Answers Chapter 1 Functions 1.1 Functions, Domain, and Range 1. a) Yes, no vertical line will pass through more than one point. b) No, an vertical line between = 6 and = 6 will pass through two points.. a) function 6 0 6 = + 1 b) not a function 0 c) function 6 = 6 8 = ( + 1) 5 0 6 b) domain { 6, 5,, }, range {}, function; for ever element of the domain there is onl one corresponding element of the range c) domain {5}, range {,, 0,, }; not a function; the single element of the domain corresponds with more than one element of the range. a) not a function; there are more range values than domain values b) function; each domain value has onl one range value 5. a) domain { R, 7 7}, range { R, 7 7} b) domain { R, 5}, range { R, 0} 6. Answers ma var. 7. a) A = + 60 b) domain { R, 0 < < 0}, range { R, 0 < < 50} 8. a) range {8} b) range { 7, 18, 11, 6, } c) range { 8,, 8 5,, 8 7 } 9. a) domain { R}, range { R} b) domain { R}, range { R, 1} c) domain { R, }, range { R} d) domain { R, }, range { R, 0} 10. a) the graph is not a function; there are fewer elements in the domain than in the range b) the graph is a function; for each value in the domain there is eactl one value in the range 11. a) i) domain { R, }, range { R, }; not a function + = 16. a) domain {,,, 5, 6}, range {, 6, 8, 10, 1}; function; for ever element of the domain there is onl one corresponding element of the range 0 Functions 11 Eercise and Homework Book MHR 187
ii) domain { R, }, range { R, 0 }, function b) 0 = 16 0 c) iii) domain { R, }, range { R, 0}; function d) 1 0 e) 1 = 16 b) Together, the semicircles in graphs ii) and iii) form the circle in graph i). 1. a) domain { R}, range { R, 7}; function b) domain { R, or }, range { R}; not a function 1. Functions and Function Notation _ 1. a) 16 5, 7 _ 5, 17 10 ; f : + 5 b) 5, 5, 5 ; f : 6 + c) 5, 1, _ 5 ; f : ( + 1) d) 7, 7, 7; f : 7 e) 1 7, 1 1 5, ; f : 1 + 1 f) 7, 1, ; f :. a) 1 _ f) 15. =. a) a = ; f() = + b) a = 6; f() = 6 + c) a = 1.5; f() = 1.5 + d) a = 6.; f() = 6. + 5. a) 1 0 1 18 1 10 6 Domain Range 188 MHR Chapter 1 978-0-07-01875-5
b) c) d) 1 Domain 5 6 7 Domain 6 9 1 15 Domain Range 1 5 Range 7 Range 6. a) function; each element of the domain corresponds with onl one element of the range b) not a function; some elements of the domain correspond with more than one element of the range c) function; each element of the domain corresponds with onl one element of the range d) function; each element of the domain corresponds with onl one element of the range 7. a) {( 8, ), (, 1), ( 1, ), (, 5), (, 9)} b) {(0, ), (, 8), (5, 7), (11, )} c) {( 5, ), (, ), (0, ), (7, 7), (7, )} d) {( 5, 0), (, ), (, 5), (, 6), (0, 7)} e) {( 1, ), (, ), (5, ), (9, )} 8. a) not a function; some elements of the domain correspond with more than one element of the range b) function; each element of the domain corresponds with onl one element of the range c) not a function; some elements of the domain correspond with more than one element of the range d) function; each element of the domain corresponds with onl one element of the range e) function; each element of the domain corresponds with onl one element of the range 9. a) f : 7 + 1 b) g : + 7 5 c) h : b 9b + 9 1 d) r : k 5k 10. a) Substitute l = 1.8 in each equation. Earth: _ T = 1.8.7 On Earth the period is approimatel.7 s. Moon: _ T = 5 1.8 6.7 On the moon the period is approimatel 6.7 s. Pluto: _ T = 8 1.8 10.7 On Pluto the period is approimatel 10.7 s. b) Substitute T = in each equation and solve for l. Earth: 5 = l.5 = l Square both sides. 6.5 = l On Earth the length of the pendulum is 6.5 m. Moon: 5 = 5 l 1 = l 1 = l On the moon the length of the pendulum is 1 m. Pluto: 5 = 8 l 5 8 = l Square both sides. _ 5 = l or l 0. 6 Functions 11 Eercise and Homework Book MHR 189
On Pluto the length of the pendulum is approimatel 0. m. c) domain {l R, l 0}, range {T R, T 0} d) Period (s) T 0 6 8 0 16 1 8 0 8 1 16 Length (m) e) all are functions; each -value corresponds with one -value f) T : l l T : l 5 l T : l 8 l 11. a) 60 000 b) i) 76 55 ii) 77 77 c) 00 d) Yes 1. a) b) T = 8 T = 5 T = 0 8 1. a) Earth: t : h 80 h.9 ; Jupiter: t : h 80 h 1.8 b) In each case, the epression under the radical sign cannot be less than zero. Since the denominators are both constants, 80 h 0. Both h and t must be positive. 80 h 0 and h 0 h 80 and h 0 0 h 80 For both relations, the domain is {h R, 0 h 80} and the range is {t R, t 0}. c) Yes; both relations are square root functions. For each relation, ever value of the domain has eactl one value in the range. d) Substitute h = 10 in each equation and solve for t. Earth: t(10) = _ 80 _ 10.9.8 Jupiter: t(10) = _ 80 _ 10 1.8. On Earth, the object reaches a height of 10 m after.8 s. On Jupiter, the object reaches a height of 10 m after. s. 15. a) Answers ma var. Sample answer: f() = b) f() = 0.5 + 16. a) Function Domain Range f() = { R} { R} g() = { R} { R, 0} c) = + 1 d) f( ) = 57, f(0) = 1, f(6) = 97 1. a) domain {i R, i 0}, range {A R, 0 A 500} b) Yes c) i) $6.8 ii) $8.67 d) i) 19.5% ii) 11.8% h() = { R} { R, 0} q() = { R} { R} p() = { R, 0} { R, 0} 190 MHR Chapter 1 978-0-07-01875-5
b) h() = q() = g() = p() = 0 f() = 6 c) all are functions d) f() = g() = h() = ; q() = p() = 1 e) f(0) = h(0) = q(0) = 0; f(1) = g(1) = p(1) = ; g( 0.77) = h( 0.77) 0.59; h(0.5) = q(0.5) = 0.5; q( ) = p( ) = 1; h(1.6) = p(1.6) 1.6; f( 1) = p( 1) = _ 17. a) g() = + ; h() = 5 b) i) ii) 8 iii) iv) 7 18. f() = + 5 1. Maimum or Minimum of a Quadratic Function 1. a) = ( ) 16 b) f() = ( + 8) 68 c) f() = ( + 5 ) + d) g() = ( 1 ) + 7 e) = ( ) f) = ( 7 ) _ 89. a) (, ); minimum b) (, 5); maimum c) ( 1, 8); maimum d) ( 5, ) ; minimum e) (, ); minimum f) ( 1, 1); maimum. a) (1, ); minimum b) (, ); maimum c) (, 5); maimum d) (1, ); minimum e) (6, 0); minimum f) (5, 11); maimum. Answers ma var. 5. $850 over cost 6. a) Let represent the number of $1.5 price decreases and D() represent the amount of the donation, in dollars. D() = (ticket price) (number of tickets sold) D() = (1.5 1.5)(10 + 8) Since the equation is in factored form, solve D() = 0. 0 = (1.5 1.5)(10 + 8) 1.5 1.5 = 0 or 10 + 8 = 0 = 5 or = 1 The maimum occurs at the average of the two zeros. 5 _ 1 = = 6 Tickets sold = 10 + 8 = 10 + 8(6) = 15 The maimum donation occurs when 15 tickets are sold. b) ticket price = 1.5 1.5 Substitute = 6. = 1.5 1.5 6 =.75 The ticket price that maimizes the donation is $.75. c) D(6) =.75 15 = 610 The maimum donation is $610. 7. a).7 m b) after 0.8 s 8. 10 m b 0 m 9. a) Let represent the larger number and let represent the smaller number. = 8, so = 8 Let p() represent the product of the two numbers. p() = = ( 8) = 8 Complete the square to determine the maimum product. p() = 8 = 8 + 16 16 = ( ) 16 The maimum product is 16. b) The maimum product occurs when =. = 8 = The two numbers that produce the maimum product are and. Functions 11 Eercise and Homework Book MHR 191
