Solution manual for university calculus early transcendentals 3rd edition by hass

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) (hh)( 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 [ −