =. The degree of f ( is, so it will have at most five real zeros and four turning points. = + + = ( + + = ( + ( + So, the zeros are -, -, and.. real zeros and turning points; ± 7. real zeros and turning points; and - 9. real zeros and turning points;, -, and. real zeros and turning points;, -, and a. The degree is, and the leading coefficient is. Because the degree is even and the leading coefficient is positive, lim f ( = and lim f ( =. - b., -, (multiplicit: c. Sample answer: (-,, (-, -, (.,., (, d. a. The degree is, and the leading coefficient is -. Because the degree is even and the leading coefficient is negative, lim - f ( = - and lim f ( = -. b., - (multiplicit: -, c. Sample answer: (-, -, (-, -, (,, (, - d. f( = ( + ( 7a. The degree is, and the leading coefficient is -. Because the degree is odd and the leading coefficient is negative, lim - f ( = and lim f ( = -. 7b.,, - (multiplicit: 7d. 7c. Sample answer: (-,, (-, -, (,, (, - f( = ( ( + 9a. The degree is, and the leading coefficient is. Because the degree is odd and the leading coefficient is positive, lim f ( = - and lim f ( =. 9b.,, - - 9c. Sample answer: (-, -9, (-,, (, -, (, 9d. f( = a. The degree is, and the leading coefficient is. Because the degree is even and the leading coefficient is positive, lim f ( = and lim f ( =. b. (multiplicit:, -, - c. Sample answer: (-,, (-, -7, (, -, (, d. f( = ( + ( - f( = + a. Sample answer: f ( = b.. ft. Sample answer: f ( = Sample answer: f ( = a. According to the data, the values are increasing as increases, therefore, lim f ( =. R Selected Answers

19 9b. Sample answer: f ( = ; The line is not a good fit. There are man outling points. [, ] scl: b [, ] scl: 9c. The leading coefficient is., so lim f ( =. Sample answer: The prediction was accurate because the leading coefficient is positive, as, f (.. es. No; undefined at =. Sample answer: f ( = Sample answer: f ( = Sample answer: f ( = +. Sample answer: f ( = - -. Sample answer: f ( = n is odd; a n is positive. 7. n is even; a n is negative. 9. f ( = f ( = f ( = a. Sample answer: Enter (,., (,, (,., (, 7., (, 7.9, (, 79. into STAT EDIT; the linear regression (r =.97 is f( =. +.9 b. Find f ( for = since is ears since 9. Graph the regression equation using a graphing calculator. Use the CALC feature on the calculator to find f (. f( = ( ( ( + ( + a. degree = b. -, - (multiplicit:, c. f ( =. ( + ( + ( - a. degree = b. -, (multiplicit:, - c. f ( = ( - ( + ( +. real zeros and turning points; no real zeros 7 The degree of the function is, so f has at most distinct real zeros and at most - or turning points. To find the real zeros, solve the related equation f ( = b factoring = - ( - + = - ( - ( - = - ( - = The epression above has factors, but solving for ields onl distinct real zeros: = and =. Both of the zeros have multiplicit. 9a. g( f( f( g( h( h( [-, ] scl: b [-, ] scl: [-, ] scl: b [-, ] scl: g( f( g( [, 7] scl: b [, ] scl: f ( = 7.%. c. Sample answer: If f ( =, then =. +.. =. 9. =. So 9. ears after 9, about, % of the population will live in metropolitan areas. 77. Sample answer: f ( = -( + ( + ( - ( - ( + f( = ( + ( + ( ( ( Sample answer: f ( = -( - ( - ( + ( + f( h( [-, ] scl: b [-, ] scl: h( f( g( [-, ] scl: b [-, ] scl: h( [-, ] scl:. b [-, ] scl:. h( g( f( [-, ] scl:. b [-, ] scl:. 9b. Sample answer: Ver close to the origin, the function approimates the behavior of the term of lower degree, or h(. 9c. Sample answer: As and, the function approimates the behavior of the term of higher degree, or g(. 9d. Sample answer: As and, the function approimates the behavior of function a and, ver close to the origin, the function approimates the behavior of function b. 9. Sample answer: No; a polnomial function f ( cannot have both an absolute maimum and absolute minimum because as, f ( will approach either or -. If f ( as, then an absolute maimum is impossible. If f ( - as, then an absolute minimum is impossible. 9. Rearranging the terms gives f ( = Notice how the first set of three terms has a common factor of and the second set of three terms has a For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R9

20 common factor of. After factoring using the Distributive Propert, f ( = ( ( - -. Now notice how the factors inside the parentheses are identical. Using the Commutative Propert, f ( = ( + ( - -. After factoring the second factor, f ( = ( + ( - ( +. The function is now completel factored and the zeros of the function are -,, and -. These were determined b setting each factor equal to zero and solving for. 9. Sample answer: There is one turning point at the absolute maimum, one at the relative minimum, and one at the relative maimum. Therefore, the minimum degree is + or no solution The graph of g( is the graph of f ( reflected in the -ais and translated units left and units down; g( = -( a. $ 7b. Sample answer: The compan s competition might offer a similar product at a lower cost F. H Lesson -. ( + ( - ( +. ( + ( + ( -. -( - ( + ( - 7. ( + 7( - ( Because -, c =. Set up the snthetic division as follows, using a zero placeholder for the missing -term in the dividend. Then follow the snthetic division procedure The quotient is ( 9.,, students. 7.,. -7, , 9. no, no. es, es; f ( = ( - ( - ( - ( +. no, no. es, no; f ( = ( + ( seconds 9. ( n + n - ( n -. ( n - ( n + ( n a. v( = b. Volume (ft v( Volume of Cla v( = Removed Amount (ft 7c. = d. about. ft 9. f ( = ( + ( - ( 7( + Using the Factor Theorem, ( - is a factor if f ( =. f ( = ( - ( + ( - ( + = ( - ( + ( - ( + = = Since f ( =, ( - is a factor of the polnomial.. True; sample answer: The Remainder Theorem states that if h( is divided b - (-, then the remainder is r = h(-, which is even; negative 7. odd; negative 7a. h = v 7b. Yes; the pump can propel water to a height of about ft. 7. (-, 77. (, 79. (-, -. K. H Lesson -. ±, ±, ±, ±, ±, ±, ±9, ±, ±, ±, ±, ±, ±, ±, ±, ±, ±9, ±;,,, -. ±, ±, ±, ±, ±, ±, ±, ±; -,, -,. ±, ±, ±, ±, ±, ±, ±, ±, ±, ±, ±, ±, ±, ±, ±, ±, ±, ±, ±, ±, ±, ±, ±, ± ;, 7. ±, ±, ±, ±, ±, ±, ±9, ±, ±, ±, ±, ±7, ±, ±, ±, ±7, ±, ±, ±, ±; (multiplicit:, 9. in., in., 9 in.. -,, - (multiplicit:., -, (multiplicit:.,, - (multiplicit: 7. das 9. Sample answer: [-, ];,, -, 7]; -,,, -,. Sample answer: [-,. Sample answer: [-, 7]; (multiplicit:,. Sample answer: [-, ]; (multiplicit:, -, (multiplicit: 7. or positive zeros, negative zero 9. or positive zeros, or negative zeros., or positive zeros, negative zero. Sample answer: f ( = Sample answer: f ( = Sample answer: f ( = Sample answer: f ( = Sample answer: f ( = a. g( has possible rational zeros of ±, ±, ±, ±. B using snthetic division, it can be determined that = - is a rational zero B using snthetic division on the depressed polnomial, it can be determined that = - is a rational zero The remaining quadratic factor ( - + ields no rational zeros. Use the quadratic formula to find the zeros. -b ± b - ac -(- ± (- = - (( a ( = ± = ± (Continued on the net page R Selected Answers

21 So, g( written as a product of linear and irreducible quadratic factors is g( = ( + ( + ( + ( - -. b. g( written as a product of linear factors is g( = ( + ( + ( - + ( - -. c. The zeros are -, -, -, +. a. f( = ( + ( - ( - b. f( = ( - ( - ( - + i( - - i c.,, - i, + i 7a. h( = ( + ( - ( + ( - 7b. h( = ( + ( - ( + ( - 7c. -,,, - 9., -,, i, -i; h( = ( - ( + ( - ( + i ( - i. -, -,, + i, - i; g( = ( + ( + ( - ( - + i( - - i.,, -, i, -i; f( = ( - ( + ( - ( + i( - i a. V(l = l - l b. = l - l c. base: in. b in.; height: in. 7. Sample answer: f ( = - 9. Sample answer: f ( = +. g( = ( - ( - ( + i ( - i;,, ±i. -. no rational zeros 7 Solve d =. ( for d =... =. ( =. (. Use a graphing calculator to graph =. (. and solve for =. Using the CALC menu, find the zeros of the function. Lesson -. D = {, -, }; =, = -, =. D = {, -, }; =, = -. D = {, -, }; =, = -, = 7. D = {, -, }; =, = -, = 9. asmptotes: = -, =, = ; -intercepts: -, ; -intercept: ; D = { -,, } f( = ( + ( - ( + ( - = = -. asmptotes: = -, =, = ; -intercept: -; D = { -,, }. asmptotes: = -, =, = ; g( = -intercept: -; -intercept: - ; D = { -,, } = = = - ( + ( + ( + ( - f( = ( - ( + = = - = = [, ] scl: b [-, ] scl:. The zeros of the function occur at = and =.. A vertical deflection of. feet will occur when a weight is placed feet or about. feet from the left piling. 9. ( + ( ( + ( Julius; sample answer: Angie did not divide b the factors of the leading coefficient. 7a. The graphs of -f( are the graphs of f ( reflected in the -ais. The zeros are the same. 7b. The graphs of f (- are the graphs of f ( reflected in the -ais. The zeros are opposites. 77. False; sample answer: The third-degree polnomial has three real zeros and no nonreal zeros. 79.,,. -, -, -. Sample answer: If a polnomial has an imaginar zero, then its comple conjugate is also a zero of the polnomial. Therefore, an polnomial that has one imaginar zero has at least two imaginar zeros The degree is, and the leading coefficient is. Because the degree is even and the leading coefficient is positive, lim - f ( = and lim f ( =. 9. It appears that f ( has a relative minimum of - at = and a relative maimum of at = It appears that f ( has a relative maimum of at = and = - and a relative minimum of - at = - and - at =. 9. E 97. A. asmptotes: = -, = -; -intercepts: -, = -, ; -intercept: ; D = { -, -, } 7. asmptote: = ; -intercept: ; -intercept: - ; D = { } 9a. D = {z z, z } 9b. horizontal asmptote: = ; -intercept: ; -intercept: - h( = ( - ( = - = - f( = + + For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R