10. a) 1 and 1 b) 8 11. 8.1 cm 1. a) 66.5 m b) 1.5 s c) 55 m 1. a) $15.50 b) $805 1. a).5 m b) 55 m 15. a) = _ 1 ( 10) 16 b) 0. m 16. a) 10.15 b).5 _ 17. ν g + h 0 18. a) minimum = 1.875 V; maimum = 17 V b) minimum:.5 min; maimum: 5 min 19. b must be an even integer 0. b must be divisible b 8 1. Skills You Need: Working With Radicals 1. a) 1 b) 0 c) d) 16 1 e) 15 6 f) 6 g) 6 5 h) 9 0. a) 6 b) 7 c) 1 d) 5 e) 6 f) 5 5 _ g) 6 h) 1 i) j) 6 5. a) 1 b) 1 _ c) 11 d) 5 + 1 6. a) b) 7 c) 5 d) 1 e) 5 f) 5 5 g) 6 9 h) 7 7 _ 5. a) 5 10 b) 0 c) 70 d) 0 e) 16 f) 8 _ g) 0 10 h) 16 6. a) 6 b) 15 _ c) 6 5 10 d) 15 6 1 _ e) 0 5 f) 1 + 0 6 g) 6 10 8 15 h) 6 7 _ 7. a) 1 5 7 b) + 6 c) d) 5 8 e) 67 f) 8. a) b) 9. a) A = 1 bh _ = 1 ( 5 )( 15 ) _ = ( 75 ) = 5 = ( 5 ) = 15 _ b) A = l w = 7 6 ( 8 ) _ = 1 8 = 1 16 = 1 ( ) = 8 c) A = h (b 1 + b ) = _ 0 _ ( 17 + 8 ) = _ 5 ( 9 + 16 ) = 5 ( 7 + ) = 5 ( 11 ) _ = 11 15 d) A = s = ( 8 + 5 ) = ( 7 + 9 6 ) = ( 7 + 6 ) = ( 7 ) + ( 7 ) ( 6 ) + ( 6 ) _ = (7) + 1 + 9(6) _ = 8 + 1 _ 10. 1 5 ; eplanations ma var _ 11. 6 1 cm 1. + = ( ) = 08 = 10 = The side length of the square game board is cm. Divide b to determine the number of small squares along each side. = 8 Each side has 8 squares, so there are 6 squares in total. 19 MHR Chapter 1 978-0-07-01875-5
1. 100 6 According to BEDMAS, subtract first, then take the square root. _ = 6 = 6 _ 100 6 According to BEDMAS, take the square root then subtract. = 10 8 = The order of the operations is reversed, so the answers are not the same. _ 1. a) _ + 11 b) 7 6 c) 1 6 8 d) 5 + _ 7 e) + _ 5 9 15. a) i) 5 ii) 5 7 iii) 1 b) i) 15 ii) 10 iii) 6 7 16. a) 5 m b) _ 17 c 7ab c) 8a 5ab + 10mn n d) b ab + 9c d 1.5 Solving Quadratic Equations 1. a), b), c) 5, 5 d) 9, e) 5, f), 5. a) = or 1 b) = ± _ 6 c) = ± _ d) = _ 7 ± 1 6 e) = ± 5. a) no roots b) two roots c) two roots d) one root e) no roots f) no roots. a) = 5 ± _ 17 b) = ± c) = ± 1 d) = 10 ± 10 e) = ± 1 5 5. a) two distinct real roots b) two equal real roots c) no real roots d) two distinct real roots e) two equal real roots 6. Methods ma var. a), b) ± 9 c) 5 ± 1 d) 0, e).78, 1.15 f), 1 g) 5 ± 7 h), 1 7. a) k = ± b) k > or k < c) < k < 8. a) = f() 1 1 0 1 6 8 8 6 5 6 7 1 b) f() c) 8 6 0 6 8 f() 8 6 0 6 8 f() = + 5 + 6 f() = + 5 + 6 = d) 0.5,.5 e) (0., ), (.6, ) Functions 11 Eercise and Homework Book MHR 19
9. a) k = ±9, ±15 b) k = ±, ±9 c) Answers ma var. Possible values of k are, 6, 1. 10. Find the values of t when the height is 0. Solve h(t) = 0..9t + 1.8t + 1.5 = 0 Use the quadratic formula. t = 1.8 ± 1.8 (.9 )( 1.5 ) (.9 ) 0.068 or.517 Time must be positive, so the ball will be in the air for approimatel.5 s. 11. a) approimatel 50 km/h b) approimatel 70 km/h c) 100 km/h 1. Let represent one of the numbers and let represent the other number. The sum of the numbers is, so + =. Isolate : = ➀ The sum of the squares of the numbers is 06, so + = 06. ➁ Substitute ➀ in ➁. + ( ) = 06 Epand and simplif. + (576 8 + ) = 06 8 + 70 = 0 Divide b. + 15 = 0 ( 9)( 15) = 0 = 9 or = 15 The two numbers are 9 and 15. 1. a) 98 m b) 10 s c) 0 s d) 8 m 1. 9 cm, 1 cm 15. 5 cm 16. 11 and 1 17. 7. m b 7. m 18. 7, 8, 9 or 7, 8, 9 19. 18.6 A and 5. A 0. a) = ±, ±1 b) = 1 ± c) = 1 ± _ 7 6 1. a) = 1 ± b) =. a > + and a <, a 0 1.6 Determine a Quadratic Equation Given Its Roots 1. a) f() = a( 1)( + ) f() = ( 1)( + ) f() = ( 1)( + ) f() 1 1 10 8 6 0 f() = ( 1)( + ) f() = ( 1)( + ) 8 b) f() = a( + )( + 6) f() = ( + )( + 6) f() = ( + )( + 6) 8 1 f() 10 8 6 0 f() = ( + )( + 6) f() = ( + )( + 6) c) f() = a( 5)( + ) f() 18 1 6 0 6 1 18 8 6 6 8 f() = ( 5)( + ) f() = ( 5)( + ) 6 8 f() = ( 5)( + ) f() = ( 5)( + ) 19 MHR Chapter 1 978-0-07-01875-5
. a) f() = a( + ) b) f() = a( + 9 + 18) c) f() = a( 10). a) = ( )( + ) b) = 0.5( + )( 5) c) = ( ). a) = b) = 0.5 + 1.5 + 5 c) = 1 8 5. a) = + 1 + b) = 1 6 c) = + 5 d) = 0.5 + 6 6. a) = ( ) 7 b) = ( + 1 ) _ + 9 c) = ( + 1) 8 d) = ( + ) + 8 7. a) = _ 1 _ + 9 8 _ b) 9 m or 6.15 m 8 c) 1 m d) = _ 1 + 7 8 e) The graph of = _ 1 + 7 is the 8 graph of = _ 1 _ + 9 8 translated 1 m to the right. f) 8. a) = _ 16 b) = 1 5 ( ) c) = ( + 6) d) = ( ) e) = 5( + 1) 9. a) = 5 5 + 10 b) = 1 c) = 9 10. If m graphs pass through the given points and have the same verte and direction of opening as the given graphs, I know m equations are correct. 11. ac > 9 1. a) The width of the arch is m, so half of the width is 16 m. When the verte is on the -ais, the -intercepts are 16 and 16. A point that is 8 m from one end of the arch and 18 m high is (8, 18); another point on the other side is ( 8, 18). The height of the arch is unknown. 0 ( 8, 18) (8, 18) 16 1 8 1 8 0 8 1 b) Use the -intercepts and the coordinates of the known point, (8, 18), to determine the factored form of the equation. = a( 16)( + 16) Substitute (8, 18) to find the value of a. 18 = a(8 56) 18 = 19a a = _ f() = _ ( 16)( + 16) is the factored form of the equation. The standard form, which in this case is also the verte form, is f() = _ +. c) The verte is (0, ), so the maimum height of the arch is m. 1. a) In this situation the -intercepts are 0 and, and a point that is 8 m from one end of the arch is (8, 18). Use this information to write the factored form of the equation. = a( ) Substitute = 8 and = 18. 18 = a(8)(8 ) 18 = a(8)( ) 18 = 19a a = 18 19 Functions 11 Eercise and Homework Book MHR 195
