Solution manual for applied calculus 6th edition by waner

Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Waner Solutions Section 0.1 Section 0.1 13 3^2+2^2+1 2 = + + = + + = 14 2(4 + (-1))(2 · -4) = 2(3)(-8) = (6)(-8) = -48 (22-2) 14 2^(2^2-2) = 2 + ([4 - 2] · 9) = + (2×9) = + 18 = 21 15 20/(3*4)-1 20 -1= -1= = 3 12 + ([3 + (-5)]) + (-2) = - 2¿2 3-4 = = -1 -1 - 2(-3)2 - 2¿9 - 18 = = 6×9 -6(4 - 1)2 -6(3) -15 = = 18 54 - 2¿9 - 2(1 - 4)2 - 2(-3) = = 2 2¿16¿2 2(5 - 1) · 2(4) ·2 1-18 17 = =64 64 18 121/(1+1/10)^2 121 121 121 = = = 100 2= 1.1 1.21 ⎛1 + ⎞ 10⎠ ⎝ (2-5*(-1))/1-2*(-1) - 5·(-1) - 2·(-1) = 2+5 = +2=7+2=9 ⎡ - 2·3 ⎤ = 3⎡-2×9⎤ = 3⎡-18⎤ 19 3⎢ ⎥ ⎢ -32 ⎥ ⎣ -9 ⎦ ⎣-(4 - 1)2⎦ ⎣ ⎦ = 3×2 = 2 ⎡ 8(1 - 4) ⎤ = -⎡8(-3) ⎤ 20 - ⎢ ⎥ ⎢ ⎥ ⎣-9(5 - 1)2⎦ ⎣-9(4)2⎦ 1 8×9 ⎤ 72 ⎞ = - ⎡ = - ⎛ = - ⎛- ⎞ = 2 -144 ⎝ ⎠ ⎝ ⎠ ⎣-9×16⎦ 2-5*(-1)/(1-2*(-1)) 5·(-1) =21 - 2·(-1) 11 -5 =2+ = =23 1+2 1⎞ 2⎤ 1⎤2 ⎡ ⎝ 2⎠ ⎦ + = 3⎣1 - 4⎦ + 27 43 = 3⎡ ⎤ = 3⎡ ⎤ + = +1= 16 16 ⎣4⎦ ⎣16⎦ 21 3⎡1 - ⎛- ⎣ 2¿(-1) 2¿1 = = =1 2 10 + · 32 = + 4×9 = + 36 = 38 2 22 3⎡ - ⎛ ⎞ ⎤ + = 3⎡ - ⎤ + ⎣9 ⎝3⎠ ⎦ ⎣9 9⎦ -3 -1 = 3⎡ ⎤ + = 3⎡ ⎤ + ⎣ ⎦ ⎣ ⎦ = 3⎡ ⎤ + = + = ⎣9⎦ 11 2·42 + = 2¿16 + = 32 + = 33 12 1-3·(-2)2¿2 = - 3¿4¿2 = - 24 = -23 23 (1/2)^2-1/2^2 ©2014 Ce ted to a publicly accessible website, in whole or in part =2 =4 17 10*(1+1/10)^3 3 = 10⎛1 + ⎞ = 10(1.1) 10⎠ ⎝ = 10×1.331 = 13.31 12-(-3) 12 - (1 - 4) = 16 - 2(5 - 1) · - 15 =1 = 15 2·(-1)2 / = 4-2 16 2-(3*4)/10 12 =2- = =25 10 =2 Full file at h pplied-Calculus-6th-Edition-by-Waner Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Waner Solutions Section 0.1 = ⎡1⎤2 - 12 = - = ⎣2⎦ 4 ⎡2⎤ ⎣3⎦ 24 2/(1^2)-(2/1)^2 2 = - ⎡ ⎤ = - = -2 1 1⎦ ⎣ = (2/3)/5 Note that we use only (round) parentheses in technology formulas, and not brackets 25 3¿(2-5) = 3*(2-5) 38 26 + 27 37 = 4+5/9 or 4+(5/9) 28 4-1 = (4-1)/3 29 3-1 = (3-1)/(8+6) 8+6 = 2/(3/5) 39 34-5¿6 = 3^(4-5)*6 Note that the entire exponent is in parentheses = 3/(2-5) 2-5 Note 3/2-5 is wrong, since it corresponds to ⎡3⎤ ⎣5⎦ - Note 3-1/8-6 is wrong, since it corresponds to - 40 = 2/(3+5^(7-9)) 3+57-9 ⎤-3 = 3*(1+4/100)^(-3) 100⎦ ⎣ Note that we use only (round) parentheses in technology formulas, and not brackets 41 3⎡1 + + 4⎤-3 42 3⎡ ⎣ 100 ⎦ = 3((1+4)/100)^(-3) - 30 + 43 32x-1 + 4x - = 3^(2*x-1)+4^x-1 Note that the entire exponent of is in parentheses = 3+3/(2-9) 2-9 31 - 44 2x - (22x)2 = 2^(x^2)-(2^(2*x))^2 4+7 = 3-(4+7)/8 45 22x -x+1 = 2^(2x^2-x+1) Note that the entire exponent is in parentheses 4¿2 32 = 4*2/(2/3) or (4*2)/(2/3) 33 [3] 46 22x -x + = 2^(2x^2-x)+1 - xy2 = 2/(3+x)-x*y^2 3+x 47 4e-2x = 4*e^(-2*x)/(2-3e^(-2*x)) 2-3e-2x or 4(*e^(-2*x))/(2-3e^(-2*x)) or (4*e^(-2*x))/(2-3e^(-2*x)) 3+x = 3+(3+x)/(x*y) 34 3+ xy 48 60 x2-1 = 3.1x^3-4x^(-2)-60/(x^2-1) 35 3.1x3 - 4x-2 - e2x + e-2x e2x - e-2x = (e^(2*x)+e^(-2*x))/e^(2*x)-e^(-2*x)) [ ( )] 2 36 2.1x-3 - x-1 + 49 - - x2-3 = 2.1x^(-3)-x^(-1)+(x^2-3)/2 ©2014 Ce ted to a publicly accessible website, in whole or in part + = 3(1-(-1/2)^2)^2+1 Note that we use only (round) parentheses in technology formulas, and not brackets Full file at h pplied-Calculus-6th-Edition-by-Waner Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Waner Solutions Section 0.2 Section 0.2 19 -x2x-3y = -x2-3y = -x-1y = - y x 33 = 27 20 -xy-1x-1 = -x1-1y-1 = -x0y-1 = - (-2) = -8 -(2 · 3)2 = -(22 · 32) = -(4 · 9) = -36 or -(2 · 3)2 = -(62) = -36 21 x3 = x3-4 = x-1 = x x4 (4 · 2)2 = 42 · 22 = 16 · = 64 or (4 · 2)2 = 82 = 64 22 y5 = y5-3 = y2 y3 23 x 2y = x2-(-1)y2-1 = x3y x-1y 24 x-1y = x-1-2y1-2 = x-3y-1 = xy x 2y 2 (-2)2 -2 ⎛ ⎞ = = ⎝ ⎠ 32 33 27 ⎛ ⎞ = = ⎝2⎠ (-2) (xy-1z3)2 x2(y-1)2(z3)2 = = x2-2y-2-1z6-2 = x2yz2 x2yz2 z4 y-3z4 = y 1 = = =8 (-2)3 -8 -3 -2-3 = - 25 1 =8 23 26 -2 y ⎛1⎞ = = = = 16 1/16 (1/4) 1/4 ⎝4⎠ x2yz2 x2yz2 2-(-1) 1-(-1) 