22 9c. Car Wash Profit (thousands p(z - p(z = z z + 7z + Week. asmptotes: = -, = ; -intercept: ; D = -, h( = - + = + = -. asmptotes: =, = ; -intercept: -7; -intercept: - 7 ; D = {, } = g( = = z = - g( = = 9. asmptotes: = -; = ; hole: (, ; -intercept: - ; -intercept: ; D = { -,, } = - = g( = ( + ( - ( - ( + a. = d + b. cm. ±. no i solution Sample answer: ( - f( = ( - ( - a. There is a vertical asmptote at the real zero of the denominator r =. There is a horizontal asmptote at r = or r =, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polnomials are equal.. asmptote: = ; -intercepts: -, -, ; -intercept: ; D = {, } g( = = ( - ( + 7 g( can be written as g( = ( - ( + ( + or +. The function is undefined at b( =, so D = { -, -,, }. There is a vertical asmptote at = -, the real zero of the simplified denominator. There is a horizontal asmptote at =, because the degree of the denominator is greater than the degree of the numerator. Since the simplified numerator has no real zeros, there are no -intercepts. The -intercept is because g( =. There are holes at (-, - and (, because the original function is undefined when = - and =. Graph the asmptotes and intercept. Then find and plot points. [, ] scl: b [, ] scl: b. Evaluate the function for each value of r to determine r. r = ( r = ( = 9 = r = ( r = ( = 7 =.7 r = ( r = (7 7 = =. r 7 r c. No; sample answer: As r approaches infinit, r approaches. This suggests that the average speeds reached during the first leg of the trip have no bounds. Being able to reach an infinite speed is not reasonable. The same holds true about r as r approaches from the right. 7. Sample answer: f ( = , ,.,.7 R Selected Answers

23 a. b. c. Function f ( = + + h ( = + g ( = + Horizontal Asmptote = = = [-, ] scl: b [-, ] scl:. [-, ] scl: b [-, ] scl:. [-, ] scl: b [-, ] scl:. Function f( = + + h( = + g( = + Real Zeros of Numerator, -, d. Sample answer: When the degree of the numerator is less than the degree of the denominator and the numerator has at least one real zero, the graph of the function will have = as an asmptote and will intersect the asmptote at the real zeros of the numerator. 9. Sample answer: The test intervals are used to determine the location of points on the graph. Because man rational functions are not continuous, one interval ma include -values that are vastl different than the net interval. Therefore, at least one, and preferabl more than one, point is needed for ever interval in order to sketch a reasonabl accurate graph of the function.. ±, ±, ±, ±, ±9, ±; -. no; es; f ( = ( - ( - -. es; no; f ( = ( ( or f ( = ( ( + ( + 7. es; no; f ( = ( + ( or f ( = ( + ( f( = ( + 7 7a. ( - ( - = 9 f( = - 7b. 7c. ft b ft i ( = C 79. C Lesson -. [-, ]. -, - [,. (-, - (, 7. (-, - (, 9. (-,.. {}. {-} ,,,,; [,,,] or,, 9. (-, - + < < ( + < < < Let f ( =. The zeros and undefined points of the + inequalit are the zeros of the numerator, = -, and denominator, = -. Create a sign chart. Then choose and test -values in each interval to determine if f ( is positive or negative. (- (+ ( The solutions of < are -values such that f ( is + negative. From the sign chart, ou can see that the solution set is (-, - (-,.. -, (,. -, (, 7. (-, 7 + (, 9. < ; to people. (, ] [,. (-, -] [,. (, - (-, (, 7. [-, ] 9a. l( - l or -l + l - 9b. [, ]; The length of the plaing field is at least and at most feet. 9c. < l( - l ; (, ] [, ; Sample answer: The area of the plaing field must be greater than square feet but at most square feet, so < l or l <.. 9,. -,. + - < 7. (-, - (, 9. (-, -] -, - [, ]. (-, - (-, - (-,. (-, - (, (, a. 7 + t 9. b. 7; Sample answer: Since t 7., Jarrick will have to spend 7 minutes studing for the test. 7. units 9. (-k, k. (-,. Neither; sample answer: The epression is undefined at, but it is positive everwhere that it is defined, so the solution set is (, (,.. a < and a + b > c + d For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R

24 7 Because ( - a n is alwas an even degree, b definition, the end behavior is to approach as ±. The absolute minimum occurs at - a = or = a. Since an other value of ( - a to an even power must be positive in the reals, ( - a n > for all values of ecept when = a, since ( n > is false. Thus, the solution set is (-, a (a,. 9. Sample answer: There will be instances when - is a negative value. When this occurs, we are multipling an inequalit b a negative and the inequalit smbol will need to be reversed. The problem occurs here because - could be positive or negative. 7. D = { -, }; = - 7. about.7 cm or 9 cm 7. a 9a + 7a a., 79b., 79c. ;. F Chapter Stud Guide and Review. leading coefficient. repeated zeros. Factor Theorem 7. Vertical 9. Oblique. D = (-,, R = [, ; intercept: ; lim f ( = and lim f ( = ; - continuous for all real numbers; decreasing: (-, ; increasing: (,. f( = f( = -9 D = (-, (,, R = (-, (, ; no intercepts; lim f ( = and lim - f ( = ; infinite discontinuit at = ; decreasing: (-, and (, es; f( = ( + ( - 9. es; es; f( = ( - ( +. ±, ±, ±, ±; -. ±, ±, ±, ±, ±, ±, ±, ± ;, -.,, 7., -,, 9.,, i, -i; f( = ( - ( - ( - i( + i. D = {, -, }; =, = -, =. D = { -, -9, }; = -, = -9, =. asmptotes: = -, = ; -intercept: ; - intercept: - ; D = { -, } = 7. asmptotes: = -, =, = ; -intercepts: and -7; = = - -intercept: ; D = {,, } ( + 7 f( = ( + ( 9. =, =, =, = ; -intercepts: and -; D = {,,, } f( = + = - f( = + =. f( = D = [.,, R = [-, ; -intercept:.; f ( = ; continuous on [., lim ; increasing: (., lim f ( = - ; lim f ( = - ; - The degree is even and the leading coefficient is negative.. lim f ( = and lim f ( = ; The degree is even and the - leading coefficient is positive.. real zeros and turning points;,, and 7. real zeros and turning points; -, -,, and 9a. lim f ( = ; lim - f( = 9b. (multiplicit,, - (multiplicit 9c. Sample answer: (-,, (,, (, -, (, - 9d. f( = ( ( + R Selected Answers = = = = 7. (, 7. (, 7 (, 7a. Chapter. (-, - (, 7. (-, - 9. [,.] scl:. b [, ] scl: Connect to AP Calculus 7b. v(t =. t -. 7c. about. mi/h 7d. about.9 second 77.. f t 79. m b m or m b m. cakes.. unit s.. unit s ; Sample answer: The approimation is less than the actual area. As more rectangles are used, the approimation approaches the actual area.. Sample answer: unit s 7. Yes; sample answer: For this eample, using right endpoints for the rectangles would produce rectangles that overlap the curve and include area that lies above the curve. This would produce a higher approimation. However, this is not alwas the case. If the curve is smmetrical, like the previous eample, the approimations will be the same regardless of the

25 endpoint used. 9. Sample answer: The more rectangles that are used, the better the approimation of the area. Smaller rectangles fit the desired region better than larger rectangles, thus producing more accurate approimations. a. rectangles: unit s ; rectangles:.7 b. Sample answer: Approimating the area using or rectangles probabl does not give a good representation of the actual area. The nature of the curve prevents rectangles of widths of and. units from reall fitting well beneath it. c. Yes; sample answer: The graph is not smmetrical. Using right endpoints to determine the height of the rectangles would produce different rectangles with different areas. 9a. Elwood Ave. Oak St. Maple St.. mi 9b.. mi ft CHAPTER Trigonometric Functions Chapter Get Read.... no vertical asmptote at = -. =, =, =. no asmptotes Lesson -. sin θ =, cos θ = 7, tan θ =, csc θ = , sec θ = 9, cot θ = 7. sin θ = 9 97, cos θ = , tan θ = 9, csc θ = 97, sec θ = 97, cot θ = 9 9. sin θ =, cos θ =, tan θ =, csc θ = 9, 9 9 sec θ = 9, cot θ = 7. sin θ =, cos θ =, tan θ =, csc θ =, sec θ =, cot θ = 9. cos θ =, tan θ =, csc θ =, sec θ =, cot θ =. sin θ =, cos θ =, csc θ =, sec θ =, cot θ =. sin θ =, tan θ =, 9 csc θ = 9, sec θ = 9, cot θ =. sin θ =, cos θ =, tan θ =, csc θ =, sec θ = 7. sin θ = 77, cos θ =, tan θ = 77, csc θ = , cot θ = ft An acute angle measure and the adjacent side length are given, so the tangent function can be used to find the length of the opposite side. tan θ = opp Tangent function adj tan = θ =, opp =, and adj = tan = Multipl each side b. 7. = Use a calculator. So, the ravine is about 7. feet wide. a. 7 ft b. 7 ft. about. ft a. Because the angle of depression from the lighthouse to a ship and the angle of elevation from that ship to the lighthouse are congruent, ou can label the angles of elevation at each ship as shown. Label the distance from the lighthouse to the first ship and the distance between the two ships ft 7 Ship 7 Ship b. From the smaller right triangle, ou can use the tangent function to find the value of. tan θ = opp adj Tangent function tan 7 = θ = 7, opp =, and adj = tan 7 = Multipl each side b. = tan 7 Divide each side b tan 7..7 Use a calculator. From the larger right triangle, ou can use the tangent function to find. tan 7 = θ = 7, opp =, and adj = + + ( + tan 7 = Multipl each side b +. + = Divide each side b tan 7. tan 7 = - Subtract from each side. tan 7 = -.7 =.7 tan 7 9. Use a calculator. Therefore, the ships are about 9 feet apart. 7. B = 7, b., c P, Q 7, r.. K 9, j., k.. H =, f 9., h 7. a. ft ft b For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R

26 7. ft ft about.9 7. > 77. < 79. >. about 7.9 in. shipwreck about ft. Sample answer: For an acute angle θ of a right triangle, sin θ = opp hp, cos θ = adj hp, tan θ = opp, and cot θ = adj adj opp. opp Using these definitions, sin θ cos θ = hp adj = opp = tan θ. adj adj hp Similarl, cos θ sin θ = hp opp = adj opp = cot θ. hp. The trigonometric functions are transcendental functions because the cannot be epressed in terms of algebraic operations. For eample, there is no wa to find the value of θ in = cos θ b adding, subtracting, multipling, or dividing a constant and θ or raising θ to a rational power. 7 Sample answer: Given a triangle with hpotenuse c and legs a and b, let sin θ = a c and cos θ = b c. B the Pthagorean Theorem, c = a + b. (sin θ + (cos θ = ( a c + ( b c = a c + b c = a + b = c c or Therefore, (sin θ + (cos θ =. 9. False; sample answer: In ABC, the measure of angle B is less than the measure of angle A, cos B.79, and cos A.. Therefore, cos B > cos A, and thus the statement is false. c a. Lesson - f( = [, ] scl: b [, ] scl: D = (-,, R = (-, ; no -intercept, -intercept: -9; horizontal asmptote at = ; lim = -, lim - = ; increasing on (-, 99. b. = c.,. G. G n is an integer n ; Sample answer:, n ; Sample answer:, n; Sample answer:, Sample answer: Since the cosine function is the reciprocal of the secant function, the sine function is the reciprocal of the cosecant function, and the tangent function is the reciprocal of the cotangent function, ou can find the value of a secant, cosecant, or cotangent function on a graphing calculator b finding one divided b the reciprocal of the function D = (-,, R = (-, ; no -intercept, -intercept: - 9 ; horizontal asmptote at f( = - - = ; lim =, lim - = - ; decreasing on (-, -. + n; Sample Answer: 7 7, - - R Selected Answers