a = _ The factored form of the equation is _ = ( ). In standard form the equation is _ = +. b) The maimum height occurs at the verte, which is halfwa between the -intercepts. The -coordinate of the verte is = 16. Substitute this value into the equation and solve for. _ = + _ = ( 16 ) + ( 16 ) = The maimum height of the arch is m. The height is the same as the one found in question 1, which makes sense since onl the orientation has changed, not the size of the arch represented b the function. 1. Yes. Eplanations ma var. 15. a) f() = ( )( + 1)( ); f() = + 10 16 b) f() = a( 1)( + )( ) c) f() = + + _ 1.7 Solve Linear-Quadratic Sstems 1. a) (, ), (, 1) b) (, 9 ), ( 5 5, ) c) ( 5, 8 ), ( 1, ) d) ( 7, 15), (, 6). Answers ma var.. a) no intersection b) two points of intersection c) one point of intersection d) two points of intersection. a) b) c) d) 5. a) _ 1 b) c) 5 d) 8 8 6. Answers ma var. 7. (, 1), ( 5, 57) 8. To determine if the two paths intersect, set the equations equal. 8 + 70 + 56 800 = 960 + 15 000 8 + 1680 88 00 = 0 Divide b 8. 10 + 11 05 = 0 Factor or use the quadratic formula. ( 105) = 0 = 105 Substitute = 105 into = 960 + 15 000 and solve for. = 960(105) + 15 000 = 00 The paths will intersect at (105, 00). 9. Answers ma var. 10. a) k > b) k = c) k < 11. Equate the epressions and simplif. k + + 10 = 5 + k + 8 + 7 = 0 b ac = 8 (k)(7) = 6 8k Use the discriminant; a = k, b = 8, c = 7. a) Two points of intersection occur when the discriminant is positive. 6 8k > 0 6 > 8k _ k < 16 7 196 MHR Chapter 1 978-0-07-01875-5
b) One point of intersection occurs when the discriminant is zero. 6 8k = 0 6 = 8k _ k = 16 7 c) No points of intersection occur when the discriminant is negative. 6 8k < 0 6 < 8k _ k > 16 7 1. ( 16.8,.), (.5, 0.7) 1. Answers ma var. Sample answer: The line = is a vertical line that intersects or cuts through the parabola such that part of the line is above the parabola and part of it is below. 1. a) 8 Chapter 1 Review 1. a) Yes; vertical line test is satisfied b) No; vertical line test is not satisfied. a) Yes; each domain value has onl one range value b) No; the domain value has four range values. a) domain { R}, range { R, } b) domain { R, }, range { 0, R}. Answers ma var. 5. a) range { 5, 1,,, 11} b) range { 7, 5, 1} 6. a) 16 8 0 16 1 8 = + 11 = 18 0 8 b) 1 = 7 5 0 6 8 10 b) (9, 18) 15. 10:0 < < 1:0, or between 10:0 a.m. and :0 p.m. 16. a) 7.1 s b) 11.8 m 17. 90 m b 160 m 18. = 10 19. Two parabolas ma have two, one, or no points of intersection. Equations and sketches will var. 0. a) Estimates will var. Sample answer: ( 8, 6), (9.6,.8) b) Estimates will var. Sample answer: (.5, 0.9), (1.5, ) 1. There are no real roots. c) 9 6 0 6 = 1 8 6 = 1 + 9 0 6 8 Functions 11 Eercise and Homework Book MHR 197
d) 0 c) ( 6, 1); minimum 8 6 0 6 8 10. a) ( 1, ); minimum b) (, ); minimum 11. a) $5 000 b) $000 _ 1. a) 7 b) 15 1. a) 7 b) 1 6 + 6 7 c) 16 7 d) 6 _ 0 e) 8 96 96 f) 10 g) 5 1. A = 18 6 8 = ( )( + 1) 7. a) function; each domain value has a single range value 7 5 1 1 Domain 8 7 6 1 Range b) not a function; the values and in the domain each have two values in the range 1 Domain 1 Range 8. a) the width of the pool b) domain { R, > 0.}, range {w R, 0}, where w is the width of the pool c) Yes; each element of the domain corresponds with one element of the range d) w = 18 ft 9. a) (, 9); maimum b) (, 16 ) ; maimum 15. = 1 or 1 6 16. = ± 7 17. two distinct real roots 18. 68. m b 11.6 m 19. a) = + + 5 b) = + 10 0. Answers ma var. Sample answer: h =.8( 5) + 10 1. = 5( + 1). ( 5, 17 ), (1, ). one. 5. after 8. s Chapter 1 Math Contest 1. B. D. A. B 5. C 6. A 7. C 8. B 9. D 10. C 198 MHR Chapter 978-0-07-01875-5
Chapter Transformation of Functions.1 Functions and Equivalent Algebraic Epressions 1. a) No. The functions do not appear to be equivalent. b) Yes. The functions appear to be equivalent. c) No. The functions do not appear to be equivalent. d) No. The functions do not appear to be equivalent. e) No. The functions do not appear to be equivalent. f) Yes. The functions appear to be equivalent.. a) f(6) = ; g(0) = b) Answers ma var. c) f(0) = 1; g(0) = 10 d) s(0) = ; t(0) = 0 e) f(1) = 0; g(0) = 5 f) Answers ma var.. a) b). a), b), 5. a) Yes; 5 _ b) No; g() = + 6 ; 6; + c) No; g() = + 1; 5 d) Yes; ; 1 6. a) _ 1 + ;, 7 _ b) + ;, _ c) + 5 + ;, _ d) + + 5 ; 5, 9 _ e) + 1 ; 1, 1 f) ( + 1) ; 1, g) + 1 1 ; 1, 1 h) 17 ;, 0, 7. a) 7 19 0 1 9 10 91 b) undefined undefined 0 0 _ 7 5 _ 10 75, Functions 11 Eercise and Homework Book MHR 199
c) 0 1 6 0 undefined 7 10 7 6 0 d) 9 11 0 9 10 11 8. a) Area of large circle is A = πr The diameter of the small circle is 8 cm, so the radius is cm. The area of the small circle is A = π() = 16π Subtract the area of the small circle from the area of the large circle. Shaded Area = πr 16π b) Shaded Area = πr 16π = π(r 16) = π(r )(r + ) c) Since this function represents area, it is restricted to positive r-values that result in positive area. The domain is { R, > }. 9. a) V = π( + 1) ( ); V = π( 7 ) b) SA = π( + 1) + π( + 1)( ); SA = π(6 + 1) c) 56.5 m ; 5.0 m d) { R, > } 10. a) V() = ( + )( 0.5)( + 1) b) SA() = ( + )( 0.5) + ( 0.5)( + 1) + ( + )( + 1) c) 5.875 m, 57.875 m ; 8 m, 15 m ; 19.65 m, 18.5 m 11. Epand and simplif each function. f() = + [ 1 ( 1)( + 1) ] = + [ 1 ( 1)] = + 1 ( 1) = + 1 ( + 1) = + 1 1 + 1 = 1 + 1 + 1 g() = [ 1 ( + 1)] = 1 ( + 1) = 1 ( + + 1) = 1 + 1 + 1 f() and g() are equivalent epressions. 1. The graph of f() is the line = + 1, with slope, -intercept 1, and two holes, one at (, 1 ) and the other at (, 9).. Skills You Need: Operations With Rational Epressions 1. a) 16 ; 0 b) 8 ; 0 c) 1b ; b 0 d) _ 1 ; 0, 0 e) 16ab; a 0, b 0 f) 18pr ; p 0, q 0, r 0. a) ; 6 b) 9;, 0 _ c) 8 6 ;, 6 _ 7 d) + ;, 1, 0 _ e) ; 6, 0, 5 + 6 _ f) + 1 ; 8,, + 00 MHR Chapter 978-0-07-01875-5
_ 1 g) ;, 1, 1, _ h) ; 6,, + 6. a) ; 1, 0 b) ; c) 5 ; 1, 10, 5 + 10 d) + 15 ; 6, 0 e) 1 6 ; 0, 9 5( + 1) f) 5 ;, 0, 5 _ g) ;, 1, + 1 _ h) + 7 + 5 ; 5,, 1,. a) 5 1 + 75 b) 1 c) 1 8 ; 0 d) 8 + 9a ab ; a 0, b 0 _ a 1b e) 8a b ; a 0, b 0 f) + 7a 10a ; a 0 ( 7) g) ( )( + ) ;, 9( + 1) h) ( + )( 5) ;, 5 0 + 7 i) ( )( + 5) ; 5, _ 5. a) + ( 6)( ) ;, 6 b) 8 7 ( 5)( + ) ;, 5 c) + ( )( + ) ;, d) + ( 1)( + ) ;, 1 _ 10 e) ; 8, 6, 5, ( + 8)( + 6) 7 f) ; 7, 5, ( )( + 5) _ 1 g) ;,, 1 + _ h) 7 + 1 ; 7,, 5, ( + 7)( 10) 10 6. a) Use the formula for time: t = d v. The total distance is 0 km, so half the total distance is 0 km. Let t 1 represent David s time for the first half of the race and let t represent David s time for the second half of the race. Let t represent the total time for the race. The time for the first half of the race is _ represented b t 1 = 0. The time for the second half of the race is _ 0 represented b t = 8. t = t 1 + t _ = 0 + _ 0 8 0( 8) + 0 = ( 8) 0 160 + 0 = ( 8) _ 0 160 = ( 8) The total time taken for the race is given _ 0 160 b t = ( 8). b) Substitute = 5 in t. _ 0(5) 160 t = 5(5 8) 1. David completed the race in approimatel 1. h, which is 1 h and 18 min. 7. a) _ 5 ; 5 _ b) + ; c) a 5 a ; a 5 d) b b ; b e) 5 + ; f) 1 ; g) 5b 8 b + ; _ h) c + 1 5c 1 ; 1 5 Functions 11 Eercise and Homework Book MHR 01
