THÔNG TIN TÀI LIỆU
CHAPTER FUNCTIONS 1.1 FUNCTIONS AND THEIR GRAPHS domain = (−∞, ∞); range = [1, ∞) domain = [0, ∞); range = (−∞, 1] domain = [−2, ∞); y in range and y = x + 10 ≥ y can be any positive real number range = [0, ∞) domain = (−∞, 0] ∪ [3, ∞); y in range and y = x − 3x ≥ y can be any positive real number range = [0, ∞) domain = (−∞, 3) ∪ (3, ∞); y in range and y = 4− t , now if t < − t > 4− t > 0, or if t > 3−t < 3−t < y can be any nonzero real number range = (−∞, 0) ∪ (0, ∞) domain = (−∞, − 4) ∪ (− 4, 4) ∪ (4, ∞); y in range and y = ≥ −4 < t < −16 ≤ t − 16 < − 16 t − 16 , t − 16 now if t < −4 t − 16 > , or if t > t − 16 > t − 16 t − 16 > 0, or if > y can be any nonzero real number range = (−∞, − 18 ] ∪ (0, ∞) (a) Not the graph of a function of x since it fails the vertical line test (b) Is the graph of a function of x since any vertical line intersects the graph at most once (a) Not the graph of a function of x since it fails the vertical line test (b) Not the graph of a function of x since it fails the vertical line test base = x; (height) + 2x = x height = 23 x; area is a( x) = 12 (base)(height) = 12 ( x ) 23 x = 43 x ; perimeter is p( x) = x + x + x = 3x ( ) () 10 s = side length s + s = d s = d ; and area is a = s a = 12 d 2 11 Let D = diagonal length of a face of the cube and = the length of an edge Then + D = d and ( 3/2 ( ) 2 D = 2 3 = d = d The surface area is 6 = d3 = 2d and the volume is 3 = d3 ) = d 3 12 The coordinates of P are x, x so the slope of the line joining P to the origin is m = xx = ( x > 0) ( ) Thus, x, x = ( m2 , m x ) 25 13 x + y = y = − 12 x + 45 ; L = ( x − 0)2 + ( y − 0)2 = x + (− 12 x + 54 ) = x + 14 x − 45 x + 16 = x2 − 25 = x + 16 20 x − 20 x + 25 16 = 20 x − 20 x + 25 14 y = x − y + = x; L = ( x − 4)2 + ( y − 0)2 = ( y + − 4)2 + y = ( y − 1)2 + y = y4 − y2 + + y2 = y4 − y2 +1 Copyright 2016 Pearson Education, Inc Chapter Functions 15 The domain is (−∞, ∞) 16 The domain is (−∞, ∞) 17 The domain is (−∞, ∞) 18 The domain is (−∞, 0] 19 The domain is (−∞, 0) ∪ (0, ∞) 20 The domain is (−∞, 0) ∪ (0, ∞) 21 The domain is (−∞, −5) ∪ (−5, −3] ∪ [3, 5) ∪ (5, ∞) 22 The range is [2, 3) 23 Neither graph passes the vertical line test (a) (b) Copyright 2016 Pearson Education, Inc Section 1.1 24 Neither graph passes the vertical line test (a) Functions and Their Graphs (b) x+ y =1 y = 1− x x + y =1⇔ or or ⇔ x + y = −1 y = −1 − x 25 26 x y − x , x ≤ 27 F ( x) = x + x, x > x y 0 , x < 28 G ( x) = x x, ≤ x 29 (a) Line through (0, 0) and (1, 1): y = x; Line through (1, 1) and (2, 0): y = − x + x, ≤ x ≤ f ( x) = − x + 2, < x ≤ 2, 0, (b) f ( x) = 2, 0, ≤ x 0: 2x − − 4x > >0 >0 2x 2x x > since x is positive; x2 − x − x < 0: 2x − − 4x > < 0 2x x < −2 since x is negative; sign of ( x − 4)( x + 2) ( x − 4)( x + 2) 2x 3x + < x − x < −5 Thus, x ∈ (−∞, −5) solves the inequality 3( x + 1) Case −1 < x < 1: x 3− < x 2+ x − < 3x + > x − x > −5 which is true if x > −1 Thus, x ∈ (−1, 1) solves the inequality Case < x : x 3− < x 2+ 3x + < x − x < −5 which is never true if < x, so no solution here In conclusion, x ∈ (−∞, −5) ∪ (−1, 1) 69 A curve symmetric about the x-axis will not pass the vertical line test because the points (x, y) and ( x, − y ) lie on the same vertical line The graph of the function y = f ( x) = is the x-axis, a horizontal line for which there is a single y-value, 0, for any x 70 price = 40 + x, quantity = 300 − 25x R ( x) = (40 + x)(300 − 25 x) 71 x + x = h2 x = h = 22h ; cost = 5(2 x) + 10h C (h) = 10 ( ) + 10h = 5h ( 2h 2+2 ) 72 (a) Note that mi = 10,560 ft, so there are 8002 + x feet of river cable at $180 per foot and (10,560 − x) feet of land cable at $100 per foot The cost is C ( x) = 180 8002 + x + 100(10,560 - x) (b) C (0) = $1, 200, 000 C (500) ≈ $1,175,812 C (1000) ≈ $1,186,512 C (1500) ≈ $1, 212, 000 C (2000) ≈ $1, 243, 732 C (2500) ≈ $1, 278, 479 C (3000) ≈ $1,314,870 Values beyond this are all larger It would appear that the least expensive location is less than 2000 feet from the point P 1.2 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS D f : −∞ < x < ∞, Dg : x ≥ D f + g = D fg : x ≥ R f : −∞ < y < ∞, Rg : y ≥ 0, R f + g : y ≥ 1, R fg : y ≥ D f : x + ≥ x ≥ −1, Dg : x − ≥ x ≥ Therefore D f + g = D fg : x ≥ R f = Rg : y ≥ 0, R f + g : y ≥ 2, R fg : y ≥ D f : −∞ < x < ∞, Dg : −∞ < x < ∞, D f /g : −∞ < x < ∞, Dg /f : −∞ < x < ∞, R f : y = 2, Rg : y ≥ 1, R f /g : < y ≤ 2, Rg /f : 12 ≤ y < ∞ D f : −∞ < x < ∞, Dg : x ≥ 0, D f /g : x ≥ 0, Dg /f : x ≥ 0; R f : y = 1, Rg : y ≥ 1, R f /g : < y ≤ 1, Rg /f : ≤ y < ∞ Copyright 2016 Pearson Education, Inc Section 1.2 Combining Functions; Shifting and Scaling Graphs (a) (d) ( x + 5) − = x + 10 x + 22 (g) x + 10 (b) 22 (e) (h) ( x − 3)2 − = x − x + (c) x + (f ) −2 (a) − 13 (d) 1x (b) (c) x 1+ − = x−+x (f ) 34 (e) (g) x − (h) 1 x +1 +1 = x+2 x +1 = x +1 x+2 ( f g h)( x) = f ( g (h( x))) = f ( g (4 − x)) = f (3(4 − x )) = f (12 − x) = (12 − 3x) + = 13 − 3x ( f g h)( x) = f ( g (h( x))) = f ( g ( x )) = f (2( x ) − 1) = f (2 x − 1) = 3(2 x − 1) + = x + ( ( )) = f ( f g h)( x) = f ( g (h( x))) = f g 1x x = + f ( x + 4x )= x + 4x +1 = 5x + 1 + 4x 2−x ( − x )2 +2 2−x − 3x 3− x = f f = = = − 2x 2 −x 3− x − 3− x ( − x) + 1 ( ) 10 ( f g h)( x) = f ( g (h( x))) = f g 2− x 11 (a) ( f g )( x) (d) ( j j )( x) (b) ( j g )( x) (e) ( g h f )( x) (c) ( g g )( x) (f ) (h j f )( x) 12 (a) ( f j )( x) (d) ( f f )( x) (b) ( g h)( x) (e) ( j g f )( x) (c) (hh)( x) (f ) ( g f h)( x) ( ( )) f (x) ( f g )( x) (a) x − x x−7 (b) x + 3x 13 g(x) (c) x 3( x + 2) = x + x2 − x −5 (d) x x −1 x x −1 (e) x −1 + 1x x x x (f ) x x x −1 x −1 x −1 = x − (xx − 1) = x 14 (a) ( f g )( x) = |g ( x)| = (b) ( f g )( x) = x −1 g ( x) − = x x+ g ( x) − g (1x) = x x+ − x x+ = g (1x) x 1+ = g (1x) , so g ( x) = x + (c) Since ( f g )( x) = g ( x) = | x |, g ( x) = x (d) Since ( f g )( x) = f x = | x |, f ( x) = x (Note that the domain of the composite is [0, ∞).) ( ) Copyright 2016 Pearson Education, Inc 10 Chapter Functions The completed table is shown Note that the absolute value sign in part (d) is optional g(x) f(x) ( f g )(x) x −1 | x| x −1 x +1 x −1 x x x +1 x2 x | x| x | x| x 15 (a) f ( g (−1)) = f (1) = (d) g ( g (2)) = g (0) = 16 (a) (b) (c) (d) (e) (b) g ( f (0)) = g (−2) = (e) g ( f (−2)) = g (1) = −1 (c) f ( f (−1)) = f (0) = −2 (f) f ( g (1)) = f (−1) = f ( g (0)) = f (−1) = − (−1) = 3, where g (0) = − = −1 g ( f (3)) = g (−1) = −(−1) = 1, where f (3) = − = −1 g ( g (−1)) = g (1) = − = 0, where g (−1) = −(−1) = f ( f (2)) = f (0) = − = 2, where f (2) = − = g ( f (0)) = g (2) = − = 1, where f (0) = − = ( ( )) = f ( − 12 ) = − ( − 12 ) = 52 , where g ( 12 ) = 12 − = − 12 (f ) f g 12 17 (a) ( f g )( x) = f ( g ( x)) = ( g f )( x) = g ( f ( x)) = +1 x x +1 = 1+ x x (b) Domain ( f g ): (−∞, − 1] ∪ (0, ∞), domain ( g f ): (−1, ∞) (c) Range ( f g ): (1, ∞), range ( g f ): (0, ∞) 18 (a) ( f g )( x ) = f ( g ( x)) = − x + x ( g f )( x ) = g ( f ( x)) = − | x | (b) Domain ( f g ): [0, ∞), domain ( g f ): (−∞, ∞) (c) Range ( f g ): (0, ∞), range ( g f ): (−∞, 1] g ( x) 19 ( f g )( x) = x f ( g ( x)) = x g ( x) − = x g ( x) = ( g ( x) − 2) x = x ⋅ g ( x) − x g ( x) − x ⋅ g ( x) = −2 x g ( x) = − 2−xx = x2−x 20 ( f g )( x ) = x + f ( g ( x )) = x + 2( g ( x))3 − = x + ( g ( x))3 = 21 (a) y = −( x + 7)2 (b) y = −( x − 4) 22 (a) y = x + (b) y = x − x+6 g ( x) = x+6 23 (a) Position (b) Position (c) Position (d) Position 24 (a) y = −( x − 1) + (b) y = −( x + 2)2 + (c) y = −( x + 4)2 − (d) y = −( x − 2) Copyright 2016 Pearson Education, Inc 28 Chapter Functions 17 [ − 5, 1] by [ − 5, 5] 18 [ − 5, 1] by [ − 2, 4] 19 [ − 4, 4] by [0, 3] 20 [ − 5, 5] by [ − 2, 2] 21 [ −10, 10] by [ − 6, 6] 22 [ −5, 5] by [ − 2, 2] 23 [ − 6, 10] by [ − 6, 6] 24 [ − 3, 5] by [ − 2, 10] 25 [−0.03, 0.03] by [−1.25, 1.25] 26 [−0.1, 0.1] by [−3, 3] Copyright 2016 Pearson Education, Inc Section 1.4 Graphing with Software 27 [−300, 300] by [−1.25, 1.25] 28 [−50, 50] by [−0.1, 0.1] 29 [−0.25, 0.25] by[−0.3, 0.3] 30 [−0.15, 0.15] by [−0.02, 0.05] 31 x + x = + y − y y = ± − x − x + The lower half is produced by graphing y = − − x − x + 32 y − 16 x = y = ± + 16 x The upper branch is produced by graphing y = + 16 x 33 34 Copyright 2016 Pearson Education, Inc 29 30 Chapter Functions 35 36 1.5 EXPONENTIAL FUNCTIONS Copyright 2016 Pearson Education, Inc Section 1.5 Exponential Functions 10 31 11 162 ⋅16−1.75 = 162 +( −1.75) = 160.25 = 161/4 = +1 12 91/3 ⋅ 91/6 = − 23 14 35/3 32/3 16 (13 ) 18 ( ) ( 12 ) 19 ( ) = 33 2 /2 = = 44.2 −3.7 = 40.5 = 41/2 = = 132/2 = 13 17 ⋅ = (2 ⋅ 7) = 14 24 1/ (2 44.2 43.7 15 (251/8 )4 = 254/8 = 251/2 = 1/2 13 = 31 = 1/2 2 = 91/2 = ) = ( ⋅ 12 1/2 ) = ( 36 = 162 = 1/2 ) = 61/2 20 ( ) = (61/ )2 32 = 69 = 23 21 Domain: (−∞, ∞); y in range y = x As x increases, e x becomes infinitely large and y becomes a smaller 2+ e and smaller positive real number As x decreases, e x becomes a smaller and smaller positive real number, ( ) y < 12 , and y gets arbitrarily close to 12 Range: 0, 12 22 Domain: (−∞, ∞); y in range y = cos(e−t ) Since the values of e−t are (0, ∞) and −1 ≤ cos x ≤ Range: [−1, 1] Copyright 2016 Pearson Education, Inc 32 Chapter Functions 23 Domain: (−∞, ∞); y in range y = + 3−t Since the values of 3−t are (0, ∞) Range: (1, ∞) 24 If e2 x = 1, then x = Domain: (−∞, 0) ∪ (0, ∞); y in range y = 1−e x If x > 0, then < e2 x < ∞ −∞ < y < If x < 0, then < e x < < y < ∞ Range: (−∞, 0) ∪ (3, ∞) 25 26 x ≈ 2.3219 x ≈ 1.3863 27 28 x ≈ −0.6309 x ≈ −1.5850 29 Let t be the number of years Solving 500,000(1.0375)t = 1, 000, 000 graphically, we find that t ≈ 18.828 The population will reach million in about 19 years 30 (a) The