2-1 y z = x 3y 2z -1 -1 = -1 -1 = x (xyz ) x y z (xy-2z)3 x3y-6z3 xy-2z 27 ⎛ -1 ⎞ = -1 = -3 = x3-(-3)y-6z3-3 = (x z) x z ⎝ x z ⎠ x x6y-6 = y -2 1 -2 = = 9/4 10 ⎛ ⎞ = (-2/3)2 4/9 ⎝ ⎠ 11 · 30 = · = 2 x4y-2 x2y-1z0⎞ = 2 = x4-2y-2-2z-2 = x2y-4z-2 = 28 ⎛ ⎝ xyz ⎠ x y z x y 4z 12 · (-2)0 = · = 13 23 22 = 23+2 = 25 = 32 or 23 22 = · = 32 14 32 = 32+1 = 33 = 27 or 32 = · = 27 -2 x-1y-2z2⎞ -1-1 -2-1 -2 -2 -3 -2 ⎛ 29 ⎝ xy ⎠ = (x y z ) = (x y z ) = 15 22 2-1 24 2-4 = 22-1+4-4 = 21 = x4y6z-4 = 16 5 -3 5 -2 =5 2-3+2-2 =5 -1 -3 xy-2 30 ⎛ -1 ⎞ ⎝x y z⎠ 3 xyz = 17 x3x2 = x3+2 = x5 18 x4x-1 = x4-1 = x3 ©2014 Ce ted to a publicly accessible website, in whole or in part x 4y z4 31 3x-4 = Full file at h pplied-Calculus-6th-Edition-by-Waner x4 = (x1-2y-2+1z-1)-3 = (x-1y-1z-1)-3 = Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Waner Solutions Section 0.2 32 -4 x = 2x 49 33 3 -2/3 x = 2/3 4x 50 34 -3/4 y = 3/4 5y 35 - 36 51 0.3 6 - x-1 = - 0.3x2 5x x-2 0.1x -4 + 3x -2 52 3 – 27 = 81 – 16 = –19 ‡ -2.668 65 ‡ 2.839 27 27 = =2 8 · 64 = 8· 64 = · = = x 0.1 + 3x 53 (–2)2 = 4=2 37 = 54 (–1)2 = 1=1 38 ‡ 2.236 55 (1 + 15) = 16 = 16 56 (3 + 33) = 36 = 36 =2 4 = 40 = 41 16 = 42 = 43 = 5 39 44 25 = =3 57 a2b2 = a =3 16 a2 = b2 58 =2 46 25 - = =2 = a b 59 (x + 9)2 = x + (x + > because x is positive) 60 ( x + 9)2 = x + 3 61 x3(a3 + b3) = x3 (Notice: Not x(a + b).) 45 + =2 b = ab a2 b = 62 16 = + = x4 = a 4b 4 3 a3 + b3 = x a3 + b3 4 x a 4 = b4 x ab 16 = - = 47 + 16 = 25 = 48 25 – 16 = 9=3 ©2014 Ce ted to a publicly accessible website, in whole or in part 63 Full file at h pplied-Calculus-6th-Edition-by-Waner 4xy3 = x 2y 4y2 = x y2 x = 2y x Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Waner Solutions Section 0.2 4(x2 + y2) = c2 64 x2 + y2 c = x2 + y2 c (Notice: Not 2(x + y)/c.) 65 = 31/2 66 = 81/2 68 69 70 71 72 73 x = x2/3 82 1/3 xy = (xy ) x x x x = x2 = x2-1/2 = x3/2 x1/2 = x = x1-1/2 = x1/2 x1/2 3x 8x x - 2/5 -3/2 - x 3/2 = 4x 2x - = 3 x -3/2 -3/5 = x - x x2y = (x2y)1/2 x2 2/5 = 3 - x 81 x3 = x3/2 67 1/2 5x x + = 1/2 8 5x x x 1/2 -1/3 -1/2 x - x + x 80 3/2 3/5 8x 3x = (x2 + 1)3 (x + 1) = 4(x2 + 1)1/3 (x2 + 1)3 (x2 + 1)-3 - (x2 + 1)-1/3 83 3 = x-2 5x2 74 2 = x3 5x-3 (x2 + 1)7 84 = -3 3(x + 1) 2 (x + 1)3 - (x2 + 1)7/3 75 3x-1.2 - 2.1 = x-1.2 - x-2.1 3x 85 22/3 = 76 x2.1 1.2 2.1 = x - x -1.2 3 3x 86 34/5 = 2x x0.1 4 77 + 1.1 = x - x0.1 + x-1.1 3 3x 87 x4/3 = 4x2 x3/2 + - = x2 + x3/2 - x-2 78 3 3x 88 y7/4 = 22 34 x y 1/2 3x x + = 79 4 x 3x x 1/2 -1/2 -3/2 x - x + x 3 1/2 1/3 1/5 - 1/2 + 1/2 = 3x 3xx ©2014 Ce ted to a publicly accessible website, in whole or in part 89 (x y ) = 90 x-1/3y3/2 = y3/2 y3 1/3 = x x Full file at h pplied-Calculus-6th-Edition-by-Waner x y + 1/3 = 2x Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Waner Solutions Section 0.2 -1/3 91 - 92 106 -1/4 3 x = - 1/4 = 2x x 3.1 x - x 0.2 3x1/2 0.2 = + -1/2 = 2/3 + 7 7x x x 11 9(1 - x)7/3 9 96 = -7/3 = 4(1 - x) (1 – x)7 1/3-1+2/3-1/3 x3/2 = x3/2-5/2 = x-1 = x x5/2 102 y5/4 = y5/4-3/4 = y1/2 = y3/4 103 x1/2y2 = x1/2+1/2y2-1 = xy x-1/2y 104 1/3 2/3 -1/3 ⎛y⎞ = ⎛y⎞ ⎝x⎠ ⎝x⎠ 10 = 10 ‡ ±0.3162 =2 -1/3 –32 = -2 115 x1/2 - = 0, x1/2 = 4, x = 42 = 16 = 1/3 116 x1/3 - = 0, x1/3 = 2, x = 23 = 117 - 1 = 0, = , x2 = 1, x = ± = ±1 x2 x 6 - = 0, = , 2x4 = 6x3, 2x = 6, x3 x x x x=3 y 2/3 118 119 (x - 4)-1/3 = 2, x - = 2-3 = x-1/2y = x-1/2-2y1-3/2 = x-5/2y-1/2 x2y3/2 x 105 ⎛ ⎞ ⎝y⎠ 1 = 0, x2 = ,x=± 10 10 =± 114 x4 - 81 = 0, x4 = 81, x = ± 81 = ±3 99 32/33-1/6 = 32/3-1/6 = 31/2 = 101 110 x2 - 113 x5 + 32 = 0, x5 = -32, x = 21/a 98 2/a = 21/a-2/a = 2-1/a = 1/a 2 =2 4 = 0, x2 = , x = ± 9 112 x2 - (2 - 3x)2 = 0, x2 = (2 - 3x)2, x = ±(2 - 3x); if x = - 3x then 4x = 2, x = 1/2; if x = -(2 - 3x) then -2x = -2, x = So, x = or 1/2 97 4-1/247/2 = 4-1/2+7/2 = 43 = 64 100 