27 7.. m 9.. d..9 mi a. Convert to radian measure, and then use the arc length formula s = rθ. = ( radians =. radians Substitute r = and θ =.. s = rθ = (.. or about feet b. Let r represent the distance the first rider is from the center of the carousel and r represent the distance the second rider is from the center. s = r θ - r θ = (. - (..9 or about feet.. rev 7. m 9.. in. a. 9 rad min min to rad b. mi/h to mi/h.. in. 9. d min 7.. ft 9..7 i n First, convert to radians. θ = ( = 7 radians Use the area formula for a sector to find the radius. A = r θ 9 = r ( 7 A = 9 and θ = 7 9 = 7 9 r Multipl. r = Solve for r. 7 r = Take the positive square root of each side. 7 r.99 or about 7 ft. in. a. Quadrant I: < θ < < θ < c. Quadrant III: < θ < d. Quadrant IV: b. Quadrant II: < θ < 7. about 7 mi 9.. radians or about 9.. radians or about.9 a. about. in. b. about. ft Let θ represent the sector with a central angle of and θ represent the sector with a central angle of 7. First, convert each angle to radian measure. = ( radians or.9 7 = 7 ( radians or. Then use the area formula for a sector to find the area of the entire shaded region. A = r θ + r θ = ( (.9 + ( (. i n 7a. about. mi/h 7b. about mi/h 9. The angle does not have a complement since it is greater than or 9 ; supplement: 7. Complements and supplements are not defined for negative angles. 7. Sarah; sample answer: The formula for the length of an intercepted arc is s = rθ. Therefore, the perimeter of the sector is the sum of the length of the intercepted arc and twice the radius, or P = rθ + r. Because P = r, using substitution, r = rθ + r. Upon simplifing, θ = radians. 7. Decrease; sample answer: If the radius decreased, then the linear speed would also decrease because the linear speed is directl proportional to the radius. 77. Increase; sample answer: The equation for linear speed can also be written as v = rω. If the angular speed increased, then the linear speed would also increase because the linear speed is directl proportional to the angular speed. 79a. The perimeter would double; sample answer: The perimeter of a sector of a circle P is equal to the sum of the arc length s and two times the radius r, so P = s + r. Because s = rθ, if the radius doubled, the arc length would become a new arc length s = (rθ, which is equal to s = (rθ or s = s. So, the perimeter would be P = s + (r or P = (s + r. Therefore, the perimeter would double. 79b. The area would quadruple; sample answer: Because A = r θ, if the radius doubled, the area would become a new area A = (r θ, which is equal to A = r θ or A = ( r θ. Therefore, the area would quadruple.. cos θ =, tan θ =, csc θ =, sec θ =, cot θ =. sin θ = 9, cos θ = 7, tan θ = 9 7, csc θ =, sec θ =. ln - ln 7. ln - ln ±, ±, ±, ±, ±, ±; -,, 9. f ( - as -, f ( as 9. h ( as -, h( as 9. { - <, }; [-, 97. B 99. C Lesson -. sin θ =, cos θ =, tan θ =, csc θ =, sec θ =, cot θ =. sin θ = -, cos θ = -, tan θ =, csc θ = -, sec θ = -, cot θ =. sin θ = -, cos θ =, tan θ = -, csc θ = -, sec θ =, cot θ = - 7. sin θ =, cos θ = - 7 7, tan θ = -, csc θ = 7, sec θ = - 7, cot θ = undefined O - 7 For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R7

28 cos θ =, csc θ =, sec θ =, cot θ =. sin θ =,. cos θ =, tan θ = -, csc θ = -, sec θ =, cot θ = - 7. sin θ = -, cos θ =, tan θ = -, csc θ = -, cot θ = - 9. sin θ = -, cos θ =, csc θ = -, sec θ =, cot θ = - Let the center of the carousel represent the origin on the coordinate plane and Zoe s position after the rotation have coordinates (,. The seat rotates, so the reference angle is - or. Because the final position of the seat corresponds to Quadrant III, the sine and cosine of are negative. cos θ = r cos = -cos = sin θ = r sin = sin = - = - = - = - = Therefore, the coordinates of Zoe s seat after a rotation of are (-, - or (-., undefined t... θ -.. or. 7., 9., 7. sin θ = -, cos θ = -, tan θ =, csc θ = -, sec θ = -, cot θ = 7. (-, 7. (-, - 77a. m 77b. 7: a.m. and 7: p.m. 79 a. On the unit circle, cos θ = at and. Since the cosine function has a period of, cos θ will also equal at multiples of and. Therefore, cos ( n = when n is an odd integer. b. The cosecant function is the reciprocal of the sine function. So, csc θ will be undefined when sin θ =. On the unit circle, sin θ = at,, and multiples of those angles. Therefore, csc ( n is undefined when n is an even integer.. False; sample answer: The epression tan (-t = -tan t means that tangent is an odd function. The tangent of an angle depends on what quadrant the terminal side of the angle lies in.. Sample answer: The sine function is represented b the -coordinate on the unit circle. Comparing sin t and sin (-t for different values of t, notice that the -coordinate is positive for sin t and is negative for sin (-t. For instance, on the first unit circle, sin t = b and sin (-t = -b. Now find -(sin t to verif the relationship. -(sin t = -(b or -b, which is equivalent to sin (-t. Thus, -sin t = sin (-t. b -b t -t sin t = b sin (-t = -b d -d sin t = d t -t sin (-t = -d. Sample answer: Since tan t = sin t, we can analze -tan t and cos t tan (-t b first looking at tan t and tan (-t on the unit circle for a given value of t. Regardless of the sign of t, the value of cosine remains the same. However, the value of sine is positive for t but negative for -t. This results in tan t = a b, but tan (-t = - a b. Now find -tan t to verif the relationship. -tan t = - ( b a or - a b which is equivalent to tan (-t. Thus, -tan t = tan (-t. t -t sin t = a cos t = b sin (-t = -a cos (-t = b t -t tan t = a b tan (-t = - a b a. 9b. about. in ±, ±; ±, ±, ±, ± ; -,,. ±, ±, ±, ± ; -.. J Lesson -. The graph of g( is the graph of f ( compressed verticall. The amplitude of g( is. g( = sin f( = sin -. The graph of g( is the graph of f ( epanded verticall. The amplitude of g( is f( = cos g( = cos R Selected Answers

29 . The graph of g( is the graph of f ( compressed horizontall. The period of g( is. g( = sin a. amplitude =.; period = ; phase shift =.7; vertical shift = 7. b. Sample answer: =. cos ( c. about 7. ft. = - and. = - 7. none and O - f( = sin 7. The graph of g( is the graph of f ( epanded horizontall. The period of g( is. - f( = cos g( = cos 9. Sample answer: =. sin t. Sample answer: =. sin t. Sample answer: =. sin t. amplitude = ; period = ; frequenc = ; phase shift = - ; vertical shift = = cos ( + 9 First, find the period of =. sin (t + c. Period = b Period formula = or b = Since the period is, the function will reach a maimum height at radians. The phase shift is the difference between the horizontal position of the function at and 7 radians, which is radians. Substitute and b into the phase shift formula to find c. Phase shift = - c Phase shift formula b phase shift = and b = = - c - = c Multipl each side b -. Therefore, the equation is =. sin (t -.. Sample answer: = sin. Sample answer: = cos +. Sample answer: = cos 7. Sample answer: = cos 9a. 9b. The graph of = cos oscillates between the graphs of = and = [-, ] scl: b [-, ] scl: 7. amplitude = ; period = ; frequenc = ; phase shift = ; vertical shift = - O = sin - 9. amplitude = ; period = ; frequenc = ; phase shift = - ; vertical shift = 7 = sin ( + + 9c. [-, ] scl: b [-, ] scl: 9d. The graph of = sin oscillates between the graphs of = and = -. 9e. The graph of = f ( sin or = f ( cos will oscillate between the graphs of = f ( and = -f(.. True; sample answer: The graph of = cos is a horizontal translation of graph of = sin. Therefore, a cosine function can be written from an sine function using the same amplitude and period b appling the necessar phase shift. The graph of = cos completes one ccle on the interval [, ]. Therefore, the graph of = cos will complete ccles on [, ]. Since cos has two zeros per ccle and there are ccles, there will be ( or zeros.. Sample answer: Although the pulse of light can be represented as a function with a period, it is not a sinusoidal function because the distance the pulse of light is from the ground changes at a constant rate. As a result, the graph of this function would resemble the graph below. - For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R9

30 ft d 7. The damping factor is ; the amplitude of the function is decreasing as approaches from both directions. t 7 7. sin θ = -, cos θ = 7 7, tan θ = -, csc θ = - 7, sec θ =, cot θ = - 9. sin θ =, cos θ =, tan θ =, csc θ =, sec θ =, cot θ = real zeros and turning points; -,, and 7. real zeros and turning points; -,, and 9. es; f ( = - -. no. C. C Lesson = tan = - cot. = csc = sec ( = cot ( = - tan ( - = sec ( = csc ( - 9. The damping factor is ; the amplitude of the function is decreasing as approaches from both directions.. The damping factor is ; the amplitude of the function is decreasing as approaches from both directions.. The damping factor is e. ; the amplitude of the function is decreasing as approaches -.. The damping factor is ; the amplitude of the function is decreasing as approaches from both directions. [-, ] scl: b [-, ] scl: [-, ] scl: b [-, ] scl: [-, ] scl: b [-, ] scl: [-, ] scl: b [-, ] scl: [-, ] scl: b [-.,.] scl:. 7a. Sample answer: = e -t cos t 7b. about. second = sec + = csc R Selected Answers

31 . a. - - ft - - = cot ( + θ Since d is the length of the hpotenuse and ou are given the length of the side opposite θ, ou can use the sine function to write d as a function of θ. sin θ = opp hp sin θ = d d sin θ = d = or d = csc θ sin θ b. d c. Substitute θ = into the function ou found in part a. d = csc θ = csc d = csc θ =. ft θ 7..7 and , -.,., and and.7 a.. N b. T =. sec θ c. d. total rope: about.9 m; tension: about 79.7 N e. θ 97. ; about N we can find the values of b, c, and d and substitute them into the general form of the tangent function. period = phase shift = - c b b = = - c ± = b = - c = b - = c ± = b vertical shift = - Therefore, a tangent function with a period, a phase shift of units to the right, and a vertical shift of unit down is given b = tan ( - - or = tan ( = cot ( - +, = cot ( To find the -intercept, let t =. -ct = ke cos wt Original equation -c( = ke cos (w Substitute t =. = ke cos Multipl. = k(( e =, cos = = k Simplif. 9. True; sample answer: Since = csc =, asmptotes will sin occur for values of when sin =. Since = cot = cos sin, asmptotes will also occur for values of when sin =.. Sample answer: = csc ( + ; = -cot ( -. amplitude = ; period = ; frequenc = ; phase shift = ; vertical shift = = sin ( - + [, ] scl: b [, ] scl:. b 7. a 9... amplitude = ; period = ; frequenc = ; phase shift = ; vertical shift = = cos ( + [-, ] scl: b [-, ] scl:. [-, ] scl: b [-, ] scl:. The epressions are The epressions are not equivalent. equivalent for all real numbers. The general form of the tangent function is = a tan (b + c + d, where the amplitude is equal to a, the period is equal to b, the phase shift is equal to - c, and the vertical shift is equal to b d. Since we are given the period, phase shift, and vertical shift, - 7. sin θ = 7, tan θ =, csc θ = 7, sec θ = 7 7, cot θ = 9. about.% 7. ( + ( + ( - 7. ( - ( + ( + 7. C 77. B For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R