8. a) The dimensions of the bo are l = 10, w = 100, and h =. V = lwh V() = (10 )(100 ) b) SA() = lw + lh + wh SA() = (10 )(100 ) + (10 ) + (100 ) c) The side length of the square must be positive, so > 0. The shortest side is 100 cm, so < 50 cm, since it would be impossible to cut two squares with sides larger than 50 cm from a side that is onl 100 cm long. The domain is {, 0 < < 50 }. _ V SA = _ (10 ) ( 100 ) ( 10 )( 100 ) + ( 10 ) + ( 100 ) ( 60 )( 50 ) = _ ( 60 )( 50 ) + ( 60 ) + ( 50 ) = ( 000 60 50 + ) 000 6 50 + + 60 + 50 = _ 110 + 000 + 000 e) The denominator cannot be _ zero. The restrictions are ± 000, or ±5.77. 9. a) 6 + 1 10 b), 1 5 10. a) b) f() = ( )( + ) c) Answers ma var. Sample answer: The two graphs are the same. The restrictions for both graphs are and. These are the vertical asmptotes of the graph. _ 19 11. ( 1) ;,,, 1, 0,, 1, _ 1. a) + ; 0, 0, _ b) + ; 0, 0, ± _ 1. a) a ( a + ) ; a b) a + a + 5 ; a, a 5. Horizontal and Vertical Translations of Functions 1. a) f () = r() = f () s() = f ( + 5) 0 0 5 1 1 6 _ 9 1 1 b) c) r() translates the points units down. s() translates the points 5 units to the left.. a) A ( 8, ), B ( 5, 1), C (, 1), D (1, 10), E (, 10), F (6, 8), G (8, 8) b) A ( 8, 7), B ( 5, 9), C (, 9), D (1, 0), E (, 0), F (6, ), G (8, ) c) A ( 5, ), B (, 5), C (1, 5), D (, ), E (6, ), F (9, ), G (11, ) d) A ( 9, ), B ( 6, 5), C (, 5), D (0, ), E (, ), F (5, ), G (7, ) e) A ( 7, 6), B (, ), C ( 1, ), D (, 1), E (, 1), F (7, 11), G (9, 11) f) A ( 10, 7), B ( 7, 1), C (, 1), D ( 1, ), E (1, ), F (, 5), G (6, 5). a) f() = ; = f() 7; translate the graph of f() down 7 units; f(): domain { R}, range { R}; g(): domain { R}, range { R} 0 MHR Chapter 978-0-07-01875-5
b) f() = ; = f() + ; translate the graph of f() up units; f(): domain { R}, range { R, 0}; g():domain { R}, range { R, } f) f() = ; = f( + ); translate the graph of f() left units; f(): domain { R, 0}, range { R, 0}; g(): domain { R, }, range { R, 0} c) f() = ; = f() + 9; translate the graph of f() up 9 units; f(): domain { R, 0}, range { R, 0}; g(): domain { R, 0}, range { R, 9} g) f() = 1 ; g() = f( 8); translate the graph of f() right 8 units; f(): domain { R, 0}, range { R, 0}; g(): domain { R, 8}, range { R, 0} d) f() = ; = f( 5) ; translate the graph of f() right 5 units; f(): domain { R}, range { R, 0}; g(): domain { R}, range { R, 0} e) f() = 1 ; = f() + ; translate the graph of f() up units; f(): domain { R, 0}, range { R, 0}; g(): domain { R, 0}, range { R, 0}. a) translate f() left units and up 1 unit; g() = f( + ) + 1; f(): domain { R}, range { R, 0}; g(): domain { R}, range { R, 1} b) translate f() right units and down 7 units; g() = f( ) 7; f(): domain { R, 0}, range { R, 0}; g(): domain { R, }, range { R, 7} c) Answers ma var. Sample answer: translate f() down 6 units; g() = f() 6; f(): domain { R}, range { R}; g(): domain { R}, range { R} Functions 11 Eercise and Homework Book MHR 0
5. a) translate f() right units and up units; g() = f( ) + ; f(): domain { R, 0}, range { R, 0}; g():domain { R, }, range { R, } b) translate f() left units and down units; g() = f( + ) ; f(): domain { R}, range { R, 0}; g(): domain { R}, range { R, } c) translate f() left units and up 5 units; g() = f( + ) + 5; f(): domain { R, 0}, range { R, 0}; g(): domain { R, }, range { R, 5} 6. a) b() = + b) h() = 5 c) m() = + 9 d) n() = 10 e) r() = + 10 f) s() = 6 g) t() = 7. a) b() = ( + ) b) h() = 5 c) m() = + 9 d) n() = ( ) 7 e) r() = ( + ) + 6 f) s() = ( + ) 8 g) t() = ( 5) + 1 8. a) b() = ( _ + ) b) h() = 5 c) m() = + 9 d) n() = 7 _ e) r() = + + 6 _ f) s() = _ + 8 g) t() = 5 + 1 _ 1 9. a) b() = + b) h() = 1 5 c) m() = 1 + 9 _ 1 d) n() = 7 _ 1 e) r() = + + 6 _ 1 f) s() = + 8 _ 1 g) t() = 5 + 1 10. a) False. The order does not matter. For eample, if the base function = is translated left units and then up 1 unit, the image of the point (, ) is (0, 5). Similarl, if (, ) is translated up 1 unit and then left units, the image point is also (0, 5). This will be true for all points on =. b) True. Consider the point (, ) on the graph of =. A translation of units right results in the image point (, ), which is on the graph of =. If the same point is translated units down, the resulting point is (, 0), which is also on the graph of =. Thus, when translated horizontall the image points of the points on the line = lie on the line =. Similarl, when translated verticall the image points of the points on the line = lie on the line =. 11. When f() = is translated 5 units to the left the transformed function is g() = f( + 5) g() = + 5 a) Equivalent; the transformed function is g() = f() + 5 = + 5 b) Not equivalent; the transformed function is g() = f() 5 = 5 c) Not equivalent; the transformed function is g() = f( ) + = ( ) + = + 1 d) Equivalent; the transformed function is g() = f( + ) + = ( + ) + = + 5 0 MHR Chapter 978-0-07-01875-5
e) Not equivalent; the transformed function is g() = f( ) + 1 = ( ) + 1 = 1. a) base function: f() = ; transformed function: g() = + b) base function: f() = ; transformed function: g() = ( + 1) + c) base function: f() = 1 ; transformed _ 1 function: g() = + d) base function: f() _ = ; transformed function: g() = 1 + 1. a) domain { R, 0}, the number of units of the product; range { R, 00}, the cost associated with producing number of units _ b) g() = 8 + 00 c) translate right 8 units d) domain { R, 8}, range { R, 00} 1. a) translate 6 units left and 5 units up f () g() = f ( + 6) + 5 f() = g() = + 11 f() = g() = ( + 6) + 5 _ f() = g() = + 6 + 5 f() = 1 g() = _ 1 + 6 + 5 b) translate units right and units down g () h() = g ( - ) - g() = + 11 h() = + g() = ( + 6) + 5 h() = ( + ) + g() = + 6 + 5 h() = + + _ 1 g() = + 6 + 5 h() = _ 1 + + c) p() = f( + ) + ; Answers ma var.. Reflections of Functions 1. a) g() f() 0 f(): domain { R}, range { R}; g(): domain { R}, range { R} b) c) d) e) f() 0 g() f(): domain { R}, range { R, 0}; g(): domain { R}, range { R, 0} 0 f() g() f(): domain { R, 1}, range { R, 0}; g(): domain { R, 1}, range { R, 0} g() 0 f() f(): domain { R, 1}, range { R, 0}; g(): domain { R, 1}, range { R, 0} g() f() 8 0 8 Functions 11 Eercise and Homework Book MHR 05