2 2 2/3 111 x2 - (1 + 2x)2 = 0, x2 = (1 + 2x)2, x = ±(1 + 2x); if x = + 2x then -x = 1, x = -1; if x = -(1 + 2x) then 3x = -1, x = -1/3 So, x = -1 or 1/3 3 5/2 = 4(1 - x) (1 – x)5 1/3 -1 2/3 -1/3 1/3 109 x2 - ± x 95 1/3 108 x2 - = 0, x2 = 1, x = ± = ±1 3.1 11 -1/7 11 94 -4/3 x = 3.1x4/3 - 1/7 = x 7x 1/3 ⎛y⎞ = ⎛y⎞ ⎛y⎞ = ⎛y⎞ ⎝x⎠ ⎝x⎠ ⎝x⎠ ⎝x⎠ 107 x2 - 16 = 0, x2 = 16, x = ± 16 = ±4 3/2 x x = 5 93 0.2x-2/3 + ⎛x⎞ ⎝y⎠ x=4+ 1/3 33 = 8 120 (x - 4)2/3 + = 5, (x - 4)2/3 = 4, x - = ±43/2 = ±8, x = ± = -4 or 12 ⎛y⎞ = ⎛y⎞ ⎝x⎠ ⎝x⎠ ©2014 Ce ted to a publicly accessible website, in whole or in part , Full file at h pplied-Calculus-6th-Edition-by-Waner Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Waner Solutions Section 0.3 Section 0.3 = 6x3 - x2 + 2x + 1 x(4x + 6) = 4x2 + 6x 19 (x2 - 2x + 1)2 = (x2 - 2x + 1)(x2 - 2x + 1) = x2(x2 - 2x + 1) - 2x(x2 - 2x + 1) + (x2 2x + 1) = x4 - 2x3 + x2 - 2x3 + 4x2 - 2x + x2 - 2x + = x4 - 4x3 + 6x2 - 4x + (4y - 2)y = 4y2 - 2y (2x - y)y = 2xy - y2 20 (x + y - xy)2 = (x + y - xy)(x + y - xy) = x(x + y - xy) + y(x + y - xy) - xy(x + y - xy) = x2 + xy - x2y + xy + y2 - xy2 - x2y - xy2 + x2y2 = x2 + y2 - 2x2y - 2xy2 + 2xy + x2y2 x(3x + y) = 3x + xy (x + 1)(x - 3) = x2 + x - 3x - = x2 - 2x - 21 (y3 + 2y2 + y)(y2 + 2y - 1) = y3(y2 + 2y - 1) + 2y2(y2 + 2y - 1) + y(y2 + 2y - 1) = y5 + 2y4 - y3 + 2y4 + 4y3 - 2y2 + y3 + 2y2 - y = y5 + 4y4 + 4y3 - y (y + 3)(y + 4) = y + 3y + 4y + 12 = y2 + 7y + 12 (2y + 3)(y + 5) = 2y2 + 3y + 10y + 15 = 2y2 + 13y + 15 22 (x3 - 2x2 + 4)(3x2 - x + 2) = x3(3x2 - x + 2) 2x2(3x2 - x + 2) + 4(3x2 - x + 2) = 3x5 - x4 + 2x3 - 6x4 + 2x3 - 4x2 + 12x2 - 4x + = 3x5 - 7x4 + 4x3 + 8x2 - 4x + 8 (2x - 2)(3x - 4) = 6x2 - 6x - 8x + = 6x2 -14x + (2x - 3)2 = 4x2 - 12x + 23 (x + 1)(x + 2) + (x + 1)(x + 3) = (x + 1)(x + + x + 3) = (x + 1)(2x + 5) 10 (3x + 1)2 = 9x2 + 6x + 11 ⎛x + 1⎞ = x2 + + x ⎠ x 12 ⎛y - 1⎞ = y2 - + y ⎠ y ⎝ ⎝ 24 (x + 1)(x + 2)2 + (x + 1)2(x + 2) = (x + 1)(x + 2)(x + + x + 1) = (x + 1)(x + 2)(2x + 3) 25 (x2 + 1)5(x + 3)4 + (x2 + 1)6(x + 3)3 = (x2 + 1)5(x + 3)3(x + + x2 + 1) = (x2 + 1)5(x + 3)3(x2 + x + 4) 13 (2x - 3)(2x + 3) = (2x)2 - 32 = 4x2 - 26 10x(x2 + 1)4(x3 + 1)5 + 15x2(x2 + 1)5 · (x3 + 1)4 = 5x(x2 + 1)4(x3 + 1)4[2(x3 + 1) + 3x(x2 + 1)] = 5x(x2 + 1)4(x3 + 1)4(5x3 + 3x + 2) 14 (4 + 2x)(4 - 2x) = 42 - (2x)2 = 16 - 4x2 15 ⎛y - ⎝ 1⎞⎛ 1 y + ⎞ = y2 - ⎛ ⎞ = y2 - y ⎠⎝ y⎠ y ⎝y⎠ 27 (x3 + 1) x + - (x3 + 1)2 x + = (x3 + 1) x + [1 - (x3 + 1)] = -x3(x3 + 1) x + 16 (x - x2)(x + x2) = x2 - (x2)2 = x2 - x4 28 (x2 + 1) x + - 17 (x + x - 1)(2x + 4) = (x + x - 1)2x + (x2 + x - 1)4 = 2x3 + 2x2 - 2x + 4x2 + 4x = 2x3 + 6x2 + 2x - [x + - 18 (3x + 1)(2x2 - x + 1) = 3x(2x2 - x + 1) + 1(2x2 - x + 1) = 6x3 - 3x2 + 3x + 2x2 - x + 29 ©2014 Ce ted to a publicly accessible website, in whole or in part 2 (x + 1)3 = (x + 1) ] = x+1 · x + [x + - (x + 1)] = (x - x) x + = x(x - 1) x + [1 + Full file at h pplied-Calculus-6th-Edition-by-Waner (x + 1)3 + (x + 1)5 = (x + 1)2 ] = (x + 1)3 · (x + 1)3 (1 + x + 1) = Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Waner Solutions Section 0.3 (b) (2x + 3)(3x + 2) = 0; 2x + = or 3x + = 0; x = -3/2 or -2/3 (x + 2) (x + 1)3 30 (x2 + 1) (x + 1)4 3 (x + 1)4 [x2 + - 3 (x + 1)7 = 42 (a) 6y2 + 17y + 12 = (3y + 4)(2y + 3) (b) (3y + 4)(2y + 3) = 0; 3y + = or 2y + = 0; y = -4/3 or -3/2 (x + 1)3 ] = (x + 1)4 [x2 + - (x + 1)] = 43 (a) 12x2 + x - = (3x - 2)(4x + 3) (b) (3x - 2)(4x + 3) = 0; 3x - = or 4x + = 0; x = 2/3 or -3/4 (x2 - x) (x + 1)4 = x(x - 1) (x + 1)4 31 (a) 2x + 3x2 = x(2 + 3x) (b) x(2 + 3x) = 0; x = or + 3x = 0; x = or -2/3 44 (a) 20y2 + 7y - = (4y - 1)(5y + 3) (b) (4y - 1)(5y + 3) = 0; 4y - = or 5y + = 0; y = 1/4 or -3/5 32 (a) y2 - 4y = y(y - 4) (b) y(y - 4) = 0; y = or y - = 0; y = or 45 (a) x2 + 4xy + 4y2 = (x + 2y)2 (b) (x + 2y)2 = 0; x + 2y = 0; x = -2y 33 (a) 6x3 - 2x2 = 2x2(3x - 1) (b) 2x2(3x - 1) = 0; x = or 3x - = 0; x = or 1/3 46 (a) 4y2 - 4xy + x2 = (2y - x)2 (b) (2y - x)2 = 0; 2y - x = 0; y = x/2 34 (a) 3y3 - 9y2 = 3y2(y - 3) (b) 3y2(y - 3) = 0; y = or y - = 0; y = or 47 (a) x4 - 5x2 + = (x2 - 1)(x2 - 4) = (x - 1)(x + 1)(x - 2)(x + 2) (b) (x - 1)(x + 1)(x - 2)(x + 2) = 0; x - = or x + = or x - = or x + = 0; x = ±1 or ±2 35 (a) x2 - 8x + = (x - 1)(x - 7) (b) (x - 1)(x - 7) = 0; x - = or x - = 0; x = or 48 (a) y4 + 2y2 - = (y2 - 1)(y2 + 3) = (y - 1)(y + 1)(y2 + 3) (b) (y - 1)(y + 1)(y2 + 3) = 0; y - = or y + = or y2 + = 0; y = ±1 (Notice that y2 + = has no real solutions.) 36 (a) y + 6y + = (y + 2)(y + 4) (b) (y + 2)(y + 4) = 0; y + = or y + = 0; y = -2 or -4 37 (a) x2 + x - 12 = (x - 3)(x + 4) (b) (x - 3)(x + 4) = 0; x - = or x + = 0; x = or -4 38 (a) y2 + y - = (y - 2)(y + 3) (b) (y - 2)(y + 3) = 0; y - = or y + = 0; y = or -3 39 (a) 2x2 - 3x - = (2x + 1)(x - 2) (b) (2x + 1)(x - 2) = 0; 2x + = or x - = 0; x = -1/2 or 40 (a) 3y2 - 8y - = (3y + 1)(y - 3) (b) (3y + 1)(y - 3) = 0; 3y + = or y - = 0; y = -1/3 or 41 (a) 6x2 + 13x + = (2x + 3)(3x + 2) ©2014 Ce ted to a publicly accessible website, in whole or in part Full file at h pplied-Calculus-6th-Edition-by-Waner Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Waner Solutions Section 0.4 Section 0.4 6x(x2 + 1)2(x3 + 2)3 - 9x2(x2 + 1)3(x3 + 2)2 = (x3 + 2)6 2 3 3x(x + 1) (x + 2) [2(x + 2) - 3x(x + 1)] = (x3 + 2)6 3x(x2 + 1)2(-x3 - 3x + 4) = (x3 + 2)4 2 -3x(x + 1) (x + 3x - 4) (x3 + 2)4 12 x - 2x + (x - 4)(2x + 1) 2x - 7x - · = = x+1 x-1 (x + 1)(x - 1) x2 - (2x - 3)(x + 3) 2x - x + · = = (x - 2)(x + 1) x-2 x+1 2x2 + 3x - x2 - x - 2 x - 2x + + = x-1 x+1 (x - 4)(x - 1) + (x + 1)(2x + 1) 3x2 - 2x + = (x + 1)(x - 1) x2 - (x2 - 1) x2 + 13 x +1 x x3 – - x2 x - x2 - (x - 1) x2 - x + = = x+1 x+1 x+1 x+1 (x2 - 1)(x2 + 1) - x4 (x2 + 1) x2 + x -1 3x4 x3 – = x(x3 - 1) - 3x4 (x3 - 1) x3 – = -2x4 - x (x3 – 1)3 = -x(2x + 1) (x3 – 1)3 x -1 (x - 1)(x - 1) - (x - 2) = = x-1 x-2 (x - 2)(x - 1) x - x - 2x + x2 - 3x + 1 (x + y)2 x2 x2 - (x + y)2 x2 - x2 - 2xy - y2 15 = = y yx2(x + y)2 yx2(x + y)2 -y(2x + y) -(2x + y) = = yx (x + y)2 x2(x + y)2 x-1 +x-1= +x-1= x x ⎛ ⎞ ⎝x - 1⎠ x - + x(x - 1) x2 - = x x = (x2 + 1)3 14 x2 + -1 = 2x - x + + = x+1 x-2 3x2 - (2x - 3)(x + 1) + (x - 2)(x + 3) = (x - 2)(x + 1) x -x-2 x4 1 (x + y)3 x3 x3 - (x + y)3 = = 16 y yx3(x + y)3 x3 - x3 - 3x2y - 3xy2 - y3 -y(3x2 + 3xy + y2) = = yx3(x + y)3 yx3(x + y)3 2 -(3x + 3xy + y ) x3(x + y)3 2x2 2x2 - 1 = = ⎛x -2 2⎞ x - x - x - x - ⎝ x ⎠ ⎛x - 1⎞ ⎛x - + x⎞ 2x - + = y⎠ x ⎝ x ⎝ xy xy ⎠ = x2y 2 y 2x - x⎞ y ⎛2x - + x⎞ y2(3x - 3) 10 ⎛ + = y⎠ xy x ⎝ y x ⎝ y ⎠ = y(3x - 3) 3xy - 3y = = x x (x + 1)2(x + 2)3 - (x + 1)3(x + 2)2 = (x + 2)6 (x + 1)2(x + 2)2[(x + 2) - (x + 1)] (x + 1)2 = (x + 2)6 (x + 2)4 11 ©2014 Ce ted to a publicly accessible website, in whole or in part Full file at h pplied-Calculus-6th-Edition-by-Waner Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Waner Solutions Section 0.5 Section 0.5 19 -x2 - 2x - = 0, -(x + 1)2 = 0, x = -1 x + = 0, x = -1 20 2x2 - x - = 0, (2x - 3)(x + 1) = 0, x = 3/2, x - = 1, x = -x + = 0, x = 2x + = 1, 2x = -3, x = -3/2 1 22 - x2 - x + = 0, x2 + x - = 0, (x + 2)(x - 1) = 0, x = -2, 4x - = 8, 4x = 13, x = 13/4 21 x2 - x - = 0, x2 - 2x - = 0, (x + 1)(x - 3) = 0, x = -1, 3 x + = 0, x = -1, x = -4/3 4 23 x2 - x = 1, x2 - x - = 0, x = 1± by the quadratic formula 7x + 55 = 98, 7x = 43, x = 43/7 24 16x2 = -24x - 9, 16x2 + 24x + = 0, (4x + 3)2 = 0, x = - 3/4 3x + = x, 2x = -1, x = -1/2 x + = 2x + 2, -x = 1, x = -1 , x = 2x - 1, x2 - 2x + = 0, (x x 1)2 = 0, x = 25 x = - 10 x + = 3x + 1, -2x = 0, x = 11 ax + b = c, ax = c - b, x = (c - b)/a 12 x - = cx + d, (1 - c)x = d + 1, x = 26 x + = d+1 1-c , (x + 4)(x - 2) = 1, x2 + x-2 2x - = 1, x2 + 2x - = 0, x = -2 ± 40 = -1 ± 10 by the quadratic formula 13 2x2 + 7x - = 0, (2x - 1)(x + 4) = 0, x = -4, 27 x4 - 10x2 + = 0, (x2 - 1)(x2 - 9) = 0, x2 = or x2 = 0, x = ±1, ±3 14 x2 + x + = 0, ∆ = -3 < 0, so this equation has no real solutions 28 x4 - 2x2 + = 0, (x2 - 1)2 = 0, x = ±1 15 x2 - x + = 0, ∆ = -3 < 0, so this equation has no real solutions 29 x4 + x2 - = 0, x2 = formula, x = ± 16 2x - 4x + = 0, ∆ = -8 < 0, so this equation has no real solutions 17 2x2 - = 0, x2 = ,x=± 18 3x2 - = 0, x2 = 1 ,x=± 3 -1 ± -1 ± by the quadratic 30 x3 + 2x2 + x = 0, x(x2 + 2x + 1) = 0, x(x + 1)2 = 0, x = 0, -1 31 x3 + 6x2 + 11x + = 0, (x + 1)(x + 2)(x + 3) = 0, x = -1, -2, -3 32 x3 - 6x2 + 12x - = 0, (x - 2)3 = 0, x = 33 x3 + 4x2 + 4x + = 0, ©2014 Ce ted to a publicly accessible website, in whole or in part 10 Full file at h pplied-Calculus-6th-Edition-by-Waner Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Waner Solutions Section 0.5 (x + 3)(x2 + x + 1) = 0, x = -3 (For x2 + x + = 0, ∆ = -3 < 0, so there are no real solutions to this quadratic equation.) 34 y3 + 64 = 0, y3 = -64, y = 35 x3 - = 0, x3 = 1, x = 36 x3 - 27 = 0, x3 = 27, x = –64 = -4 1=1 27 = 37 y3 + 3y2 + 3y + = 0, (y + 2)(y2 + y + 1) = 0, y = -2 (For y2 + y + = 0, ∆ = -3 < 0, so there are no real solutions to this quadratic equation.) 38 y3 - 2y2 - 2y - = 0, (y - 3)(y2 + y + 1) = 0, y = (For y2 + y + = 0, ∆ = -3 < 0, so there are no real solutions to this quadratic equation.) 39 x3 - x2 - 5x + = 0, (x - 1)(x2 - 5) = 0, x = 1, ± 40 x3 - x2 - 3x + = 0, (x - 1)(x2 - 3) = 0, x = 1, ± 41 2x6 - x4 - 2x2 + = 0, (2x2 - 1)(x4 - 1) = 0, [or (2x2 - 1)(x2 - 1)(x2 + 1) = 0; in any case, think of the cubic you get by substituting y for x2], x = ±1, ± 42 3x6 - x4 - 12x2 + = 0, (3x2 - 1)(x4 - 4) = 0, [or (3x2 - 1)(x2 - 2)(x2 + 2) = 0], x = ± , ± 43 (x2 + 3x + 2)(x2 - 5x + 6) = 0, (x + 2)(x + 1)(x - 2)(x - 3) = 0, x = -2, -1, 2, 44 (x2 - 4x + 4)2(x2 + 6x + 5)3 = 0, (x - 2)4(x + 1)3(x + 5)3 = 0, x = -5, -1, ©2014 Ce ted to a publicly accessible website, in whole or in part 11 Full file at h pplied-Calculus-6th-Edition-by-Waner Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Waner Solutions Section 0.6 Section 0.6 3 (x2 - x) (x + 1)4 = 0, x(x - 1) (x + 1)4 = 0, x = -1, 0, 1 x4 - 3x3 = 0, x3(x - 3) = 0, x = 0, 13 (x + 1)2(2x + 3) - (x + 1)(2x + 3)2 = 0, (x + 1)(2x + 3)(x + - 2x - 3) = 0, (x + 1)(2x + 3)(-x - 2) = 0, x = -2, -3/2, -1 x6 - 9x4 = 0, x4(x2 - 9) = 0, x = 0, ±3 x4 - 4x2 = -4, x4 - 4x2 + = 0, (x2 - 2)2 = 0, x =± 14 (x2 - 1)2(x + 2)3 - (x2 - 1)3(x + 2)2 = 0, (x2 - 1)2(x + 2)2(x + - x2 + 1) = 0, -(x2 - 1)2(x + 2)2(x2 - x - 3) = 0, x = -2, -1, 1, x4 - x2 = 6, x4 - x2 - = 0, (x2 - 3)(x2 + 2) = 0, x = ± (1 ± (x + 1)(x + 2) + (x + 1)(x + 3) = 0, (x + 1)(x + + x + 3) = 0, (x + 1)(2x + 5) = 0, x = -1, -5/2 13)/2 (x + 1)2(x + 2)3 - (x + 1)3(x + 2)2 = 0, (x + 2)6 (x + 1)2(x + 2)2[(x + 2) - (x + 1)] (x + 1)2 = 0, = (x + 2) (x + 2)4 0, (x + 1)2 = 0, x = -1 15 (x + 1)(x + 2)2 + (x + 1)2(x + 2) = 0, (x + 1)(x + 2)(x + + x + 1) = 0, (x + 1)(x + 2)(2x + 3) = 0, x = -1, -2, -3/2 16 6x(x2 + 1)2(x2 + 2)4 - 8x(x2 + 1)3(x2 + 2)3 = 0, (x2 + 2)8 2 2 2x(x + 1) (x + 2) [3(x + 2) - 4(x + 1)] = 0, (x2 + 2)8 -2x(x2 + 1)2(x2 - 2) = 0, (x2 + 2)5 (x2 + 1)5(x + 3)4 + (x2 + 1)6(x + 3)3 = 0, (x2 + 1)5(x + 3)3(x + + x2 + 1) = 0, (x2 + 1)5(x + 3)3(x2 + x + 4) = 0, x = -3 (Neither x2 + = nor x2 + x + = has a real solution.) 