32 Lesson = sin ( = arccos = arctan. = arcsin.. 9. Since is the length of the side opposite θ and ou are given the length of the side adjacent to θ, ou can use the tangent function to write θ as a function of. tan θ = opp adj tan θ = θ = tan - or arctan b. Substitute = and = into the function ou found in part a. - θ = ta n - θ = ta n - θ = ta n - θ = ta n Let u = arccos, so cos u =. You can draw a right triangle, where the side adjacent to u is equal to and the hpotenuse is equal to. The domain of the inverse cosine function is restricted to Quadrants I and II, so u must lie in one of those quadrants. Using the Pthagorean Theorem, ou can find the length of the side opposite u to be -. Now, find tan u. tan u = opp adj = - or - So, tan (arccos = The graph of g( is the graph of f ( translated unit to the right and units down.. The graph of g( is the graph of f ( epanded verticall and translated units down. The graph of g( is the graph of f ( compressed horizontall and translated units up. a. about b. about ft. 7. D = { - }; 9. D = { }; R = { } R = { } = tan ( [-, ] scl: b [-, ] scl: [-, ] scl: b [-, ] scl:. D = { -, }; R = { } 7 a. m θ [-, ] scl: b [-, ] scl: R Selected Answers

33 . tan (si n - or cot (co s - a. Sample answer: For f f, the domain is [, ] and the range is [, ]. For f f, the domain is (, and the range is -,. b. Sample answer: f f f f f - f f - f f f f f [-, ] scl: b [-, ] scl: [-, ] scl: b [-, ] scl: c. Sample answer: For g g -, the domain is [-, ] and the range is [-, ]. For g - g, the domain is (-, and the range is [, ].The graph of g g - should be the line = for -. The inverse propert of trigonometric functions states that on the closed interval [-, ], cos (co s - =. The graph of g - g should be the line = for. Once the graph reaches, it will turn and decrease until it reaches the -ais at the same rate. When it reaches the -ais, it will turn again and increase until it reaches. It will continue to do this as approaches infinit. d. g g g g [-, ] scl: b [-, ] scl: [-, ] scl: b [-, ] scl: e. Sample answer: For f ( = tan (ta n -, due to the inverse propert of trigonometric functions, for all values of, f ( =. This should result in the line = for all real numbers. The graph of g( = ta n - (tan will be different because tan is undefined for multiples of. As a result, asmptotes for multiples of can be epected. We can also epect a range of -, due to the definition of arctan. f ( = tan(ta n - g ( = ta n - (tan 7. - = sin - - (., = cos - - (., Notice from the graphs of = si n - and = co s - that when =., si n - = and co s - =. So, si n - + co s - is + or. When =, si n - = and co s - =, so si n - + co s - is + or. When = -, si n - = - and co s - =, so si n - + co s - is - + or. Therefore, on the interval [-, ] it appears that si n - + co s - =. The graph of = si n - + co s - supports this conjecture = sin - + cos - 9. Sample answer: The function is odd. The definition of an odd function states for ever in the domain of f, f (- = -f(. If we let si n - = u, we have = sin u. From Lesson -, we know that the sine function is odd, so - = sin (-u. From here, we can get si n - (- = -u. Graphicall, it can be seen that for ever in the domain of f, f (- = -f(. 7. Sample answer: The function is odd. The definition of an odd function states for ever in the domain of f, f (- = -f(. If we let ta n - = u, we have = tan u. From Lesson -, we know that the tangent function is odd, so - = tan (-u. From here, we can get ta n - ( - = -u. Graphicall, it can be seen that for ever in the domain of f, f (- = -f(. 7. = tan θ O - - θ - = csc θ - [-, ] scl: b [-, ] scl: [-, ] scl:. b [-, ] scl: [ f g]( = - ; [ g f ]( = - ; [ f g]( =.7. questions. H 7. F For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R

34 Lesson -7. A =, a =., b = 9.. K =, j =., l =.. T =, t = 9., u = about. ft 9 a. Let B =, C = 7, and a = 9. 9 mi 7 First, find A. m A = or Since the measures of A, B, and a are known, ou can use the Law of Sines to find b. sin = sin 9 b 9 sin b = or. mi sin So, the entire trip is 9 +. or. miles. b. If the direct flight distance is c, sin = sin 7 9 c 9 sin 7 c = sin =.9 Therefore, the distance of a direct flight to the destination is about.9 miles.. no solution. B =, C =, c =.7. B = 9, C =, c =.9 7. B = 9, C =, c = 9. B =, C = 7, c =.7 and B = 7, C =, c = 7.. B = 7, C = 9, c =. and B =, C = 9, c =.. B = 7, C =, c =. and B =, C =, c =... mi 7. B = 9, C = 99, a =.9 9. Q = 7, R =, p =.. R = 9, S = 9, T =. C =, D =, b =... ; 7. ; c m 9.. f t..9 d a.. step s b.,7 f t.. m m 7.. f t i n. about. ft or about. ft To find the total area of the figure, find the sum of the areas of FJE and FGH. For FJE: s = (a + b + c = ( = 9. Area = s(s - a(s - b(s - c = 9.(9. - (9. - ( m m For FGH: s = (7 + + = Area = ( - 7( - ( - 7. m m So, the total area is or.7 m m... f t 7. South Ba is about.7 mi from the boat and Steep Rock is about.9 mi from the boat mm.. i n..9 m. Neither; for the acute case, h = sin or.7. Because a < b and h < a, there are two solutions. 7. Sample answer: The domain of inverse cosine includes angles measured to degrees. The domain of inverse sine includes angles measured -9 to 9 degrees. 9. A is acute. A is obtuse. A b c C h B C a h h Sample answer: Let h be an altitude of either triangle shown above. From the definition of the sine function, sin A = h b or h = b sin A, and sin B = h a or h = a sin B. Equating the two values of h, b sin A = a sin B, or sin a A = sin B, where sin A b and sin B. When an altitude h is drawn from verte B to side AC (etended in the obtuse triangle, sin A = h c or h = c sin A, and sin C = h a or h = a sin C. Equating the two values of h, c sin A = a sin C, or sin a A = sin c C. B the Transitive Propert of equalit, sin A a = sin B = sin C b c. 7 Draw a diagram to represent the situation. mi. B X mi. A To find the amount of time that it takes to complete one full orbit, we need to find the measure of the arc intercepted b points B and X. From the information given, m XAC = 9 + or and the length of XC is + or 9 miles. Now that we have two side lengths and an angle measure, we can find m C in XAC. b A a c h mi. B C A X 9 mi. First, use the Law of Sines to find X. C sin = sin X 9 sin = 9 sin X sin = sin X 9 - sin sin ( 9 = X.7 X So, ACX is - ( +.7 or about 9.7. Because point B is directl above point A, ACX BCX. From geometr, m BCX = m XB. Therefore, m XB 9.7. Find the amount of time that it takes to complete one full orbit or. 9.7 = 9.7 = ( (. Convert. minutes to hours. h. min min. h R Selected Answers

35 Therefore, one full orbit takes approimatel. hours or hours and minutes f ( = -; the amplitude of the function is decreasing as approaches from both sides.. + n, n is an integer; Sample answer: 7, i n. 7. [, ] scl: b [, ] scl: 79. f ( = ( - ; the amplitude of the function is decreasing as approaches from both sides. [, ] scl: b [, ] scl:. Latitude Distance,.,7.9 7,9.,.7 9 The distances range from about, mi to mi.. Sample answer: f ( = Sample answer: f ( = E 9a. θ = cos t θ. -. t 9b. The period is or about. seconds. This represents the amount of time it takes for the pendulum to complete one full swing or ccle. The amplitude is. This represents the maimum angular displacement of the pendulum. The frequenc is or about.9. This represents the number of swings the pendulum completes per second. 9c. The maimum angular displacement is.. 9d. The midline represents when the pendulum is vertical and there is no angular displacement. 9e. about. +.n seconds and. +.n seconds Chapter Stud Guide and Review. true. false, angle of depression. true 7. false, amplitude 9. false, arccosine. sin θ =, cos θ =, tan θ =, csc θ =, sec θ =, cot θ = sin θ =, tan θ =, csc θ =, sec θ =, cot θ =. cos θ =, csc θ =, sec θ =, tan θ = -, cot θ = -.. undefined 7. The graph of g( is the graph of f ( stretched g( = sin verticall. The amplitude of g( is, and the period is. 9. The graph of g( is the graph of f ( compressed verticall. The amplitude of g( is, and the period is amplitude = ; period = ; frequenc = ; phase shift = ; vertical shift = = cos ( 9 f( = sin f( = sin g( = sin For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R

36 . amplitude = ; period = ; frequenc = ; phase shift = - ; vertical shift = = cos ( = tan = sec ( = cot ( = sec ( B =, C =, c =.. B = 9, C = 7, c = 9.. A = 7, B =, C = 7 7a. ft 7b..7 9a. /min or./s 9b. 7, /min 7a. 7b. ft 7c.. min 7a. h min 7b. : p.m. 7.. ft 77a. 77b. unit s Chapter. d 7 Connect to AP Calculus t. Sample answer: If one of the cars slowed down, the distance between the two cars would grow at a slower rate. Therefore, the average rate of change of the distance between the two cars would be smaller. If one of the cars sped up, the distance between the two cars would grow at a faster rate. Therefore, the average rate of change of the distance between the two cars would be larger.. For t =, A 79 c m. For t =, A c m. For t =, A 77 cm. For t =, A 7 c m. For t =, A 9 c m. 7. For t =, m approaches cm /sec. For t =, m approaches 7 cm /sec. For t =, m approaches c m /sec. 9. Sample answer: In the first activit, the graph resembled a linear function. Therefore, the average rate of change could be found b using an two points since the rate of change remained constant. In the second activit, the graph resembled a quadratic function. The difference quotient had to be used to approimate the rate of change at each individual value of t. Appendi: Trigonometr and the Unit Circle Guided Eample, - -,, - cos ( Guided Eample ; ; sin ( ; ; sin ( - 7 ; Guided Eample ; ; ; ; ; ; opposite A, adjacent to A, ; Questions a. 7 b. (-.7,. 9..,,,,, 7. -, -,,, -, - 9a. no 9b. These values do not satisf the Pthagorean Identif:.7 +. =. a. (cos, sin b. (-.99,. a. positive b. positive a. undefined b. zero 7a. 7b. 7c. about 7. mph B the first part the Logarithmic Graph Similarit Theorem, the graphs of = and = log are congruent. B the second part, the graphs of = log and + log are similar. Therefore the graph of = log is similar to the graph of =. a. 99 to 997, 99 to 999, to, to b. 99 to 997, to c. 7. mwh d.. mwh R Selected Answers