f(): domain { R, 7}, range { R, 0 }; g(): domain { R, 7 }, range { R, 0 }. a) 0 f() h() f(): domain { R}, range { R}; h(): domain { R}, range { R} b) f() e) 8 0 f() 8 h() f(): domain { R, 7}, range { R, 0 }; h(): domain { R, 7}, range { R, 0}. A reflection in the -ais is represented b k() = f( ). a) f() k() 0 0 h() f(): domain { R}, range { R, 0}; h(): domain { R}, range { R, 0} c) h() 0 f() f(): domain { R, 1}, range { R, 0}; h(): domain { R, 1}, range { R, 0} d) 0 f() h() f(): domain { R, 1}, range { R, 0}; h(): domain { R, 1}, range { R, 0} f(): domain { R}, range { R}; k(): domain { R}, range { R} b) 0 f() k() f(): domain { R}, range { R, 0}; k(): domain { R}, range { R, 0} c) 0 f() k() f(): domain { R, 1}, range { R, 0}; k(): domain { R, 1}, range { R, 0} 06 MHR Chapter 978-0-07-01875-5
d) e) 0 k() f() f(): domain { R, 1}, range { R, 0}; k(): domain { R, 1}, range { R, 0} 0 k() f() 6. a) i) (0, 8) ii) (, 0) and (, 0) iii) There are no invariant points. b) Answers ma var. Sample answer: The graph of = a, {a R, a 0}, will have (0, 0) as an invariant point under each tpe of reflection. 7. a) No b) Yes; f() c) Yes; f( ) d) Yes; f( ) e) No f) Yes; f( ) 8. a) Use a graphing calculator. f(): domain { R, 7}, range { R, 0 }; k(): domain { R, 7 }, range { R, 0}. a) g() = 1 9 b) g() = ( + 8) 17 c) g() = ( + 1) 11 _ 1 d) f() = 6 + 5 Since g() = f( ), replace with in f() and multipl f() b 1. g() = [ _ 1 ( ) 6 + 5 ] 1 = [ 6 + 5 ] 1 = [ ( + 6 ) + 5 ] Common factor 1 in the denominator. 1 = 5 Multipl each ( + 6 ) term in the brackets b 1. _ 1 = + 6 5 Simplif. _ 1 Therefore, g() = + 6 5. e) g() = + + 19 _ 1 f) g() = 8 g) g() = 18 5. a) g() = f( ); reflection in -ais b) g() = f(); reflection in -ais c) g() = f( ); reflection in - and -aes b) g() = f( ) g() = ( + ) c) translate f() 6 units to the right d) f( 6) = ( + 6) = ( ) g() = ( + ) Common factor 1. = [ 1( )] = ( 1) ( ) Appl the laws of eponents for powers. = ( ) = f( 6) A reflection of f() in the -ais is equivalent to a translation of 6 units right, f( 6). e) This would not be true for reflections in the -ais because the direction of opening of the parabola would change, whereas translations do not change the direction of opening. f) This would work for functions that have a vertical line as ais of smmetr. In this case reflection in the -ais can Functions 11 Eercise and Homework Book MHR 07
be obtained b appling a translation. For eample, the quadratic function in part a) has a vertical line that passes through the verte (, ) as an ais of smmetr. A cubic function such as f() = ( + ) does not have a vertical line of smmetr, and so a reflection in the -ais cannot be epressed as a translation. 9. a) i) f() = 1 ii) f() = b) i) translate 9 units right and units up ii) translate 5 units left and 7 units down _ 1 c) i) k() = 9 ; p() = _ 1 + 9 + ; _ 1 q() = + 9 _ ii) k() = + 5 + 7; p() = + 5 7; q() = + 5 + 7 10. a) i) g() = ( +) ii) g() = _ ( +) iii) g() = _ ( +6) b) i) a reflection of f() = in the -ais and a translation of units left ii) a reflection of f() = in the -ais and a translation of units left iii) a reflection of f() = in the -ais and a translation of 6 units left c) Yes; eplanations ma var; g() = ( a ) 11. a) f( ) = 16 ; f( ) = 16 ; f( ) = 16 ; f() = f( ) and f() = f( ) are equivalent b) invariant points: (, 0) and (, 0) 0 f() = 16 f( ) = 16 6 f() = 16 f( ) = 16 c) f() = 16 and f( ) = 16 : domain { R, }, range { R, 0 } f() = 16 and f( ) = 16 : domain { R, }, range { R, 0} 1. a) All points are invariant under each reflection. b) onl true for circles with centre (0, 0).5 Stretches of Functions 1. a) f () = b) g() = f() h() = f ( ) 16 1 9 9 0 0 0 0 9 16 1 9 c) g() represents a vertical compression of f() b a factor of ; h() represents a horizontal stretch of f() b a factor of 08 MHR Chapter 978-0-07-01875-5
. a) c) g() is a horizontal stretch b a factor of 6 of f() = b) c) d) g() is a vertical stretch b a factor of 7 of f() = 1 d) e) g() is a horizontal compression b a factor of _ 1 5 of f() =. a) a = 8; g() is a vertical stretch b a factor of 8 of f() b) k = 6; g() is a horizontal compression b a factor of 1 of f() 6 c) a = ; g() is a vertical compression b a factor of of f() d) k = 1 ; g() is a horizontal stretch b a 9 factor of 9 of f(). a) g() is a vertical stretch b a factor of 1 of f() = f) g() is a vertical compression b a factor of 1 of f() = 8 5. a) g() = b) g() is a horizontal compression b a factor of 1 of f() = b) g() = _ Functions 11 Eercise and Homework Book MHR 09
c) g() = d) g() = 5 1 ; domain { R, 10 10}, range { R, 0 5} 6. a) The base function is f(t) = t. b) vertical stretch b a factor of a c) Earth: d(t) =.9t ; Neptune: d(t) = 5.7t ; Mercur: d(t) = 1.8t d) For all three planets, the domain is {t R, t 0} and the range is {d R, d 0}. 7. a) compress f() horizontall b a factor of 1 8 ; g() = _ 8 b) stretch f() verticall b a factor of 7; g() = 7 _ 8. a) g() = 5 ; domain { R, 5 5}, range { R, 0 10} b) g() = 5 ; domain { R,.5.5}, range { R, 0 5} c) g() = 1 _ 5 ; domain { R, 5 5}, range { R, 0.5} 9. a) b) g() = ; h() = () () c) d) No; simplified, h() = 8 10. The value of the parameter c is 18 in both f() and g(). In g(), the term 10 ma be written as follows: 10 = (0. 5 ) = 0.(5 ) = 0.(5) So g() = f(5), since f(5) = 0.(5) + 18. f(5) = 0.(5 ) + 18 = 10 + 18 To obtain the graph of g(), appl a horizontal compression b a factor of 1 5 to the graph of f()..6 Combinations of Transformations 1. a) a =, d = 5; verticall stretch f() b a factor of, then translate 5 units right b) a = 1, c = ; verticall compress f() b a factor of 1, then translate units up c) d = 6, c = ; translate f() 6 units left and units up 10 MHR Chapter 978-0-07-01875-5