population is given by P(t ) = 6250(1.0275)t , where t is the number of years after 1890 Population in 1915: P(25) ≈ 12,315 Population in 1940: P(50) ≈ 24,265 (b) Solving P(t) = 50,000 graphically, we find that t ≈ 76.651 The population reached 50,000 about 77 years after 1890, in 1967 31 (a) A(t ) = 6.6 t /14 ( 12 ) (b) Solving A(t) = graphically, we find that t ≈ 38 There will be gram remaining after about 38.1145 days 32 Let t be the number of years Solving 2300(1.60)t = 4150 graphically, we find that t ≈ 10.129 It will take about 10.129 years (If the interest is not credited to the account until the end of each year, it will take 11 years.) Copyright 2016 Pearson Education, Inc Section 1.6 Inverse Functions and Logarithms 33 33 Let A be the amount of the initial investment, and let t be the number of years We wish to solve A(1.0625)t = A, which is equivalent to 1.0625t = Solving graphically, we find that t ≈ 11.433 It will take about 11.433 years (If the interest is credited at the end of each year, it will take 12 years.) 34 Let A be the amount of the initial investment, and let t be the number of years We wish to solve Ae0.0575t = A, which is equivalent to e0.0575t = Solving graphically, we find that t ≈ 19.106 It will take about 19.106 years 35 After t hours, the population is P(t ) = 2t /0.5 , or equivalently, P(t ) = 22t After 24 hours, the population is P(24) = 248 ≈ 2.815 ×1014 bacteria 36 (a) Each year, the number of cases is 100% − 20% = 80% of the previous year’s number of cases After t years, the number of cases will be C (t ) = 10, 000(0.8)t Solving C(t) = 1000 graphically, we find that t ≈ 10.319 It will take 10.319 years (b) Solving C(t) = graphically, we find that t ≈ 41.275 It will take about 41.275 years 1.6 INVERSE FUNCTIONS AND LOGARITHMS Yes one-to-one, the graph passes the horizontal line test Not one-to-one, the graph fails the horizontal line test Not one-to-one since (for example) the horizontal line y = intersects the graph twice Not one-to-one, the graph fails the horizontal line test Yes one-to-one, the graph passes the horizontal line test Yes one-to-one, the graph passes the horizontal line test Not one-to-one since the horizontal line y = intersects the graph an infinite number of times Yes one-to-one, the graph passes the horizontal line test Yes one-to one, graph passes the horizontal line test 10 Not one-to-one since (for example) the horizontal line y = intersects the graph twice 11 Domain: < x ≤ 1, Range: ≤ y 12 Domain: x < 1, Range: y > Copyright 2016 Pearson Education, Inc 34 Chapter Functions 13 Domain: −1 ≤ x ≤ 1, Range: − π2 ≤ y ≤ π2 14 Domain: −∞ < x < ∞, Range: − π2 < y ≤ π2 15 Domain: ≤ x ≤ 6, Range: ≤ y ≤ 16 Domain: −2 ≤ x ≤ 1, Range: −1 ≤ y < 17 The graph is symmetric about y = x (b) y = − x y = − x x = − y x = − y y = − x = f −1 ( x) 18 (a) The graph is symmetric about y = x 19 Step 1: y = x + x = y − x = (b) y= x x= y y= y −1 Step 2: y = x − = f −1 ( x) 20 Step 1: y = x x = − y , since x ≤ Step 2: y = − x = f −1 ( x) 21 Step 1: y = x3 − x3 = y + x = ( y + 1)1/3 Step 2: y = x + = f −1 ( x) Copyright 2016 Pearson Education, Inc x = f −1 ( x ) Section 1.6 Inverse Functions and Logarithms 22 Step 1: y = x − x + y = ( x − 1)2 Step 2: y = + x = f −1 23 Step 1: y = ( x + 1)2 y = x − 1, since x ≥ x = + y ( x) y = x + 1, since x ≥ −1 x = y −1 Step 2: y = x − = f −1 ( x) 24 Step 1: y = x 2/3 x = y 3/2 Step 2: y = x3/2 = f −1 ( x) 25 Step 1: y = x5 x = y1/5 Step 2: y = x = f −1 ( x); Domain and Range of f −1: all reals; f ( f −1 ( x)) = ( x1/5 )5 = x and f −1 ( f ( x)) = ( x5 )1/5 = x 26 Step 1: y = x x = y1/4 Step 2: y = x = f −1 ( x); Domain of f −1: x ≥ 0, Range of f −1: y ≥ 0; f ( f −1 ( x)) = ( x1/4 )4 = x and f −1 ( f ( x)) = ( x )1/4 = x 27 Step 1: y = x3 + x3 = y − x = ( y − 1)1/3 Step 2: y = x − = f −1 ( x); Domain and Range of f −1: all reals; f ( f −1 ( x)) = (( x − 1)1/3 )3 + = ( x − 1) + = x and f −1 ( f ( x)) = (( x3 + 1) − 1)1/3 = ( x3 )1/3 = x 28 Step 1: y = 12 x − 72 12 x = y + 72 x = y + Step 2: y = x + = f −1 ( x); Domain and Range of f −1: all reals; f ( f −1 ( x)) = 12 (2 x + 7) − 72 = x + 72 − 72 = x and f −1 ( f ( x)) = ( ) ( 12 x − 72 ) + = ( x − 7) + = x 29 Step 1: y = 12 x = 1y x = y x Step 2: y = = f −1 ( x) Domain of f f ( f −1 ( x)) = x −1 : x > 0, Range of f −1: y > 0; ( ) x = ( ) x = x and f −1 ( f ( x)) = 1 x2 = ( 1x ) = x since x > Copyright 2016 Pearson Education, Inc 35 36 Chapter Functions 1 30 Step 1: y = 13 x3 = x = 1/3 x y y Step 2: y = = 1x = f −1 ( x); 1/3 x Domain of f −1: x ≠ 0, Range of f −1: y ≠ 0; f ( f −1 ( x)) = (x −1/3 ) = x −1 = x and f −1 ( f ( x)) = ( ) x3 −1/3 −1 = 1x =x () y +3 31 Step 1: y = xx−+23 y(x − 2) = x + xy − 2y = x + xy − x = 2y + x = y −1 Step 2: y = 2xx−+13 = f −1 ( x); Domain of f −1: x ≠ 1, Range of f −1: y ≠ 2; f ( f −1 ( x )) = ( 2xx−+13 )+3 = (2 x +3)+3( x −1) = x = x ( 2xx−+13 )−2 (2 x +3)−2( x −1) y x x −3 32 Step 1: y = ( and f −1 ( f ( x)) = ( xx −+32 )+3 = 2( x+3)+3( x−2) = x = x ( xx −+32 )−1 ( x +3)−( x −2) ) x − = x y x − 3y = x y x − x = 3y x = 3y y −1 ( ) Step 2: y = x3−x1 = f −1 ( x); ( ) Domain of f −1: (−∞, 0] ∪ (1, ∞), Range of f −1: [0, 9) ∪ (9, ∞); f(f −1 ( x3−x1 ) ; ( x3−x1 ) −3 ( x)) = 3 f −1 ( f ( x)) = If x > or x ≤ 3x x −1 ≥0 ( x3−x1 ) = ( x3−x1 ) −3 3x x −1 3x − x −1 = 3x x −3( x −1) = 3x = x and = x x −1 x −3 ( ) x x −3 9x ( ( x− x −3 )) = 9x =x 33 Step 1: y = x − x, x ≤ y + = ( x − 1)2 , x ≤ − y + = x − 1, x ≤ x = − y + Step 2: y = − x + = f −1 ( x); Domain of f −1: [−1, ∞), Range of f −1: (−∞, 1]; f ( f −1 ( x)) = − x + ( ) ( ) − − x + = − x + + x + − + x + = x and f −1 ( f ( x)) = − ( x − x) + 1, x ≤ = − ( x − 1)2 , x ≤ = − |x − 1| = − (1 − x) = x Copyright 2016 Pearson Education, Inc Section 1.6 Inverse Functions and Logarithms y −1 y −1 34 Step 1: y = (2 x3 + 1)1/5 y = x3 + y − = x3 = x3 x = Step 2: y = x 2−1 = f −1 ( x); Domain of f −1: (−∞, ∞), Range of f −1: (−∞, ∞); f(f −1 ( x)) = 1/5 + 1 x5 −1 1/5 ( ( ) + 1) = x5 −1 = (( x5 − 1) + 1)1/5 = ( x5 )1/5 = x and f 35 −1 ( f ( x)) = 1/5 (2 x +1) −1 =3 (2 x3 +1) −1 x+b x−2+2+b 2+b = = 1+ x−2 x−2 x−2 2x + b f −1 ( x ) = x −1 y= 36 Since x ≤ b , x − 2bx − y = x = b − b2 + y x =3 x x3 =x 2y + b y −1 2b ± 4b + y f −1 ( x ) b b2 b b2 y x y f −1 ( x ) = x 37 (a) y = mx x = m m (b) The graph of y = f −1 ( x) is a line through the origin with slope m1 y x − b ; the graph of f −1 ( x ) is a line with slope and 38 y = mx + b x = m − mb f −1 ( x) = m m m b y-intercept − m 39 (a) y = x + x = y − f −1 ( x) = x − (b) y = x + b x = y − b f −1 ( x) = x − b (c) Their graphs will be parallel to one another and lie on opposite sides of the line y = x equidistant from that line 40 (a) y = −x + x = −y + f −1 ( x) = − x; the lines intersect at a right angle (b) y = −x + b x = −y + b f −1 ( x) = b − x; the lines intersect at a right angle (c) Such a function is its own inverse Copyright 2016 Pearson Education, Inc 37 38 Chapter Functions 41 (a) ln 0.75 = ln 34 = ln − ln = ln − ln 22 = ln − ln (b) ln 94 = ln − ln = ln 22 − ln 32 = ln − ln (d) ln = 13 ln = 13 ln 32 = 32 ln (c) ln 12 = ln1 − ln = − ln (e) ln = ln + ln 21/2 = ln + 12 ln (f) ln 13.5 = 12 ln13.5 = 12 ln 27 = 12 (ln 33 − ln 2) = 12 (3ln − ln 2) = ln1 − 3ln = −3ln ln 125 (b) ln 9.8 = ln 49 = ln − ln = ln − ln 5 (c) ln 7 = ln 73/2 = 32 ln (d) ln1225 = ln 352 = ln 35 = ln + ln (e) = ln − ln 53 = ln − 3ln ln 0.056 − ln 125 42 (a) (f) 43 (a) (c) ln 35+ ln 17 ln 25 = ln 5+ ln − ln ln = θ ln sin θ − ln sin5 θ = ln sin sin θ = ln ( ) ln(4t ) − ln 2 ( ) (b) ln(3 x − x) + ln 31x = ln x 3−x x = ln( x − 3) ( ) ( ) = ln 4t − ln = ln 2t − ln = ln ( ) = ln(t ) 2t 2 44 (a) ln sec θ + ln cos θ = ln[(sec θ)(cos θ)] = ln = (b) ln(8 x + 4) − ln 22 = ln(8 x + 4) − ln = ln x4+ = ln(2 x + 1) ( ) ( ) (t +1)(t −1) = ln(t − 1) (c) 3ln t − − ln(t + 1) = 3ln(t − 1)1/3 − ln(t + 1) = 13 ln(t − 1) − ln(t + 1) = ln (t +1) () 45 (a) eln 7.2 = 7.2 46 (a) eln( x 47 (a) ln e = ln e1/2 = (2) (c) 48 (a) (c) + y2 ) = x2 + y2 (b) e− ln x = = 12 (c) eln x −ln y = eln( x / y ) = (b) e− ln 0.3 = (c) eln π x −ln = eln(π x /2) = π2x eln x ( 12 ) ln e = 1 eln 0.3 x = 0.3 (b) ln(ln ee ) = ln(e ln e) = ln e = 2 ln e( − x − y ) = (− x − y ) ln e = − x − y ( ) ln esecθ = (sec θ )(ln e) = sec θ ( ) = ln x ln(e2ln x ) = ln eln x 2 x (b) ln e(e ) = (e x )(ln e) = e x = ln x 49 ln y = 2t + eln y = e2t + y = e 2t + 50 ln y = −t + eln y = e−t +5 y = e −t +5 51 ln(y − 40) = 5t eln( y − 40) = e5t y − 40 = e5t y = e5t + 40 ( ) t 52 ln(1 − 2y) = t eln(1−2 y ) = et − y = et −2 y = et − y = − e 2−1 Copyright 2016 Pearson Education, Inc x y Section 1.6 Inverse Functions and Logarithms ( ) y −1 53 ln(y − 1) − ln = x + ln x ln(y − 1) − ln − ln x = x ln x = x e y − = xe x ln ( ) = e x y −1 = e x y −1 2x 2x y = xe + x y −1 54 ln( y − 1) − ln( y + 1) = ln(sin x) ln y +1 = ln(sin x) ln(y − 1) = ln(sin x) eln( y −1) = eln(sin x ) y − = sin x y = sin x + 55 (a) e2 k = ln e2 k = ln 2k ln e = ln 22 2k = ln k = ln (b) 100e10k = 200 e10k = ln e10 k = ln 10k ln e = ln 10k = ln k = ln102 (c) 56 (a) k ln e = ln a k = ln a k = 1000 ln a ek /1000 = a ln ek /1000 = ln a 1000 1000 e5k = 14 ln e5k = ln 4−1 5k ln e = −ln 5k = −ln k = − ln54 (b) 80ek = e k = 80−1 ln e k = ln 80−1 k ln e = −ln 80 k = −ln 80 (c) 57 (a) e(ln 0.8) k = 0.8 (eln 0.8 )k = 0.8 (0.8)k = 0.8 k = e−0.3t = 27 ln e −0.3t = ln 33 (−0.3t)ln e = ln −0.3t = ln t = −10 ln (b) ekt = 12 ln e kt = ln 2−1 = kt ln e = − ln t = − lnk2 (c) 0.4 e(ln 0.2)t = 0.4 (eln 0.2 )t = 0.4 0.2t = 0.4 ln 0.2t = ln 0.4 t ln 0.2 = ln 0.4 t = ln ln 0.2 58 (a) e−0.01t = 1000 ln e −0.01t = ln1000 (−0.01t)ln e = ln 1000 −0.01t = ln 1000 t = −100 ln 1000 ln e kt = ln10−1 = kt ln e = − ln10 kt = −ln 10 t = − ln10 (b) ekt = 10 k (c) e(ln 2)t = 12 (eln )t = 2−1 2t = 2−1 t = −1 59 e t = x ln e t = ln x t = 2ln x t = 4(ln x) 2 2 60 e x e2 x +1 = et e x + x +1 = et ln e x + x +1 = ln et t = x + x + 61 (a) 5log5 = (b) 8log8 = (d) log 16 = log 42 = log 4 = ⋅1 = (e) log3 = log3 31/2 = 12 log3 = 12 ⋅1 = 12 = 0.5 (f) log 14 = log 4−1 = −1log 4 = −1⋅1 = −1 () Copyright 2016 Pearson Education, Inc (c) 1.3log3 75 = 75 39 40 Chapter Functions 62 (a) 2log = 10log10 (1/2) = (b) (c) π logπ = (d) log11 121 = log11 112 = log11 11 = ⋅1 = (e) log121 11 = log121 1211/2 = (f) log3 ( 12 ) log121 121 = ( 12 ) ⋅1 = 12 ( 19 ) = log3 3−2 = −2 log3 = −2 ⋅1 = −2 63 (a) Let z = log x z = x 22 z = x (2 z ) = x z = x (b) Let z = log3 x 3z = x (3z )2 = x 32 z = x z = x (c) log (e(ln 2) sin x ) = log 2sin x = sin x 64 (a) Let z = log5 (3x ) z = 3x 25 z = x (b) log e (e x ) = x log (2e x 65 (a) log x log3 x = ln x ln x = ÷ ln ln (c) ln x a ln a = ln a ln x ÷ log9 x log3 x = ln x ln x = ÷ ln ln log x (c) x 66 (a) (b) (c) log 10 log a b logb a x = sin x = ) = log 4(e ln a ln x ln x ln 10 ln b ln a ÷ sin x )/2 ln x ⋅ ln ln ln x = = ln a ⋅ ln x ln x ln a ln x ⋅ ln ln ln x ln x ln a = ÷ ln ln b x = = e x sin x ln ln = = (b) 2ln x ln x () ( lnln ba ) = x = ÷ ln ln ln x ⋅ ln ln ln x = 3ln ln =3 ln ln10 67 (a) − π6 (b) π 68 (a) π (b) 3π ln x ln 2 () = = =2 ln ln x ⋅ ln10 ln x ln b ⋅ ln b ln a ln a log x log8 x (c) − π3 (c) π 69 (a) arccos(−1) = π since cos(π) = −1 and ≤ π ≤ π ( ) (b) arccos(0) = π2 since cos π2 = and ≤ π2 ≤ π 70 (a) ( ) since sin ( − π4 ) = − arcsin(−1) = − π2 since sin − π2 = −1 and − π2 ≤ − π2 ≤ π2 ( )=− (b) arcsin − π and − π2 ≤ − π4 ≤ π2 71 The function g(x) is also one-to-one The reasoning: f(x) is one-to-one means that if x1 ≠ x2 then f ( x1 ) ≠ f ( x2 ), so − f ( x1 ) ≠ − f ( x2 ) and therefore g ( x1 ) ≠ g ( x2 ) Therefore g(x) is one-to-one as well Copyright 2016 Pearson Education, Inc Section 1.6 Inverse Functions and Logarithms 41 72 The function h(x) is also one-to-one The reasoning: f(x) is one-to-one means that if x1 ≠ x2 then f ( x1 ) ≠ f ( x2 ), so f ( x1 ) ≠ f (1x ) , and therefore h( x1 ) ≠ h( x2 ) 73 The composite is one-to-one also The reasoning: If x1 ≠ x2 then g ( x1 ) ≠ g ( x2 ) because g is one-to-one Since g ( x1 ) ≠ g ( x2 ), we also have f ( g ( x1 )) ≠ f ( g ( x2 )) because f is one-to-one; thus, f g is one-to-one because x1 ≠ x2 f ( g ( x1 )) ≠ f ( g ( x2 )) 74 Yes, g must be one-to-one If g were not one-to-one, there would exist numbers x1 ≠ x2 in the domain of g with g ( x1 ) = g ( x2 ) For these numbers we would also have f ( g ( x1 )) = f ( g ( x2 )), contradicting the assumption that f g is one-to-one 75 (a) y= 100 1+ 2− x x = − log ( 100 y ) − = − log ( 100 − y y ) = log ( y 100 − y ( ( ) )) = ( f f −1 )( x) = f log 100x− x ( ( 100 1+ log (100x− x ) ( f −1 f )( x) = f −1 100− x = log 100− 100− x 1+ 1+ ) ( y= 50 1+1.1− x 100 1+ 2− x = 100 y ) − → − x = log ( 100 y ) −1 ) Interchange x and y: y = log 100x− x → f −1 ( x) = log 100x− x Verify (b) ( → + 2− x = 100 → 2− x = 100 − → log (2− x ) = log y y 100 ( log 100 − x x 1+ ) ) x =x = 100 = x +100 = 100 100 − x 100 1+ 100 − x x x 100 = log = log 2− x = log (2 ) = x −x 100(1 ) 100 + − ( ) ( ) ( ) → + 1.1− x = 50 → 1.1− x = 50 − → log1.1 (1.1− x ) = log1.1 50 − → − x = log1.1 50 −1 y y y y ) ( ( ) ( ) 50 − y y x = − log1.1 50 − = − log1.1 y = log1.1 50− y y Interchange x and y: y = log1.1 50x− x → f −1 ( x) = log1.1 50x− x Verify ( ( ( f f −1 )( x) = f log1.1 50x− x ( f −1 f )( x) = f −1 ( ( 50 1+1.1− x ) )) = ) = log ( 50 ( log1.1 x 50 − x 1+1.1 1.1 50 1+1.1− x 50− 50− x 1+1.1 ) = 50 ( log1.1 50 − x x 1+1.1 ) ) = 50 1+ 50 − x x x = 50 x = x = x +50 50 − x 50 x 50 = log1.1 = log1.1 1.1− x = log1.1 (1.1 ) = x −x 50(1 1.1 ) 50 + − ( ) 76 sin −1 (1) + cos −1 (1) = π2 + = π2 ; sin −1 (0) + cos −1 (0) = + π2 = π2 ; and sin −1 (−1) + cos −1 (−1) = − π2 + π = π2 If x ∈ (−1, 0) and x = −a, then sin −1 ( x) + cos −1 ( x) = sin −1 (−a) + cos −1 (−a) = − sin −1 a + (π − cos −1 a ) = π − (sin −1 a + cos −1 a) = π − π2 = π2 from Equations (3) and (4) in the text 77 (a) (b) (c) (d) (e) (f) Begin with y = ln x and reduce the y-value by y = ln x − Begin with y = ln x and replace x with x − y = ln(x − 1) Begin with y = ln x, replace x with x + 1, and increase the y-value by y = ln(x + 1) + Begin with y = ln x, reduce the y-value by 4, and replace x with x − y = ln(x − 2) − Begin with y = ln x and replace x with −x y = ln(−x) Begin with y = ln x and switch x and y x = ln y or y = e x Copyright 2016 Pearson Education, Inc 42 Chapter Functions 78 (a) Begin with y = ln x and multiply the y-value by y = ln x () (b) Begin with y = ln x and replace x with 3x y = ln 3x (c) Begin with y = ln x and multiply the y-value by 14 y = 14 ln x (d) Begin with y = ln x and replace x with 2x y = ln 2x 79 From zooming in on the graph at the right, we estimate the third root to be x ≈ −0.76666 80 The functions f ( x) = x ln and g ( x) = 2ln x appear to have identical graphs for x > This is no accident, because x ln = eln 2⋅ln x = (eln )ln x = 2ln x 81 (a) (b) t /12 Amount = 12 () t /12 t /12 t = → t = 36 8( ) = → ( 12 ) = → ( 12 ) = ( 12 ) → 12 t /12 There will be gram remaining after 36 hours ln(2) 82 500(1.0475)t = 1000 → 1.0475t = → ln(1.0475t ) = ln(2) → t ln(1.0475) = ln(2) → t = ln(1.0475) ≈ 14.936 It will take about 14.936 years (If the interest is paid at the end of each year, it will take 15 years.) () () 83 375, 000(1.0225)t = 1, 000, 000 → 1.0225t = 83 → ln(1.0225t ) = ln 83 → t ln(1.0225) = ln 83 →t = ln ( 83 ) ln(1.0225) ≈ 44.081 It will take about 44.081 years 84 ln(0.9) y = y0 e −0.18t represents the decay equation; solving (0.9) y0 = y0 e−0.18t t = −0.18 ≈ 0.585 days Copyright 2016 Pearson Education, Inc ... 1] by [ − 2, 4] 19 [ − 4, 4] by [0, 3] 20 [ − 5, 5] by [ − 2, 2] 21 [ −10, 10] by [ − 6, 6] 22 [ −5, 5] by [ − 2, 2] 23 [ − 6, 10] by [ − 6, 6] 24 [ − 3, 5] by [ − 2, 10] 25 [−0.03, 0.03] by. .. 4] by [ − 8, 8] 13 [ − 1, 6] by [ − 1, 4] 14 [ −1, 6] by [ −1, 5] 15 [ − 3, 3] by [0, 10] 16 [ −1, 2] by [0, 1] Copyright 2016 Pearson Education, Inc 27 28 Chapter Functions 17 [ − 5, 1] by. .. − 2, 6] by [ − 250, 50] [ −1, 5] by [ − 5, 30] Copyright 2016 Pearson Education, Inc Section 1.4 Graphing with Software [ − 4, 4] by [ − 5, 5] 10 [ − 2, 2] by [ − 2, 8] 11 [ − 2, 6] by [ −
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