37 CHAPTER Trigonometric Identities and Equations Chapter Get Read. -,. -,. -,, 7. s 9. a., b., A =. k 9., J., L Lesson sec θ =, cos θ = or. tan θ =, sin θ =. cos θ =, tan θ = or. cot θ = - or , sin θ = sin 9. sin 7. cos 9. ta n. sin cos. sin. cot - cos + tan + sin tan I 7 a. I = I - csc θ = I ( - csc θ = I ( - si n θ = I cos θ b. I = I cos θ = I ( = I ( Thus, = I of the original intensit emerges. 9. sec (csc +. sec (csc -. - cos. cos (csc + 7. sec - 9. Even; the graph shown is f ( = sec. f (- = sec (- = sec = f (. Since f (- = f (, f ( = sec is an even function.. -cot. -csc a. - - b. = tan + [-, ] scl: b [-, ] scl: = sec cos - sin sec = tan sec - sin [-, ] scl: b [-, ] scl: = sin tan [-, ] scl: b [-, ] scl: c., = d. Sample answer: The graphs of and support the conjecture that. Although the table from part a and the graphs of and support the conjecture that =, it is impossible to determine whether the two equations are equal over the entire domain unless the equations are verified algebraicall ln cos 9. ln csc a. csc θ = d b. θ =. radian mλ - si n. Jenelle; si n - co s = - si n ( - co s - co s cos =. sin = ± - co s, - co s csc = ± - co s cos, tan = ± - cos cos, - co s, sec = cot = ± cos -cos 7. False; sample answer: The domain - co s does not include values of where csc is undefined, such as n. 9 According to the Pthagorean Theorem, + = r. When each side of the equation is divided b, + = r. Since tan = and sec = r, + = r is equivalent to + ta n = se c or ta n + = se c. 7. C = 7, a., b. 7. B, C 9, c. 77. A, B, b True; all of the elements of A are also elements of B. 7. D 9. B Lesson -. (se c θ - co s θ = (ta n θ co s θ = ( si n θ co s θ co s θ = si n θ. sin θ - sin θ co s θ = sin θ ( - co s θ = sin θ si n θ = si n θ. co t θ cs c θ - co t θ = co t θ (cs c θ - = co t θ co t θ = co t θ [-, ] scl: b [-, ] scl: For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R7

38 7. sec θ sin θ - sin θ cos θ = cos θ sin θ - sin θ cos θ = sin θ cos θ - si n θ sin θ cos θ = - si n θ sin θ cos θ cos θ = sin θ cos θ = cos θ sin θ = cot θ cos θ 9. + sin θ + tan θ cos θ = + sin θ + sin θ cos θ = cos θ cos θ cos θ + sin θ + + sin θ + sin θ sin θ cos θ cos θ + sin θ + si n = θ cos θ ( + sin θ + sin θ = cos θ ( + sin θ = cos θ = sec θ. - tan θ + - cot θ = + - sin θ - cos θ cos θ sin θ = + co s θ co s θ - si n θ sin θ co s θ sin θ - co s θ sin θ = + cos θ - si n θ sin θ - co s θ cos θ si n θ cos θ = cos θ - si n θ + si n θ si n θ - co s θ cos θ = co s θ - si n θ si n θ si n θ - co s θ cos θ = cos θ - si n θ + -si n θ cos θ - si n θ = cos θ - si n θ cos θ - si n θ =. (csc θ - cot θ(csc θ + cot θ = csc θ - cot θ =. - sin θ + + sin θ = + sin θ + sin θ - sin θ + - sin θ - sin θ + sin θ = + sin θ - si n θ + - sin θ - si n θ = - si n θ = cos θ = sec θ 7. csc θ - c ot θ = (cs c θ - co t θ(csc θ + co t θ = [cs c θ - (cs c θ - ][cs c θ + (cs c θ - ] = cs c θ - = (co t θ + - = co t θ + - = co t θ + 9 a. v tan θ g se c θ = v ( sin θ cos θ g ( cos θ = v sin θ g b. h = v sin θ g = ( (sin 9.7 m (9.. ta n θ - si n θ = si n θ cos θ - si n θ = si n θ cos θ - si n θ ( cos θ cos θ = si n θ cos θ - si n θ co s θ co s θ = si n θ - si n θ co s θ cos θ = si n θ ( - co s θ cos θ = si n θ si n θ cos θ = si n θ ( si n θ co s θ = si n θ ta n θ. + csc θ + = sin θ sec θ cos θ sin θ sin θ + sin θ = cos θ = sin θ + cos θ sin θ sin θ cos θ + cos θ = sin θ sin θ cos θ = + cos θ sin θ sin θ = cos θ + cot θ. + ta n + sin θ θ - ta n θ = cos θ - sin θ cos θ cos θ cos θ + sin θ cos θ = cos θ cos θ - sin θ cos θ cos θ + si n θ cos θ = co s θ - si n θ co s θ = cos θ + si n θ cos θ cos θ cos θ - si n θ = cos θ + si n θ cos θ - si n θ = cos θ - si n θ = cos θ - ( - co s θ = co s θ - 7. sec θ - cos θ = cos θ - cos θ = cos θ - cos θ cos θ R Selected Answers

39 = - co s θ cos θ = si n θ cos θ = tan θ sin θ 9. (csc θ - cot θ = ( sin θ - cos θ sin θ = ( - cos θ sin θ ( - cos θ = si n θ ( - cos θ = - co s θ ( - cos θ ( - cos θ = ( - cos θ ( + cos θ = - cos θ + cos θ + csc θ sec θ. = csc θ sec θ csc θ sec θ + = csc θ sec θ + = sin θ cos θ + (si n + co s θ = si n θ + sin θ cos θ + co s θ = (sin θ + cos θ I. I m - m cot θ + = I m ( - = I m ( - csc θ. cot θ + = I m ( - si n θ = I m co s θ Y = sec + tan [-, ] scl: b [-, ] scl: Y = sec - tan [-, ] scl: b [-, ] scl: sec - tan = cos - cos sin = - sin cos cos = + sin - sin + sin cos + sin cos = - si n cos + sin cos = cos = cos + cos sin = sec + tan Y = cot - + cot [-, ] scl: b [-, ] scl: cot - + co t = cot - csc cos si n - = si n Y = cos - sin = co s - si n = ( - si n - si n = - si n [-, ] scl: b [-, ] scl: Y = cot - tan tan + cot [-, ] scl: b [-, ] scl: cos cot - tan tan + cot = sin - cos sin sin cos + cos sin cos sin cos = - sin sin cos sin sin cos + cos sin cos = cos - si n sin cos sin + co s sin cos = cos - si n si n + co s = co s - si n. sec - sec + = sec - sec + sec - sec - = (sec - se c - = (sec - ta n = sec - tan. ln cot + ln tan cos = ln cos sin + ln cos sin cos = ln cos sin sin = ln cos Y = - sin [-, ] scl: b [-, ] scl: For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R9

40 . si n θ - co s θ - = (si n θ + co s θ(sin θ - co s θ - = (si n θ - co s θ - = - co s θ - co s θ - = - co s θ 7. se c θ - ta n θ = (se c θ - ta n θ(sec θ + ta n θ = (sec θ - tan θ(sec θ + sec θ tan θ + ta n θ (sec θ + tan θ(sec θ - sec θ tan θ + ta n θ = (sec θ - ta n θ[( + ta n θ + sec θ tan θ + ta n θ] [( + ta n θ - sec θ tan θ + ta n θ] = ( + ta n θ + sec θ tan θ( + ta n θ - sec θ tan θ = ( + ta n θ - (sec θ tan θ = + ta n θ + ta n θ - se c θ ta n θ = + ta n θ ( + ta n θ - se c θ = + ta n θ [ + (se c θ - - se c θ] = + ta n θ ( + se c θ - - se c θ = + ta n θ ( se c θ = + ta n θ se c θ = se c θ ta n θ + 9. se c cs c = (ta n + cs c = ( si n cos + ( sin = cos + si n = se c + cs c. cos h - sin h = (e + e - - (e - e - = [e + + e - - ( e - + e - ] = ( =. - tan h = - sin h cos h = cos h cos h - sin h cos h = cos h - sin h cos h = cos h. = sec h = 7. = (tan + sec ( - sin = cos [-, ] scl: b [-, ] scl: [-, ] scl: b [-, ] scl: (tan + sec ( - sin = tan - tan sin + sec - sec sin = cos sin - cos sin sin + cos - cos sin = cos sin - cos sin sin + cos - cos sin = cos sin - sin cos + cos - cos sin = si n cos + cos = - si n cos = cos cos = cos 9a. sin = 9b. The graphs of = sin 9c. and = intersect (-, (, at and over [,. = sin = [, ] scl: b [-, ] scl: 9d. The graphs of = sin and = intersect at -, 7 -,, and over (-,. = sin = [-, ] scl: b [-, ] scl: 9e. Sample answer: Since sine is a periodic function, the solutions of sin = are = + n and = + n, where n is an integer. sec cos = cos sec - tan csc sec [-, ] scl: b [-, ] scl: tan sec - csc = cos cos - sin = cos - sin cos = se c - ta n = [-, ] scl: b [-, ] scl: cos cos sin Using the Law of Sines, sin β b A = ab sin γ A = a ( a sin β sin α sin γ A = a sin β sin γ sin α a sin β sin γ A = sin ( - (β + γ A = a sin β sin γ sin (β + γ = sin α a, so b = a sin β sin α. R Selected Answers

41 . Sample answers: tan sin + cos = sec and sin + cot cos = csc ; tan sin + cos = cos sin sin + cos = si n cos + cos = - co s cos + cos = cos - cos + cos = cos = sec sin + cot cos = sin + cos sin cos = sin + co s sin = sin + - si n sin = sin + = sin = csc sin - sin. Sample answer: You could start on the left side of the identit and simplif it as much as possible. Then, ou could move to the right side and simplif until it matches the left side co t θ = tan = sec (-7,. (-, -] [, 7. (-, -7] [, 9. B 9. D Lesson n, + n, 7. + n, n. + n, + n, + n, n. + n, + n, n. + n, n 9.,. 9 a. d = v sin θ = ( sin θ = sin θ + n, + n, n + n, + n, n., 7., θ =. θ =. Sine is also positive in Quadrant II. So, -. = 9.. So if θ = 9., θ = 7.. Hence, the possible interval is [., 7. ]. b. = ( sin θ = sin θ θ = 9. or θ =. θ = 9.9 θ = 7.., 7,.,,., 7, 7., 9., 7. + n, + n, n.,., ,.9.,.,.,, 9, 7.,,, 7, 9,,, 9a..9 radians 9b.. radians. < or 7 < <. <. < or < < 7. Both; sample answer: Vija s solutions are correct; however, the are not stated in the simplest form. For eample, his solutions of = + n and = + n could simpl be stated as = + n because when n =, + n is equivalent to. 9 co s - si n cos + si n - = co s ( - si n + (si n - = co s ( - si n - ( - si n = ( - si n ( co s - = - si n = or co s - = si n = co s = sin = ± cos = ± =, =,, 7,. Sample answer: si n = sin. Sample answer: When solving an equation, ou use properties of equalit to manipulate each side of the equation to isolate a variable. When verifing an identit, ou transform an epression on one side of the identit into the epression on the other side through a series of algebraic steps.. + tan θ + sin θ + cot θ = cos θ + cos θ sin θ sin θ + cos θ cos θ = sin θ + cos θ sin θ sin θ + cos θ = sin θ cos θ sin θ + cos θ = sin θ cos θ ( f - g( = ( f g ( = - +, - 7. D 77. D + For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R