10x(x2 + 1)4(x3 + 1)5 - 10x2(x2 + 1)5(x3 + 1)4 = 0, 10x(x2 + 1)4(x3 + 1)4[x3 + - x(x2 + 1)] = 0, 10x(x2 + 1)4(x3 + 1)4(1 - x) = 0, x = -1, 0, -2x(x2 + 1)2(x2 - 2) = 0, x = 0, ± x4 2(x2 - 1) x2 + (x3 + 1) x + - (x3 + 1)2 x + = 0, (x3 + 1) x + [1 - (x3 + 1)] = 0, -x3(x3 + 1) x + = 0, x = 0, -1 10 (x2 + 1) x + - 17 x2 + x +1 2 2(x - 1)(x + 1) - x4 2 (x + 1) x + = 0, = 0, x4 - 2 (x + 1) x + (x + 1)3 = 0, - = 0, x = ± 2 x + [x + - (x + 1)] = 0, (x2 - x) x + = 0, x(x - 1) x + = 0, x = -1, 0, 11 (x + 1)3 + (x + 1)5 = 0, x + 1) = 0, (x + 2) 12 (x2 + 1) (x + 1)4 18 x3 - 4x(x - 1) - 3x4 (x + 1)3 (1 + (x + 1)3 = 0, x = -1 (x = -2 is not a solution because -2.) 4x x3 – - (x3 - 1) x3 – (x + 1)3 is not defined for x = 3x4 x3 – = 0, = 0, x4 - 4x (x3 - 1) x3 - 3 = 0, x4 - 4x = 0, x(x3 - 4) = 0, x = 0, ± (x + 1)7 = 0, 19 x - (x + 1)4 [x2 + - (x + 1)] = 0, ©2014 Ce ted to a publicly accessible website, in whole or in part 12 Full file at h pplied-Calculus-6th-Edition-by-Waner = 0, x2 - = 0, x = ±1 x = 0, x4 Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Waner Solutions Section 0.6 20 - = 0, x2 - = 0, x = ±2 x2 21 = 0, x2 - = 0, x = ±3 x x3 22 1 = 0, x + - x2 = 0, x2 x + x2 - x - = 0, x = (1 ± 23 5)/2 x (x - 4)(x - 1) - x(x + 1) x-4 = 0, = x-1 (x + 1)(x - 1) x+1 0, -6x + = 0, -6x + = 0, x = 2/3 (x + 1)(x - 1) 2x - 2x + = 0, x-1 x+1 (2x - 3)(x + 1) - (2x + 3)(x - 1) = 0, (x - 1)(x + 1) -2x = 0, -2x = 0, x = (x - 1)(x + 1) 24 3x(x + 4) + (x + 1)(x + 4) x+4 x+4 + = 0, = 3x 3x(x + 1) x+1 (x + 4)(3x + x + 1) (x + 4)(4x + 1) 0, = 0, = 0, (x 3x(x + 1) 3x(x + 1) + 4)(4x + 1) = 0, x = -4, -1/4 25 2x - 2x - = 0, x+1 x (2x - 3)(x + 1) - x(2x - 3) = 0, x(x + 1) 2x - (2x - 3)(x + - x) = 0, = 0, x(x + 1) x(x + 1) 2x - = 0, x = 3/2 26 ©2014 Ce ted to a publicly accessible website, in whole or in part 13 Full file at h pplied-Calculus-6th-Edition-by-Waner Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Waner Solutions Section 0.7 Solve the equation x + y = for y to get y = 1-x Then plot some points: x y = 1-x -2 -1 1 -1 Graph: Section 0.7 P(0, 2), Q(4, -2), R(-2, 3), S(-3.5, -1.5), T(2.5, 0), U(2, 2.5) P(-2, 2),Q (3.5, 2), R(0,-3), S(-3.5, -1.5), T(2.5, 0), U(-2, 2.5) Solve the equation y - x = -1 for y to get y = 1+x Then plot some points: x y = 4+x -2 -3 -1 -2 -1 Graph: ©2014 Ce ted to a publicly accessible website, in whole or in part 14 Full file at h pplied-Calculus-6th-Edition-by-Waner Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Waner Solutions Section 0.7 Solve the equation 2y - x = for y to get y = (1+x )/2 Then plot some points: Graph: y = (1+x )/2 x -2 2.5 -1 0.5 1 2.5 Graph: 1.3333 10 Solve the equation x y = -1 for y to get y = -1/x Then plot some points: Solve the equation 2y + x -3 -2 -1 Graph: x = for y to get y = (1- x)/2 Then plot some points: x 16 Graph: y = (1- x)/2 0.5 -0.5 -1 -1.5 y = -1/x -0.111 -0.25 -1 -1 -0.25 -0.111 11 Solve the equation xy = x +1 for y to get y = x+1/x Then plot some points: x y = x+1/x -2 -2.5 -1 -2 -1/2 -2.5 1/2 2.5 2 2.5 Solve the equation xy = for y to get y = 4/x Then plot some points: x y = 4/x -3 -1.333 -2 -2 -1 -4 ©2014 Ce ted to a publicly accessible website, in whole or in part 15 Full file at h pplied-Calculus-6th-Edition-by-Waner Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Waner Solutions Section 0.7 Graph: 18 Set the two distances equal and solve: (k + 1)2 + (k - 0)2 = 2k2 + 2k + = (k - 0)2 + (k - 2)2 ; 2k2 - 4k + ; 2k2 + 2k + = 2k2 - 4k + 4; 6k = 3; k = 19 Circle with center (0, 0) and radius 20 The single point (0, 0) 12 Solve the equation xy = 2x +1 for y to get y = 2x +1/x Then plot some points: y = 2x +1/x x -2 7.5 -1 -1/2 -1.5 1/2 2.5 8.5 Graph: 13 (2 - 1)2 + (-2 + 1)2 = 14 (6 - 1)2 + (1 - 0)2 = 2 26 15 (0 - a) + (b - 0) = a2 + b2 16 (b - a)2 + (b - a)2 = 2(b - a)2 = |b - a| 17 Set the two distances equal and solve: (1 - 0)2 + (k - 0)2 = + k2 = (1 - 2)2 + (k - 1)2 ; - 2k + k2 ; 1 + k2 = - 2k + k2; 2k = 1; k = ©2014 Ce ted to a publicly accessible website, in whole or in part 16 Full file at h