d) k = 1, c = 7; horizontall stretch f() b a factor of, then translate 7 units up e) k =, c = 8; horizontall compress f() b a factor of 1, then translate down 8 units f) a = 5, c = ; verticall stretch f() b a factor of 5, then translate down units. a) a =, k =, c = ; verticall stretch b a factor of, horizontall compress b a factor of 1, and then translate units down b) a = 5, c = 6; verticall stretch b a factor of 5, reflect in the -ais, and then translate 6 units up c) a = 1, d = 8, c = 1; verticall compress b a factor of 1, then translate 8 units right and 1 unit up d) k =, c = 6; horizontall compress b a factor of 1, reflect in the -ais, and then translate 6 units up e) a = 1, k = 1, c = 1; reflect in the -ais, horizontall stretch b a factor of, and then translate 1 unit down f) a = 5, k = 5, c = 7; verticall compress b a factor of 5, horizontall compress b a factor of 1 5, and then translate 7 units down. a) verticall stretch b a factor of and horizontall compress b a factor of 1 ; g() = _ c) horizontall stretch b a factor of, then translate 1 unit left; g() = 1 ( +1) d) verticall stretch b a factor of, reflect in the -ais, and then translate 6 units down; g() = 6. a) verticall compress b a factor of 1, reflect in the -ais, horizontall compress b a factor of 1, and then translate units left and units down; g() = 6 b) verticall stretch b a factor of, reflect in the -ais, horizontall compress b a factor of 1, and then translate 1 unit right and 6 units up; g() = 16( 1) + 6 b) verticall stretch b a factor of, reflect in the -ais, and then translate 1 unit right and 7 units up; g() = 1 + 7 c) verticall compress b a factor of 1, horizontall compress b a factor of Functions 11 Eercise and Homework Book MHR 11
1, and then translate 5 units left and units up; g() = 1 ( + 5) + d) verticall stretch b a factor of 5, reflect in the -ais, and then translate 1 unit _ 5 right and units up; g() = 1 + e) f() = 1 ; f(): domain { R, 0}, range { R, 0}; g(): domain { R, 9}, range { R, 0} 5. a) f() = ; f(): domain { R}, range { R}; g(): domain { R}, range { R} f) f() = 1 ; f(): domain { R, 0}, range { R, 0}; g(): domain { R, }, range { R, 6} b) f() = ; f(): domain { R}, range { R, 0}; g(): domain { R}, range { R, } 6. d) 1 8 b) 0 16 1 8 0 f() a) 8 8 c) c) f() = ; f(): domain { R}, range { R, 0}; g(): domain { R}, range { R, 0} d) f() = ; f(): domain { R, 0}, range { R, 0}; g(): domain { R, }, range { R, 0} 1 _ 7. a) = ; graph ii) + b) = 1 + + ; graph iv) The equation = 1 + + ma be epressed as = 1 ( _ ) +. This equation matches graph iv) because the base function for graph iv) is f() =. The graph of f() = is reflected in the -ais, 1 MHR Chapter 978-0-07-01875-5
verticall compressed b a factor of 1, and then translated right units and up units. The point (0, ) satisfies the equation. _ c) = + 1; graph i) d) = [ 1 ( 5) ] + ; graph iii) 8. a) = _ + 1; graph ii) + _ b) = + 1; graph i) c) = + 1; graph iv) + d) = 1 _ + 5 5 ; graph iii) 9. a) Since s is in the denominator, each time function is a reciprocal function, so the base function is f(s) = 1 s. b) In t 1 = _ s, a =, so t 1 is a vertical stretch b a factor of of f(s). In t = 18 s, a = 18 and d =, so t is a vertical stretch of f(s) b a factor of 18 and a translation of units right. _ 6 In t = s +, a = 6 and d =, so t is a vertical stretch of f(s) b a factor of 6 and a translation of units left. c) 10. a) domain {t R, t 0}, range {V R, 0 V 000} b) $000 c) i) $9 ii) $667 iii) $186 11. i) ii) 1. a) b) i) g() = ( + 5) d) Substitute s = 6 in each time function. t 1 = _ 6 = It will take Andrew and David h to travel across the lake. _ 18 t = 6 = 9 It will take Andrew and David 9 h to travel up the river. _ 6 t = 6 + = It will take Andrew and David h to travel down the river. ii) h() = 1 56 ( ) + 1. a) translate units right and 5 units up Functions 11 Eercise and Homework Book MHR 1
b) translate units left and 7 units up.7 Inverse of a Function 1. a) {(6, ), (1, ), ( 1, ), (, 5)}; function: domain {,,, 5}, range {6, 1, 1, }; inverse: domain {6, 1, 1, }, range {,,, 5} b) {(7, ), (5, ), (, 1), ( 6, 0)}; function: domain {,, 1, 0}, range {7, 5,, 6}; inverse: domain {7, 5,, 6}, range {,, 1, 0} c) d). a) 8 0 1 5 Domain 1 1 5 7 Range function: domain {, 1, 1,, 5, 7}, range { 8,,, 0, 1, 5}; inverse: domain { 8,,, 0, 1, 5}, range {, 1, 1,, 5, 7} 7 11 17 Domain 0 1 5 Range function: domain {0, 1,,,, 5}, range {,, 7, 11, 17, }; inverse: domain {,, 7, 11, 17, }, range {0, 1,,,, 5}. a) function: domain { 6,,, 5, 8}, range {, 0, 1,, 5}; inverse: domain {, 0, 1,, 5}, range { 6,,, 5, 8} b) function: domain { 7,, 1,, 6, 8}, range { 5, 1,,, 5, 7}; inverse: domain { 5, 1,,, 5, 7}, range { 7,, 1,, 6, 8} f 1 () 0 = function b) f() 0 f() = f 1 () not a function c) f 1 () 0 = f() not a function. a) f 1 () = 5 b) f 1 () = + c) f 1 () = + 7 d) f 1 () = + 1 5. a) f 1 _ () = ± 5 b) f 1 () = ± 7 1 MHR Chapter 978-0-07-01875-5
c) f 1 () = ± d) f 1 () = ± ( + ) 6. a) f() = ( ) 1 _ f 1 () = ± + 1 b) f() = ( 7) + 10 _ f 1 () = 7 ± + 10 c) f() = ( + ) f 1 () = ± _ + d) f() = ( + ) 5 f 1 () = ± 5 7. a) i) f 1 () = + 5 ii) iii) function b) i) f 1 () = + 1 ii) iii) function _ c) i) f 1 () = ± 8 ii) b) r(s) = S ; domain {S R, S 0}, π range {r R, r 0} 9. a) Let s represent Aubre s weekl sales and let e represent her weekl earnings. The equation that represents her weekl earnings is e = 50 + 0.08s, where 50 corresponds to the $50 she earns each week and 0.08s represents 8% of her sales. b) To find the inverse, solve the equation for s. e = 50 + 0.08s e 50 = 0.08s e 50 0.08 = s s = 1.5e 565 c) The inverse represents Aubre s weekl sales. d) Substitute 105 in s. s = 1.5(105) 565 = 7187.50 Aubre s sales for the week were $7187.50. 10. a) u = 0.89c b) c = 1.1u; the value of the Canadian dollar in U.S. dollars c) $80.00 11. a) Since time must be positive, the domain is restricted to t 0. domain {t R, t 0}, range {h R, 0 h 100} iii) not a function d) i) f 1 () = 8 ± + ii) iii) not a function 8. a) domain {r R, r 0}, range {S R, S 0} b) Isolate t to determine the inverse. h = 100.9t h 100 =.9t t _ = 100 h.9 t = 100 h.9 Time is positive, so ignore the negative root. domain {h R, 0 h 100}, range {t R, t 0} Functions 11 Eercise and Homework Book MHR 15
c) The inverse represents the time for an object to fall from a height of 100 m above the ground. d) The object hits the ground when the height is 0 m. Substitute h = 0 in the inverse function and solve for t. t = 100 0.9 =.5 The object hits the ground after.5 s. 1. a) Yes; g() is a reflection of f() in = b) Yes; g() is a reflection of f() in = c) No; g() is a reflection of f() in the -ais 1. a) f 1 () = + 5 b) f(): domain { R, 5 }, range { R, 0}; f 1 (): domain { R, 0}, range { R, 5 } c) 1. a) f 1 () = + _ b) f(): domain { R, 1 }, range { R, 0}; f 1 (): domain { R, 0}, range { R, 1 } _ 60 15. a) n(a) = 180 a b) function: domain {n N, n }, range {a R, 60 a < 180}; inverse: domain {a R, 60 a < 180}, range {n N, n } c) Chapter Review 1. a) No b) Yes c) No. a) Yes; 1, 0 +1 b) No; g() = ;, 0. a) _ 1 ; 8, + b) + 6 ;, 6. a) a b; a 0, b 0 b) 5ab ; a 0, b 0, c 0 18c c) ; 9, _ d) 8 ;,, 5 + e) ;, _ f) + 8 ;, 1, 8 8 5. a) 19 15 ; 0 _ 7a + a ab b) 1ab ; a 0, b 0 _ c) + 8 + ( 1)( + ) ;, 1 _ d) + 1 ( )( ) ;, e) 10 9 ( )( + ) ;, _ 5 f) ; 5,, ( + ) 6. a) s() = 6 b) t() = 7. i) a) s() = ( + ) 8 b) t() = ( _ 5) + 1 ii) a) s() = + 8 _ b) t() = 5 + 1 _ 1 iii) a) s() = + 8 _ 1 b) t() = 5 + 1 8. a) base function: f() = 1 ; transformed _ 1 function: g() = + 6 b) base function: f() = _ ; transformed function: g() = + + 1 9. a) No b) Yes; g() = f( ) 16 MHR Chapter 978-0-07-01875-5