42 Lesson cos nwt 9a. =. sin ( b. The new function h( represents the average of the high and low temperatures for each month tan θ 9. sin. cos V L = IwL cos ( wt + = IwL cos (wt cos - sin (wt sin = IwL [cos (wt - sin (wt ] = IwL [-sin (wt] = -IwL sin wt. ( - - ( ( ( ( sec ( - = cos ( - = cos cos + sin sin = (cos + (sin = sin = csc. cos ( - θ = cos cos θ + sin sin θ = -(cos θ + (sin θ = -cos θ 7. sin ( - θ = sin cos θ - cos sin θ = (cos θ - (-(sin θ = sin θ 9. cos (7 - θ = cos 7 cos θ + sin 7 sin θ = (cos θ + (-(sin θ = -sin θ. ; cos (α + β 7. sin α sin β.,., cos α cos β - sin α sin β = sin α cos β cos α cos β = - sin α sin β sin α cos β sin α cos β = cos α sin α - sin β cos β = cot α - tan β 9. sin (a + b + sin (a - b = sin a cos b + cos a sin b + sin a cos b - cos a sin b = sin a cos b. The function is equivalent to =. cos ( + + co s ( - = ( cos cos - sin sin + ( cos cos + sin sin = ( cos - sin + ( cos + sin = cos - sin cos + sin + cos + sin cos + sin = co s + si n = [-, ] scl: b [-, ] scl:. sin z = sin [ - ( + ] = sin cos ( + - cos sin ( + = cos ( + - [(- sin ( + ] = - [(-(sin cos + cos sin ] = sin cos + cos sin sin cos h + cos sin h - sin a. h b. [-, ] scl: b [-, ] scl: [-, ] scl: b [-, ] scl: c. cos sin (α + β 7. tan (α + β = cos (α + β sin α cos β + cos α sin β = cos α cos β - sin α sin β sin α cos β + cos α sin β cos α cos β cos α cos β = cos α cos β - sin α sin β cos α cos β cos α cos β sin α cos α + sin β cos β = sin α sin β - cos α cos β [-, ] scl: b [-, ] scl: [-, ] scl: b [-, ] scl: tan α + tan β = - tan α tan β 9. sin (α + β = cos [9 - (α + β] = cos [(9 - α - β] = cos (9 - α cos β + sin (9 - α sin β = sin α cos β + cos α sin β. sin ( + + z = sin [( + + z] = sin ( + cos z + cos ( + sin z = (sin cos + cos sin cos z + (cos cos - sin sin sin z = sin cos cos z + cos sin cos z + cos cos sin z - sin sin sin z R Selected Answers

43 sin ( - = sin cos - cos sin = (- ( - ( ( = = 9 a. sin cos = cos sin sin cos - cos sin = sin ( - = sin = = and = b. [-, ] scl: b [-, ] scl: cos ( + h - cos 7. h cos cos h - sin sin h - cos = h cos (cos h - - sin sin h = h = cos ( cos h - h - sin ( sin h h sec θ sin θ - sin θ cos θ = sin θ cos θ - sin θ cos θ = sin θ cos θ - si n θ sin θ cos θ cos θ = sin θ cos θ = cos θ sin θ = cot θ 7. 7a. Eponential; the base, + r n, is fied, but the eponent, nt, is variable since the time t can var. 7b. A(t = (. t 7c. $ ±, ±, ±, ±, ±, ±, ± and ±; -, 79. D. A Lesson -. - ; - 7 ; ; - 7 ; 7 a. d = v sin θ = sin θ 7. - ; - 9 ; ,,,. - ; ; -..,, Distance equation d = and v = (( = sin θ Solve for sin θ. = sin θ Simplif. 9 = θ - sin = 9 = θ Divide each side b., b. d = v sin θ = v sin θ cos θ = v sin θ cos θ tan θ - tan θ cos θ 7. + cos θ cos θ - cos θ cos θ.. + n, sin θ = sin θ cos θ Simplif. cos θ + cos θ cos θ 9. - cos θ - cos θ + cos θ. + cos θ + n, n 7. n, + n, n ,,. 7. cos θ + cos θ 9. sin + sin n, + n, n. n, + n, n. n, + n, n, n. ±cos 7. (cos a + cos b 9. [sin (b + θ cos l + sin l sin (θ - ] a. b. + + cos l - sin l. cos θ = cos (θ + θ = cos θ cos θ - sin θ sin θ = co s θ - si n θ. tan θ = tan (θ + θ tan θ + tan θ = - tan θ tan θ tan θ = - ta n θ 7. ta n - cos θ θ = + cos θ tan θ = ± - cos θ + cos θ tan θ = ± - cos θ + cos θ 9. i. co s θ - = ( + cos θ - = + cos θ - = cos θ ii. co s θ - = cos θ cos θ - = [cos (θ - θ + cos (θ + θ] - = (cos + cos θ - = + cos θ - = cos θ 7. [ - 7 cos θ + cos θ - cos θ cos θ] 7. ( cos θ + 7 cos θ cos θ + cos θ cos θ + cos θ cos θ cos θ 7a. f( = (sin θ cos - cos θ sin [-, ] scl: b [-, ] scl: For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R

44 7b. sin ( θ - ; sin ( θ - = ( angles identit 7c. sin θ cos g( = cos (θ - - sin (θ - - cos θ sin, difference of two f( = the th da of the ear, or around April D = [,, R = [, ; intercept: (, ; lim f ( = ; continuous on [, ; increasing on (, [-, ] scl: b [-, ] scl: 7d. cos (θ - ; cos (θ - = cos ( θ - = co s ( θ - - si n ( θ I, II, or between I and II; I 79. I, II, or between I and II; II cos θ = cos (θ = - si n θ = - (sin θ(sin θ = - ( sin θ cos θ( sin θ cos θ = - si n θ co s θ. ta n θ = sin θ co s θ - cos θ = + cos θ - cos θ = + cos θ. [sin (α + β + sin (α - β] = (sin α cos β + cos α sin β + sin α cos β - cos α sin β = ( sin α cos β = sin α cos β 7. Let = α + β and let = α - β. cos ( α + β cos ( α - β = cos cos = [cos ( + + cos ( - ] = cos ( + + cos ( - = cos ( α + β + α - β = cos α + cos β + cos ( α + β 9. Let = α + β and let = α - β. - sin ( α + β sin ( α - β = - sin sin = - [cos ( - - cos ( + ] = -cos ( - + cos ( + = -cos ( α + β = -cos β + cos α = cos α - cos β - α - β + cos ( α + β - α - β + α - β. 7. J f( = D = (-,, R = (-, ; intercept: (, ; lim - f ( = - and lim f ( = ; continuous for all real numbers; increasing on (-, Chapter Stud Guide and Review. cos θ ; reciprocal identit. tan θ; pthagorean identit. -tan θ; odd-even identit 7. cos α; double-angle identit 9. si n θ; power-reducing identit. sec θ =, cos θ = tan θ = - or. csc θ = -,. sec θ = or, sin θ = - or - 9. sec. s ec + tan sec sin θ. - cos θ + sin θ + cos θ sin θ ( + cos θ = + sin θ ( - cos θ - co s θ - cos θ sin θ ( + cos θ = + sin θ ( - cos θ sin θ sin θ = + cos θ + - cos θ sin θ sin θ = sin θ = csc θ cot θ. + csc θ + + csc θ cot θ cot θ = cot θ ( + csc θ + ( + csc θ cot θ ( + csc θ = cot θ + ( + csc θ cot θ ( + csc θ = csc θ csc θ + cs c θ cot θ ( + csc θ = c sc θ + csc θ cot θ ( + csc θ csc θ (csc θ + = cot θ ( + csc θ = sin θ sin θ cos θ = cos θ = sec θ 7. R Selected Answers

45 cot θ 7. + csc θ = co t θ ( - csc θ ( + csc θ( - csc θ = cot θ ( - csc θ - csc θ = cot θ ( - csc θ -cot θ = -( - csc θ = csc θ - sec θ + csc θ 9. cos θ + = sin θ + tan θ cos θ cos θ + sin θ cos θ sin θ + cos θ sin θ cos θ = sin θ + cos θ cos θ sin θ + cos θ = sin θ cos θ cos θ sin θ + cos θ = sin θ = csc θ sin θ sin θ. sin θ + cos θ = cos θ sin θ cos θ + cos θ cos θ tan θ = + tan θ., 9. n,. n,.,,, + n, n. + n, n ,,, + n, + n, n cos (θ + - sin (θ + = cos θ cos - sin θ sin - (sin θ cos + cos θ sin = cos θ - sin θ - sin θ - cos θ = -sin θ 7. cos ( θ - = cos θ cos + sin θ sin + cos θ cos = cos θ + sin θ + cos θ - sin θ = cos θ. + + cos ( θ , - 7 9, , - 7, , sin α 7. - cos α = sin α - cos α + cos α 7. es + cos α sin α ( + cos α = - co s α sin α ( + cos α = si n α = + cos α sin α - sin θ sin Chapter Connect to AP Calculus. Sample answer: The values state that the rate of change for f ( at is, at is, and at is -. This should be the slope of the lines tangent to f ( at those values of. The graph of f ( and the tangent lines verif this conclusion.. For =, the rate of change is. For =, the rate of change is. For =, the rate of change is. cos ( + h - cos a. m = h b. m = cos ( cos h - h - sin ( sin h h c. lim cos h - = ; lim sin h = h o h h o h d. m = cos ( - sin ( e. m = -sin f. For =, the rate of change is. For =, the rate of change is -. For =, the rate of change is. For =, the rate of change is. CHAPTER Applications of Trigonometr Lesson -. a.9. C 9.. C. 7. C = 7, b 9. cm, c 9. cm 9. C = 9, a = in., c = in... C = 9, a = mi, b = mi. A = 7, b 9. km, c 7. km 7. 9a. cm 9b. 9c. 9d.. cm. cm. not possible. B =, C = 9, b =.9 mi A 9, B, a. mi or A, B 9, a. mi 7. A 9, B, a. ft or A, B 9, a. ft 9. not possible. A., B., b. mi. A 9., A.7, +.7 > ; no second solution possible. C 7., C.7, ; two solutions possible 7. not possible, sin A > 9... million miles or 9.7 million miles a. No b..9 mi. V S =.7 km, V P =. km 7a. No 7b. about. ft 7c. sec 9. Two triangles Angles Sides Angles Sides A. a = cm A.9 a = cm B = b = cm B = b = cm C.9 c. cm C. c. cm For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R