pplied-Calculus-6th-Edition-by-Waner Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Wane SSoolluuttiioonnss SSeeccttiioonn 1 1 SSeeccttiioonn 1 1 Using the table, a f(0) = b f(2) = 0.5 Using the table, a f(−1) = b f(1) = Using the table, a f(2) − f(−2) = 0.5 − = −1.5 Using the table, a f(1) − f(−1) = − = −3 b f(−1)f(−2) = (4)(2) = b f(1)f(−2) = (1)(2) = c 3f(−2) = 3(2) = From the graph, we find a f(1) = 20 b f(2) = 30 In a similar way, we find: c f(3) = 30 d f(5) = 20 d f(3) − f(2) = 30 − 30 = From the graph, we find a f(1) = 20 b f(2) = 10 In a similar way, we find: c f(3) = 10 d f(5) = 20 e f(3) − f(2) = 10 − 10 = From the graph, we estimate a f(−3) = −1 b f(0) = 1.25 ©2014ted C to a publicly accessible website, in whole or in part c −2f(−1) = −2(4) = −8 17 Full file at ht pplied-Calculus-6th-Edition-by-Waner Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Wane SSoolluuttiioonnss SSeeccttiioonn 1 1 In a similar way, we find: c f(1) = since the solid dot is on (1, 0) d f(2) = f(2) − f(1) 1−0 e Since f(2) = and f(1) = 0, = = 2−1 2−1 From the graph, we estimate a f(−2) = b f(0) = In a similar way, we find: c f(1) = since the solid dot is on (1, 0) d f(3) = f(3) − f(1) − e Since f(3) = and f(1) = 0, = = 3−1 3−1 f(x) = x − x2 , with domain (0, +∞) a Since is in (0, +∞), f(4) is defined, and f(4) = − x =4− = 63 16 16 42 c Since −1 is not in (0, +∞), f(−1) is not defined b Since is not in (0, +∞), f(0) is not defined 10 f(x) = − x2, with domain [2, +∞) a Since is in [2, +∞), f(4) is defined, and f(4) = b Since is not in [2, +∞), f(0) is not defined − 42 = − 16 = − 31 2 c Since is not in [2, +∞), f(1) is not defined 11 f(x) = x + 10 , with domain [−10, 0) a Since is not in [−10, 0), f(0) is not defined b Since is not in [−10, 0), f(9) is not defined c Since −10 is in [−10, 0), f(−10) is defined, and f(−10) = −10 + 10 = = 12 f(x) = − x2 , with domain (−3, 3) a Since is in (−3, 3), f(0) is defined, and f(0) = − = b Since is not in (−3, 3), f(3) is not defined c Since −3 is not in (−3, 3), f(−3) is not defined 13 f(x) = 4x − a f(−1) = 4(−1) − = −4 − = −7 b f(0) = 4(0) − = − = −3 c f(1) = 4(1) − = − = d Substitute y for x to obtain f(y) = 4y − ©2014ted C to a publicly accessible website, in whole or in part 18 Full file at ht pplied-Calculus-6th-Edition-by-Waner Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Wane SSoolluuttiioonnss SSeeccttiioonn 1 1 e Substitute (a+b) for x to obtain f(a+b) = 4(a+b) − 14 f(x) = −3x + a f(−1) = −3(−1) + = + = b f(0) = −3(0) + = + = c f(1) = −3(1) + = −3 + = d Substitute y for x to obtain f(y) = −3y + e Substitute (a+b) for x to obtain f(a+b) = −3(a+b) + 15 f(x) = x2 + 2x + a f(0) = (0)2 + 2(0) + = + + = b f(1) = 12 + 2(1) + = + + = c f(−1) = (−1) + 2(−1) + = − + = d f(−3) = (−3)2 + 2(−3) + = − + = e Substitute a for x to obtain f(a) = a + 2a + f Substitute (x+h) for x to obtain f(x+h) = (x+h)2 + 2(x+h) + 16 g(x) = 2x2 − x + a g(0) = 2(0)2 − + = − + = b g(−1) = 2(−1)2 − (−1) + = + + = c Substitute r for x to obtain g(r) = 2r2 − r + d Substitute (x+h) for x to obtain g(x+h) = 2(x+h)2 − (x+h) + 17 g(s) = s2 + a g(1) = 12 + c g(4) = 42 + 1 s b g(−1) = (−1)2 + =1+1=2 = 16 + = 65 or 16.25 −1 =1−1=0 d Substitute x for s to obtain g(x) = x2 + e Substitute (s+h) for s to obtain g(s+h) = (s+h)2 + s+h f g(s+h) − g(s) = Answer to part (e) − Original function = (s+h)2 + 18 h(r) = s+h − s2 + r+4 1 1 a h(0) = = b h(−3) = = =1 0+4 (−3)+4 1 1 c h(−5) = = = −1 d Substitute x2 for r to obtain h(x2) = (−5)+4 (−1) x +4 1 e Substitute (x2+1) for r to obtain h(x2+1) = = (x2+1)+4 x2+5 f h(x2) + = Answer to part (d) + = +1 x2+4 ©2014ted C to a publicly accessible website, in whole or in part 19 Full file at ht pplied-Calculus-6th-Edition-by-Waner s x Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Wane