c) No d) Yes; g() = f() _ 10. a) g() = 1 8 b) g() = ( + ) 10 _ 1 c) g() = 7 + 11. a) a = 9; vertical stretch b a factor of 9 b) k =; horizontal compression b a factor of 1 c) a = 5 ; vertical compression b a factor of 5 d) k = 1 7 ; horizontal stretch b a factor of 7 1. a) vertical stretch of f() = b a factor of 1 b) a = 1 5, c = ; compress verticall b a factor of 1 5, then translate units down c) d = 9, c = 8; translate 9 units left and 8 units up d) k = 1, c = 10; stretch horizontall b a factor of, then translate 10 units up _ 1. a) g() = b) g() = 1 _ + 9 b) horizontal compression of f() = b a factor of 1 5 c) g() = _ 1 ( + 5) 16 c) horizontal stretch of f() = b a factor of d) g() = 7 15. a) 8 = d) vertical stretch of f() = 1 b a factor of 6 f 1 () 0 8 f() 1. a) a = 7, d = 1; stretch verticall b a factor of 7, then translate 1 unit right function: domain {, 1,, 6, 8}, range { 6,, 0,, }; inverse: domain { 6,, 0,, }, range {, 1,, 6, 8} Functions 11 Eercise and Homework Book MHR 17
b) f() 0 = f 1 () function: domain { R, 7 7}, range { R, 5}; inverse: domain { R, 5}, range { R, 7 7} 16. a) f 1 () = + 7 b) f 1 () = + 5 _ c) f 1 () = ± 1 _ d) f 1 () = ± 9 Chapter Math Contest 1. C. C. B. D 5. C 6. A 7. D 8. A 9. C 10. B 11. 8 _ b a 1. 8a 6b 1. 50; 50 = 1 + 9; 50 = 5 + 5 1. 5 15. 50 Chapter Eponential Functions.1 The Nature of Eponential Growth 1. a) First Second Da PopulationDifferencesDifferences 0 5 1 75 50 5 150 100 675 50 00 05 150 900 5 6075 050 700 b) Yes. For each additional da the ant population increases b a common factor of. c) The ratio of the first differences is. The ratio of the second differences is also. d) Answers ma var. Sample answer: Yes, the pattern will continue. e) First Second Third Fourth Da Population Differences Differences Differences Differences 0 5 1 75 50 5 150 100 675 50 00 00 05 150 900 600 00 5 6075 050 700 1800 100. a) The value 5 and the variable have different positions in each function. In particular the value 5 is the coefficient in = 5, the eponent in = 5, and the base in = 5. The variable is multiplied with the 5 in = 5, is the base in = 5 and is the eponent in = 5. The graph of = 5 is a straight line with slope 5. The graph of = 5 is a polnomial function of degree 5 and etends from quadrant to quadrant 1. The graph of = 5 is an eponential function that etends from quadrant to quadrant 1. b) = 5: domain { R}, range { R} = 5 : domain { R}, range { R} = 5 : domain { R}, range { R, > 0}. Justifications ma var.. a) a 5 a 5 = a a a a a a a a a a b) 1 c) a 0 d) a 0 = 1 5. a) B b) The constant 15 is the initial population. The power n represents the dail quadrupling. 6. a) e) 1 f) 1 18 MHR Chapter 978-0-07-01875-5
7. a) eponential: f() = 8 is of the form = a, where the base a is a constant, such that a > 0, and the eponent is a variable b) linear: f() = 11 9 can be rewritten as f() = 9 + 11, which is of the form = m + b c) none of linear, quadratic, or eponential: f() = is a square root function d) quadratic 8. Determine the first and second differences for each table of values. a) First Second Differences Differences Ratio of First and Second Differences 16 8 8 1 0.5 0 1 1 1 0.5 1 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 The table represents an eponential function. Neither the first differences nor the second differences are constant; however, the ratio of the first and second differences is constant. b) First Differences 1 10 7 1 0 1 1 5 The table represents a linear function because the first differences are constant. c) d) e) First Differences Second Differences 18 11 7 6 5 1 0 1 1 1 6 The table represents a quadratic function because the second differences are constant. First Second Differences Differences Ratio of First and Second Differences 65 15 500 5 100 00 1 5 0 80 0. 0 1 16 0. 1 0. 0.8. 0. 0.0 0.16 0.6 0. The table represents an eponential function. Neither the first differences nor the second differences are constant; however, the ratio of the first and second differences is constant. First Differences Second Differences. 0.8 1. 0. 1 0. 1 0.8 0.6 0. 0 1 0. 0. 1 0.8 0. 0. 0. 0.6 0. The table represents a quadratic function because the second differences are constant. Functions 11 Eercise and Homework Book MHR 19
9. a) Answers ma var. b) Answers ma var. First terms are 1,, 16, 6. c) i) 10 ii) 19 0 d) 5 terms e) 6 terms f) The answers differ b 1 term. This is an eponential function in which each term is multiplied b. 10. a) The initial population is ; since it doubles ever 0 minutes, multipl b for each 0-min interval. The equation is p(t) = t. b) P(t) Population 500 00 00 00 100 P(t) = t 0 1 t Time (0-minute intervals) c) Answers ma var. Sample answer: Since there are two 0-min intervals in each hour, 1 h on the graph corresponds to t =. B moving up to the curve and then across to the vertical ais, it can be estimated that at t = the population is approimatel 10 bacteria. This can be checked in the equation. To use the equation, substitute t = to obtain p() = = 18, which is ver close to 10. d) Since the graph does not etend to.5 h, it is easier to use the equation. In.5 h, there are seven 0-min intervals. Substitute t = 7 in p(t) = t : p(7) = 7 = 096 After.5 h there are 096 bacteria. 11. a) A(n) = 00(1.0 5) n b) Number of Compounding Periods (ears) Amount ($) 0 00.00 1 09.00 18.1 8. 8.50 5 9. 6 60.5 c) Number of Compounding Amount First Second Periods (ears) ($) Differences Differences 0 00.00 1 09.00 9.00 18.1 9.1 0.1 8. 9.8 0.1 8.50 10.7 0.5 5 9. 10.7 0.7 6 60.5 11.1 0.7 Neither. The first differences are not equal, nor are the second differences. d) The ratio of the first differences and second differences is approimatel 1.0. The function is eponential. e) In the equation the constant ratio is represented b the value 1.05. f) No. The amount remains the same between compounding periods and increases onl at the end of each compounding period. 1. a) $166.08 b) $1975.1 1. a) i) 500 ii) 1 500 b) r(n) = 5 n, where r is the number of residents called in each interval, n c) approimatel 5. intervals d) This is an eample of eponential growth, because the number of residents notified increases b a factor of 5 with each interval. 1. a) approimatel 1 ears; Answers ma var. Sample answer: Sstematic trial. b) 5.18 or approimatel 6 more ears (since the interest is paid at the end of the ear) 0 MHR Chapter 978-0-07-01875-5