46 . Angles A 7. Sides a = 9 cm B 9. b. cm C c = cm Angles A. Sides a = 9 cm B =. b. cm C c = cm. a.7 ft, c. ft. Rhmes to Tarrson:.7 km, Seton to Tarrson:. km 7.. mi 9. h.9 d. angle = 9 ; sides 9. cm, cm; diameter cm; it is a right triangle a. about 7 m b. about m c. about m. = sin. cm sin ; sin 9 = sin.( cm. cm 7. A = 9, B =, C =, a = cm, b 7. cm, c. cm 9., mph 7. tan - sin = sin cos - sin 7a. = 9-9 = sin cos - sin cos cos = sin - sin cos cos = sin ( - cos cos = sin sin cos = sin sin cos = sin tan 7b. units Lesson -. es. no. es 7. verified 9. B.. a 7.. A.. A., B., c. cm 7. A., C., b mi 9. B 9..9 mm 9. A mm 9 mm. A 7.9, B., C.. A 9., B., C 9.. A.. C.. mi.9 mi. mi 7. A 9.7, B.7, C. 9. C.. about mi. P 7.7 ; heading It cannot be constructed (available length,7. ft mi 9. P. cm, A, B., C. C B. A., B., C...7 cm. a = A. b.= B. c = C ; ft 9a.. = %, 9b. $,9,. about,9 km. 7 + = 9 < 9. ( a = b + c - bc cos A ( b = a + c - ac cos B, use substitutino for a and ( becomes b = ( b + c - bc cos A + c - ab cos B. Then = c - bc cos A - ac cos B, bc cos A + ac cosb = c, b cos A + a cos B = c sin = -, csc = - cos =, sec =, tan = -, cot = - Lesson -... knots 9 knots knots V V (, (, H7, I 7 ( 7. Terminal point: (, -, magnitude:. Terminal point: (-,, magnitude: a. 7a. H, I QI t b. 7 c.. 7b. 9 7c ,.., , -. a. -, 9 b., - (, 9 c. -., d., -9 (., 7a., 7b., - v r u.v r r u u (, v V V H, I ( (, H, I u (, u r v v r r QIII (, (, 9 u v (, N M J N R Selected Answers

47 7c.., 7d., - 9a. -, 9b. -, - 9c. -., 9d. -, -. True. False. True 7. u + v =, 9. u + v = -9, - u + v = -, u + v = 7, u v u r (, v r u (.,.v u u v (., r.v u v. u + v = -, -. u = i + j u - v = -7, u = 7.p = -.i -.7j 7a. p. u v u v p u v 7b. v = -., -. 7c. v = -.i, -.j 9a. 9b. w =., 9. 9c. w =.i + 9.j r u (, u v r (, v u v r (, r v u d. s = i + 7j; s., θ 7. a. p = i + j; p.9, θ. b. q = i + j; q., θ. c. r = i + j; r 9.7, θ. d. s = i + j; s., θ. 7. 7,, verified , 9, verified. 9 i - j, verified. 7 9 i - j, verified. -, 7 7, verified 7. 7 i + j, verified ,., , -., hor. comp ft/sec; vert. comp.. ft/sec 79. heading. at.7 mph. (. cm,. cm. a, b = a, b = a, b. a, b = - c, d = a - c, b - d = a + (-c, b + (-d = a, b + -c, -d = a, b + - c, d = u + (-v 7. (cku = cka, ckb = c ka, kb = c (ku c(ku = cka, ckb = kca, kcb = k ca, cb = k (cu 9. u + (-u = a, b + -a, -b = a - a, b - b =, 9. (c + ku = (c + k a, b = (c + ka, (c + kb = ca + ka, cb + kb = ca, cb + ka, kb = cu + ku 9., +, +, - +, - + -, - + -, =, 9. Answers will var, one possibilit:,., =, ± 7 ; see graph Mid-Chapter Check. sin B = b sin a a. cos B = a + c - b ac. a 9 m, B., C.. A., b., c.. A = a =. km. A =. a = 7 d B. b. km B.. b 7. d C 7.9 c =. km C 7 c = d or A = a =. km B.9 b. km C. c =. km 7. about.7 ft. 9 m 9. α 9. ; β 9. ; γ mi Reinforcing Basic Concepts w r 7. a. p = -i + j; p =, θ = b. q = i - j; q =, θ =.9 c. r = -i +.j; r =., θ =. d. s = i - j; s., θ = 7. a. p = i + j; p., θ =. b. q = i + j; q., θ.7 c. r =. i +.j; r., θ 9.9. Latitude A = B. C. Distance a =. cm a = cm a = cm. For A =, a. For A =, a. For A = 7, a 9.; es, ver close Ver close. For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R7

48 Lesson -.,. -,. -i - j 7. -.i +.j 9. -., , -7. F.; θ.. 7. kg 7..9 lb 9..7 kg. appro..79 lb. appro... N-m 7. appro. 97. ft 9. appro.,9.9 ft-lb. appro..7 lb. appro. 9. N-m. ft-lb 7. ft-lb 9. verified. verified a. 9 b. a. b. 9 7a. 7b es. no. es a..7,. b. u =.7,., u = -.7,. a. -.,. b. u = -.,., u = -., -. a..i +.7j b. u =.i +.7j, u = -.i +.j 7a. projectile is about 7 ft awa, and. ft high 7b. appro..7 sec and. sec 9a. projectile is about. ft awa, and. ft high 9b. appro.. sec and. sec 7. about 7. ft; t.9 -. =.7 sec 7. w (u + v = e, f a + c, b + d = e(a + c + f (b + d = ea + ec + fb + fd = (ea + fb + (ec + fd = e, f a, b + e, f c, d = w u + w v 7. u =, a, b = (a + (b = u = a, b, = a( + b( = 77. θ.9 ; answers will var a., B. C. ; P. m, A,9. m Lesson -. z = z + z. z = z + z i z. (cos + i sin 7. (cos + i sin 9. cos ( + i sin (. cos ( + i sin (. cis tan - ( ; cis.9. cis + tan - ( ; cis cis tan - ( 7. ;. cis. 9. cis + tan - (- ; cis.. r =, θ = z = cis ( z z 7 = + i i (, z i z z. r =, θ = z = cis ( = + i. r = 7, θ = tan - ( z = 7 cis tan - ( = 7 ( i = + i 7. r =, θ = - tan - ( z = cis - - tan = (- + i = - + i 9. r =, r =, θ =, θ = ; z = z z = - + i r =, θ = ; r r = ( = θ + θ = + =. r =, r =, θ =, θ = ; z = z z = - i r =, θ = - ; r r = = θ - θ = - = -. z z = - + i, z z = - + i. z z = - i, z z = i 7. z z = -. +.i, z = i 9. z z = + i, z z = z + i. z z = - - i, z z = - + i. z z = -,9 +.i, z =.9 +.i z. verified; verified, u + v + w = uv + uw + vw ( + i + (97 + i + (-9 + i = (7 + i + (- + i + ( + i, 9 + i = 9 + i 7a. V(t = 7 sin(t 7b. t V(t c. t.7 sec 9a. 7 cis. 9b. V i (, i 7 tan (, tan (, i R Selected Answers

49 a.. cis. b.. V a. cis. b.. V. I = cis ; Z = cis ; V = cis 7 7. I = cis. ; Z = 7 cis.9 ; V = 7 cis. 9. V = cis ; Z = cis ; I = cis. V = cis.9 ; Z =. cis.9 ; I = cis 7. cis 9.7. verified 7. z = - 7 i, z =- + 7 i 9.,, 9, 7 7. Lesson -. r = ; n = ; θ = ; -. r = ; n = ; θ = ;. r = ; n = ; θ = - ; + i 7. r = ; n = ; θ = - ; i 9. r = ; n = ; θ = ; -i. r = ; n = ; θ = ; - i. verified. verified 7. verified 9. verified. r = ; n = ; θ = ; roots:,.9 ±.9i, -.9 ±.7i. r = ; n = ; θ = ; roots:,.97 ±.i, -.7 ±.7i -. r = 7; n = ; θ = 7 ; roots: i, - i, - i 7.,. ±.9i, -. ±.7i 9. + i, - + i, -i i,.797 +,i, -. +.i, -. -.i,. -.i. =, - ± i. These are the same results as in Eample.. r = ; n = ; θ = ; roots: + i, - + i, - -i, - i 7. r = 7 ; n = ; θ = ; roots: i, i i,.79i -.9i 9. D = -, z = cis, z = cis, z = cis, z = cis 7, z = cis 9, z = cis. verified a. numerator: -7 + j, denominator: - + 7j b. + j c. verified. Answers will var i 9. z -.7, z.7, z. Note:Using sum and difference identities, all three solutions can actuall be given in eact form: - -, - +,.. tan sec + = sec - sec + (sec + (sec - = sec +. = - = sec - = cos - cos cos = - cos cos + Summar and concept review.. Angles Sides. appro.. ft A = a.9 d B = b d C = c. d. appro.. and 9.. Angles A = B = C = Angles A = B. Sides a. cm b. cm a 9 cm Sides a = 7 cm b = cm C. c.7 cm Angles A = B =.. no; 7. appor..9. appor.. m 9..,., and.7.,7.7m.. -i + j; u.; θ 9. Sides a 7 cm b cm C 9. c. cm horiz. comp.., vertical comp... -, - ; u + v.7, θ i + 9 j. QII; since the -compotent is negative and the -component is positive. 7. mi. appro , -. appro p q = -; θ ft-lb. appro. 7. lb. appro..77 ft-lb a ft;.7 ft b. appro..7 sec 7. (cos + i sin. + i For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R9

50 9.. z z = cis ( i ; z z = cis (. + j. Z. θ.7,. cis i. verified. + i, - + i,- i., - ± i 7. - i, - ± i. ± i, - ± i 9. verified Mied Review. Angles sides A = a. in B = 7 b 9. in C = c = 9 in, Area 7.9 in..9,.. appro. 7. ft 7. appro mph; heading. 9. One solution possible since side a > side b Angles A = B. C.9 sides a = m b = m c. m. No; barel touches ( tangent at a. (cos + i sin b. - + i i i.. 7. comp v u -.7, proj v u - + i 9. z = + i, z = - + i, z = - - i, z = - i Practice Test.. mi. 7. ft. Angles \, e Sides (in. A. a = B = b = C. c 7. Angles Sides (in. A. a = B = b = C. c. a. θ. b. proj v u = -., 7. c. u = -., 7., u = -., -... ft;.7 sec. cis (. cis 7 ; verified. - - i 7. verified. + i, - + i, -i 9..9 ±.9i, -.9 ±.9i.,, mi Strengthening Core Skills Eercise :. lb,. lb Eercise :.7 lb,.7 lb Eercise : es Stud Guide and Review Lesson -. ambiguous. I; II. Answers will var. Lesson -. cosines. pthagorean. B., C 9.9, a 9.; law of sines Lesson -. scalar. directed; line. Answers will var. Lesson -. equilibrium; zero. orthogonal. Answers will var. Lesson -. modulus; argument. multipl; add. (cos + i sin, z is in QIII Lesson -. r [cos (θ + i sin (θ ]; De Moivre s. comple. z = cis = cis, z = cis = cis 7, z 7 = cis = cis ; Answers will var a. No b.. mi a. No b. c.. sec a.. mi b. 7. ft 7. A,79 mi, P.7, B., M 7.. speed 7. mph bearing θ... cm to the right and. cm down from the initial point on the ceiling. F.9 N, θ. R Selected Answers