SSoolluuttiioonnss SSeeccttiioonn 1 1 19 f(x) = −x3 (domain (−∞, +∞)) Technology formula: -(x^3) 20 f(x) = x3 (domain [0, +∞)) Technology formula: x^3 21 f(x) = x4 (domain (−∞, +∞)) Technology formula: x^4 ©2014ted C to a publicly accessible website, in whole or in part 20 Full file at ht pplied-Calculus-6th-Edition-by-Waner Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Wane SSoolluuttiioonnss SSeeccttiioonn 1 1 22 f(x) = 23 f(x) = x (domain (−∞, +∞)) Technology formula: x^(1/3) x2 24 f(x) = x + (x ≠ 0) Technology formula: 1/(x^2) x (x ≠ 0) Technology formula: x+1/x 25 a f(x) = x (−1 ≤ x ≤ 1) Since the graph of f(x) = x is a diagonal 45° line through the origin inclining up from left to right, the correct graph is (I) b f(x) = −x (−1 ≤ x ≤ 1) Since the graph of f(x) = −x is a diagonal 45° line through the origin inclining down from left to right, the correct graph is (IV) ©2014ted C to a publicly accessible website, in whole or in part 21 Full file at ht pplied-Calculus-6th-Edition-by-Waner Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Wane SSoolluuttiioonnss SSeeccttiioonn 1 1 c f(x) = x (0 < x < 4) Since the graph of f(x) = x is the top half of a sideways parabola, the correct graph is (V) d f(x) = x + − (0 < x < 4) x If we plot a few points like x = 0, 1, 1, 2, and 3, we find that the correct graph is (VI) e f(x) = |x| (−1 ≤ x ≤ 1) Since the graph of f(x) = |x| is a "V"-shape with its vertex at the origin, the correct graph is (III) f f(x) = x − (−1 ≤ x ≤ 1) Since the graph of f(x) = x−1 is a straight line through (0, −1) and (1, 0), the correct graph is (II) 26 a f(x) = −x + (0 < x ≤ 4) Since the graph of f(x) = −x + is a straight line inclining down from left to right, the correct graph must be (IV) .b f(x) = − |x| (−2 < x ≤ 2) Since f(x) = − |x| is obtained from the graph of y = |x| by flipping it vertically (the minus sign in front of |x|) and then moving it units vertically up (adding to all the values), the correct graph is (VI) c f(x) = x + (−2 < x ≤ 2) The graph off(x) = x + is similar to that of y = x , which is half a parabola on its side, and the correct graph is (I) d f(x) = −x2 + (−2 < x ≤ 2) The graph of f(x) = −x2 + is a parabola opening down, so the correct graph is (III) e f(x) = − x The graph of f(x) = − is part of a hyperbola, and the correct graph is (V) x f f(x) = x − (−2 < x ≤ 2) The graph of f(x) = x2 − is a parabola opening up, so the correct graph is (II) 27 Technology formula: 0.1*x^2-4*x+5 Table of values: x f(x) 1.1 −2.6 −6.1 −9.4 −12.5 −15.4 −18.1 −20.6 −22.9 10 −25 28 Technology formula: 0.4*x^2-6*x-0.1 Table of values: x −5 g(x) 39.9 −4 −3 −2 −1 30.3 21.5 13.5 6.3 −0.1 −5.7 −10.5 −14.5 −17.7 −20.1 29 Technology formula: (x^2-1)/(x^2+1) Table of values: ©2014ted C to a publicly accessible website, in whole or in part 22 Full file at ht pplied-Calculus-6th-Edition-by-Waner Solution Manual for Applied Calculus 6th Edition by Waner Full file at https://./Solution-Manual-for-Applied-Calculus-6th-Edition-by-Wane SSoolluuttiioonnss SSeeccttiioonn 1 1 x 0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5 10.5 h(x) −0.6000 0.3846 0.7241 0.8491 0.9059 0.9360 0.9538 0.9651 0.9727 0.9781 0.9820 30 Technology formula: (2*x^2+1)/(2*x^2-1) Table of values: x −1 r(x) 3.0000 −1.0000 3.0000 1.2857 1.1176 1.0645 1.0408 1.0282 1.0206 1.0157 1.0124 x if −4 ≤ x < Technology formula: x*(x=0) (For a graphing calculator, use ≥ if ≤ x ≤ instead of >=.) 31 f(x) = a f(−1) = −1 We used the first formula, since −1 is in [−4, 0) b f(0) = We used the second formula, since is in [0, 4] c f(1) = We used the second formula, since is in [0, 4] −1 if −4 ≤ x ≤ Technology formula: (-1)*(x0) (For a graphing calculator, use x if < x ≤ ≤ instead of