15. Prize A. The value of Prize A at week 6 alone is $5 5.. The value of Prize B for all 6 weeks is onl $60 000. 16. a) 6 das; population of A is 600; population of B is 819 b) das sooner, on the th da 17. a) b) approimatel 1 h c) For h = 1, M 99.9. d) approimatel 107 h. Eponential Deca: Connecting to Negative Eponents 1. a) 1 7 b) 1 10 d) mn 1 e) 1. a) a b) 5 c) 1 a f) _ 1 b c) 9 d) 5 b 6. a) 1 b) 1 5 _ 1 c) 81 1000 d) 1 e) _ 1 f ) _ 1 16 16 6 g) 81 _ h) 5 8. a) 5 b) 1 51 c) 79 d) 1 6 e) f ) 6 5. a) a b) 1 v 7 _ m c) a 10 d) _ 6. a) 6 b) 9 16 _ c) 6 7 d) 81 65 7. a) b) 11b _ c) b8 a 1 8n1 d) 15m 6 8. a) Number of 0-da Intervals, n Amount Remaining (mg) 0 0 1 0 10 5.5 5 1.5 Half the amount remains after each 0-da interval. b) A = 0 ( 1 ) n, where n is the number of half-life periods, in 0-da intervals, and A is the amount of polonium-10 remaining, in milligrams c) Plot the points from the table. Connect them with a smooth curve. Amount (mg) A 50 0 0 0 10 0 1 A = 0( ) n 6 8 Half-Life Period The graph starts at point (0, 0), decreases b a factor of 1 with each 0-da interval, and has a horizontal asmptote, = 0. d) Since there are 7 das in a week, 10 weeks = 70 das. This represents.5 half-life periods. Substitute n =.5 into the equation to determine the amount remaining: A = 0 ( 1 ).5 =.5 Approimatel.5 mg of polonium-10 will remain after 10 weeks. n Functions 11 Eercise and Homework Book MHR 1
e) Answers ma var. Sample answer: Determine 8% of 0 mg: 8% = 0.08, so 0.08(0) mg =. mg Substitute A = 6 into the equation and solve for n:. = 0 ( 1 ) n Divide each side b 0. 0.08 = ( 1 ) n Use sstematic trial to solve for n. Observe the chart in part a). Note that when n = the amount remaining is 5 mg, and when n = the amount remaining is.5 mg. Therefore, for. mg, the value of n is between and. Tr n =.5 in ( 1 ) n : ( 1 ).5 0.088, which is a bit high. Tr n =.6 to get ( 1 ).6 0.080, which is ver close, so n =.6. Multipl to find the number of das..6(0) 7 das Therefore, it takes approimatel 7 das for polonium-10 to deca to 8% of its initial mass. 1 f) Since = 1, an equivalent wa to write the equation is A = 0( 1 ) n or A = 0( n ). Since 1 = 0.5, another equivalent wa to write the equation is A = 0(0.5) n. 9. a) In the formula 1 500 represents the initial value of the motorccle. Since the value depreciates b 0% per ear, 80% of the value remains. This amount is represented b the decimal 0.8. b) i) $10 800 ii) $58.9 c) Answers ma var. Sample answer: The equation is of the form f() = ab, where 0.8 is the constant ratio. This value is less than 1, so the initial amount is decreasing. d).1 ears, or approimatel ears 1 month 10. $70.6 11. a) The initial deposit was made 5 ears ago, so determine the amount in the account when n = 5. Substitute A = 00, i = 0.075, and n = 5 in the formula P = A(1 + i) n. P = 00(1 + 0.075) 5 95.55 Denise s initial deposit was $95.55. b) Substitute A = 00, i = 0.075, and n = in the formula P = A(1 + i) n. P = 00(1 + 0.075) 6.0 The amount in the account ears ago was $6.0. c) Use the formula A = P(1 + i) n to determine the amount in the account ears from now. Substitute P = 00, i = 0.075, and n =, and solve for A. A = 00(1 + 0.075) 85.6 Two ears from now the amount in the account will be $85.6. To determine the total interest earned, subtract the initial deposit from the amount at this point. $85.6 $95.55 = $198.08 The total interest earned up to this point is $198.08. 1. approimatel 17.% 1. a) c = 100 ( t 5 ) b) i) 16 h 6 min ii) h 1 min 1. a) T = 80 ( t 5 ) + 0 b) 15 min. Rational Eponents 1. a) 6 b) 11 c) 5 7 d) 9. a) 5 b) c) d). a) 1 5 = ( 5 ) = ( 5 ) = = 8 MHR Chapter 978-0-07-01875-5
b) ( 6) = ( 6 1 ) c) 6 5 1 6 = ( 6 = ( 6 ) = ( ) = 16 6 ) 5 = ( 6 6 ) 5 = 5 = 5 d) 65618 = ( 6561. a) 178 1 = b) 6 = 1 8 ) 5 = ( 8 6561 ) 5 = 5 = _ 1 1 178 1 = ( 178 ) = _ 1 1 _ 1 6 1 = 1 ( 6 ) 1 = ( 6 ) = 1 6 1 = 16 8 c) ( 15 ) 5 1 = 8 ( 15 ) 5 _ 1 = ( 8 1 ) 5 _ 1 15 _ 1 = 5 _ 8 ) ( = ( ( 15 ) ) _ 1 ( 5 ) 5 1 = _ 15 _ = 15 _ d) ( 10 ) 5. a) 8 1 8 5 = 1 _ ( 10 _ ) 5 _ 1 = 5 _ 10 ( 5 ) = 1 ( ) = _ 1 _ 6 7 _ = 7 6 = 8 1 + Use the product rule for eponents. = 8 = 8 b) 16 1 16 1 16 1 = 16 1 + Find a common denominator for the fractional eponents. 1 = 16 1 + = 16 Reduce the eponent. 1 = 16 _ = 16 = c) 6 1 6 1 6 6 1 = 6 + 1 6 Find a common denominator for the eponents. = 66 + 1 6 6 1 = 66 _ 1 = Take the reciprocal. 6 1 6 _ 1 = 6 6 = 1 d) 7 9 = ( ) 9 Functions 11 Eercise and Homework Book MHR
= 9 = = 9 6. a) b) a 7 6 c) a 8 b d) z _ 8 15 7. a) b _ 15 b) _ 1 a _ 1 1 c) w _ 11 d) a 1 8. 165.7 cm 9. a) C = π ( A π ) 1 b) C = ( πa ) 1 c) i) 6.8 cm ii) 108.0 cm iii) 16.1 cm d) Answers ma var. 10. a) 00 b) approimatel h 0 min c) Yes. The eponents are equivalent _ because t 0.5 = t. d) Answers ma var. 11. a) N = 1000( t ), where t is the time, in minutes b) 000 c) approimatel 18 min 1. a) i) beats per min ii) beats per min iii) 159 beats per min b) i) 5 breaths per min ii) 10 breaths per min iii) 5 breaths per min c) i) 5.08 kg ii) 0.987 kg or 987 g iii) 0.0 kg or 0. g d) i) the larger the animal, the fewer beats per min ii) the larger the animal, the fewer breaths per min iii) the larger the animal, the greater the brain mass e) Answers ma var. 1. a) f 1 () = ; 16 b) f 1 ( ) = ( + ) 5 ; c) f 1 ( ) = + ; 68.01 1. a) _ 9 0 b) _ 1 _ 7 1 _ c) 9 7m _ 1 d) 6 b 9 + 1 e) a _ 5 8 t _ 15. a) P = P 0 (10) 0 b) approimatel. times greater 16. a) No. Answers ma var. b) No. Answers ma var.. Properties of Eponential Functions 1. Graph a) matches B = ( 1 ) Graph b) matches D = Graph c) matches C = 1 ( ) Graph d) matches A = ( ). Graph a) matches C = 1 () Graph b) matches A = () Graph c) matches D = ( 1 ) Graph d) matches B = (). a) Answers ma var. b) No. The conditions are satisfied b an curve whose equation is of the form = b, where b > 1.. a) Answers ma var. b) No. The conditions are satisfied b an curve whose equation is of the form = b, where b > 1. 5. = 6 6. = 8 ( 1 ) 7. a) C b) 9.5 mg 8. a) 8 6 0 1 f() = ( ) 6 MHR Chapter 978-0-07-01875-5