51 CHAPTER Sstem of Equations and Matrices Chapter Get Read. (,. (-,. (-, - 7. dog: $, cat: $ Lesson (-, -7. (,. (, 9, 7. (-,, a. The total amount of mone raised can be represented b c + p + g =.. A linear equation that relates the cost of a pie to the cost of a cake is p - c = -, and a linear equation that relates the cost of a cake to a giant cookie is c - g =. b. Let the first column of the augmented matri represent c, the second column represent p, the third column represent g, and the fourth column represent the cost in dollars c. Perform elementar row operations to obtain a row-echelon form of the matri. Begin b switching the third and first rows so that the first row has a in the first entr.. - interchange rows R + R -R + R - -R + R R You can use substitution to find that c =.9 and p =.9. Therefore, the solution is (.9,.9,.9. So, a cake costs $.9, a pie costs $.9, and a giant cookie costs $ es 9. es. es. (-,. (-, 7. (, -, 9. (,,. Sample answer: There are an infinite number of solutions to the sstem. One of the rows in the matri is eliminated, so the solution set is (-.7s +,.s +., s. More information is needed to determine which of these solutions is valid.. (z +, 9 - z, z. (z +, z + 9, z 7. (z -, z +, z 9. (w -, w -, w -, w ; ; R + R 7 The fourth row of the second matri is different from the fourth row in the first matri. Therefore, an operation involving the fourth row was performed. The first entr in the fourth row was - in the first matri but is in the second matri. Thus, was added to this entr. Since there is no in the first column for an of the three other rows, one of the rows must have been multplied b a factor. Starting with the first row, if the row is multiplied b, the entries will become,,,, and. Adding this new row to the fourth row results in the fourth row for the second matri. Thus, the row operation performed is R + R. 9a. + = ;. +. =. 9b. (7.,.; The nurses will need to mi 7. L of % and. L of % saline.. or 7. False; Sample answer: Because the equation in the last row is =, the value of one of the variables cannot be determined. Therefore, the corresponding sstem of equations ma have infinitel man solutions.. Sample answer: In Gaussian elimination, matrices are in row-echelon form. In Gauss-Jordan elimination, matrices are in reduced row-echelon form. Matrices in row-echelon form and reduced row-echelon form ma have rows consisting entirel of s. If the do, these rows appear at the bottom of the matrices. In both forms of matrices, the first entr in an row with nonzero entries is. This is called a leading. In both forms of matrices, if there are two successive rows with nonzero entries, the leading in the higher row is farther to the left than the leading in the lower row. For a sstem with a unique solution, when the matri is in reduced row-echelon form, ever column that has a leading has s above and/or below that. 7. ta n = si n cos - cos ( + cos ( = = - cos + cos For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R

52 about. ft 7a. =,(. 7b. about,9,9, 9., 9 7. B 7. C Lesson -. -; AB = [9 -]; BA is undefined.. AB = [7 -]; BA is undefined.. AB is undefined; BA = AB = -9 ; BA is undefined Let a matri represent the number of free throws, -pointers, and -pointers that each plaer made, and let a matri represent the point value of each shot. FT -pt -pt Re Chris Jerr FT -pt -pt Perform matri multiplication to determine the total amount of points scored b each plaer. 7 = So, Re scored points, Chris scored 7 points, and Jerr scored points = ; (, -, = - ; (., -, = ; (.,., 7 = ; (,, 9. es -. no. es. es 7. singular 9. A - = - -. A - = 9 7. ; -. ; singular ; - -. singular. ; - - Find the determinate of the matri formed b the three vertices. - = = or Since the area of triangle is of the determinate of the coordinates, A = ( or unit s. 7. units 9. - a. A = b. B =, A =, B =. =., = -. c, A = =, A n =, B = B = =, B n = n - n - n - n - or B n = n - c. C = =, C = =, C = = C = =, C n = n n n d a. b. c. The are the same. d. a d g b e h c f i = a d g b e h c f i a d g b e h = (aei + bfg + cdh - (gec + hfa + idb = aei + bfg + cdh - gec - hfa - idb = a(ei - hf + b( fg - id + c(dh - ge = a(ei - hf - b(id - fg + c(dh - ge = a(ei - hf - b(di - gf + c(dh - ge = a e h If A = f i - b d g f i + c d g e h, det(a = -. n - - n = e. No; sample answer:,, R Selected Answers

53 = = ((( + ((( + ((( + ((( - [((( + ((( + ((( + (((] ( = Since -, this method does not work for a matri. 7. Suppose B and C are both inverses of square matri A. Then b definition, AB = BA = I and AC = CA = I. Then B(AC = BI = B and (BAC = IC = C. But B(AC = (BAC b the Associative Propert of Matri Multiplication. Therefore, B = C. r s t 9 Let B = u v w. z Set up three sstems of equations. AB = A B - r s = u v r - u + = s - v + = r + u + = s + v + = r + u + = s + v + = Solve the first sstem. r + = r = - ((r - u + = (+r + u + = r - u + = (+r + u + = r + = 7 ( - + = = 7 - = - = r = - ( r = r - u + = ( - u + ( = - u + = -u = u = - Solve the net sstem. s + = s = - ((s - v + = (+s + v + = s - v + = (+s + v + = s + = s + = ( - + = = - = - = t w z t - w + z = t + w + z = t + w + z = s - v + = ( - v + ( = -v + = -v = - v = s = - ( s = Solve the last sstem. t + z = t = - z (t - w + z = (+t + w + z = t - w + z = (+t + w + z = t + z = t + z = ( - z + z = - z + z = -z = -9 z = t = - ( t = t - w + z = ( - w + ( = - w + = -w = w = r B = u s v t w = - z 7. Sample answer: A =, B = 7. C(A + B = c c c c ( a a a a + b b c c = c c a + b a + b a + b a + b c (a + b + c (a + b = c (a + b + c (a + b c (a + b + c (a + b c (a + b + c (a + b c a + c b + c a + c b = c a + c b + c a + c b c a + c b + c a + c b c a + c b + c a + c b c a + c a + c b + c b = c a + c a + c b + c b c a + c a + c b + c b c a + c a + c b + c b b b For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R

54 c a + c a = c a + a a c b + c b c b + c b = CA + CB 7. c(ab = c ( a a a b + a b = c a b + a b c( a b + a b = c( a b + a b a a c a + c a c a + c a + c b + c b c b + c b b b Subst. b b a b + a b a b + a b c( a b + a b c( a b + a b = ca b + ca b ca b + ca b ca b + ca b ca b + ca b = ca ca b b ca ca b b = ( c a a a a b b b b = (cab Subst. = c(ab = c ( a a b b a a b b Subst. = c a b + a b a b + a b a b + a b a b + a b = c( a b + a b c( a b + a b c( a b + a b c( a b + a b = ca b + ca b ca b + ca b ca b + ca b ca b + ca b Def. Matri Mult. Def. Scalar Mult. Dist. Prop. Def. Matri Mult. Def. Scalar Mult. Def. Matri Mult. Def. Scalar Mult. Dist. Prop. = a cb + a cb a cb + a cb a cb + a cb a cb + a cb = a a cb cb a a cb cb = A(cB Subst. Comm. Prop. Def. Matri Mult. 77. ; sample answer: Since the number of columns of matri A must be equal to the number of rows of matri B, matri B must have rows. The product matri AB has the same number of columns as matri B. So, matri B must have columns C 9. G Lesson -. (,. (-,. (-, 7, 7. no unique solution 9. sitcoms, talk shows, and movies. (-, -. (,. no unique solution 7. (,, gallons Let = the cost of each etra gaming minute, = the cost of each etra call minute, and z = the cost of each tet message. The corresponding sstem of equations is shown below. + + z = z = =. The coefficient matri is. Calculate the 7 determinant of the matri. 7 = = (- - (- + ( or - Since the determinant does not equal, ou can appl Cramer s Rule to find,, and z..9.7 = A. 7 = =.99 A A =. = A = 7. =. - = A z = A z Each etra gaming minute is $.99, each etra call minute is $., and each tet message is $ (-,, 9. (-,, - When the determinant of a matri is, the matri is singular and has no inverse. Let A = n -. Find an equation for the determinant of A and solve for n when the determinant is. n - = n( - (- or n + n + = n = - n = - Therefore, when n = -, the sstem cannot be solved using an inverse matri.... kg of % allo,. kg of % allo, and. kg of % allo 7. AB = mm, BC = mm, AC = mm i -. X = (A + B - i - (D - -. X = (-C - D(I - A - 7. X = (B - A - (AD R Selected Answers

55 9. Neither; if the determinant of the coefficient matri is, then there is no unique solution. The sstem ma have no solution, or it ma have an infinite number of solutions. The sstem of equations shown has an infinite number of solutions.. Yes; sample answer: A = a + bc ab + bd ac + cd bc + d. Therefore, ( A - = a d - abcd + b c bc + d -ab - bd -ac - cd a + bc and A - = ad - bc d -b d -b -c a = ad - bc ad - bc -c a, so ad - bc ad - bc ( A - = a d - abcd + b c bc + d -ab - bd -ac - cd a + bc. Thus, ( A - = ( A -. a. Sample answer: Gauss-Jordan elimination can be used to solve an sstem of linear equations. It is possible to perform row operations on an matri. b. Sample answer: Inverse matrices can onl be used to solve sstems with square coefficient matrices because matri multiplication can onl be done if the number of columns of the first matri is equal to the number of rows of the second matri. c. Sample answer: Cramer s Rule uses determinants to solve sstems, and because it is onl possible to find the determinant of a square matri, this method can onl be used to solve square sstems.. AB = , BA = es 9a. Sirius 9b. Sirius:., Vega:. 9c. Vega. F. H Lesson ( ( ( a. Decompose s( =. First, divide the + numerator b. + + s( = = + The fraction is proper, so we can rewrite the epression as partial fractions with constant numerators, A, B, and C, and denominators that are the linear factors of the original denominator. This time, however, we have repeated factors in the denominator. + = ( ( + Because the factor has a multiplicit of, include partial fractions with denominators of,, and = A + + B + C + Multipl each side b the LCD, = A + + B + C = A(( + + B( C = A(( + + B( + + C ( + + = A + A + B + B + C ( + + = A + C + A + B + B ( + + = (A + C + (A+ B + B Equate the coefficients on the left and right side of the equation to form a sstem of equations. In other words, the coefficients of the -terms on the left side of the equation must equal the coefficients of the -terms on the right side. A + C = A + B = + B = Use an method to solve the sstem. B = A + B = + A + = + A = A + C = + C = C = - Replace A, B, and C with,, and - in the partial fraction decomposition b. = A + B + = + C [-, ] scl: b [-, ] scl: ( ( ( ( , - +, A = t, B = r - ( + t - 9. A = -t, B =, C = t +, D = r - Rewrite the epression as partial fractions with constant numerators A, B, C, and D and denominators that are linear factors of the original denominator. + ( - ( + + = A ( - + B ( + + C ( + + D ( + For Homework Help, go to Hotmath.com connected.mcgraw-hill.com R