MULTIPLE CHOICE Choose the one alternative that best completes the statement or answers the question Find the average velocity of the function over the given interval 1) y = x2 + 8x, [5, 8] 128 B) A) 16 1) C) 21 63 D) 2) y = 2x3 + 5x2 + 7, [-5, -1] A) - 10 3) y = C) 32 D) - 128 2x, [2, 8] 3) A) 4) y = 2) B) B) C) - 10 D) , [4, 7] x-2 A) - 4) 10 5) y = 4x2 , 0, B) C) D) 5) A) C) B) C) B) D) - 10 6) y = -3x2 - x, [5, 6] A) -34 7) h(t) = sin (2t), 0, A) 7) B) 8) g(t) = + tan t, D) -2 π π A) - 6) π C) π D) - π π π , 4 8) B) π C) D) - π Use the table to find the instantaneous velocity of y at the specified value of x 9) x = x y 0 0.2 0.02 0.4 0.08 0.6 0.18 0.8 0.32 1.0 0.5 1.2 0.72 1.4 0.98 A) B) 1.5 C) 0.5 9) D) 10) x = x y 0 0.2 0.01 0.4 0.04 0.6 0.09 0.8 0.16 1.0 0.25 1.2 0.36 1.4 0.49 A) 1.5 10) B) 0.5 C) D) 11) x = x y 0 0.2 0.12 0.4 0.48 0.6 1.08 0.8 1.92 1.0 1.2 4.32 1.4 5.88 A) 11) B) C) D) 12) x = 12) x y 10 0.5 38 1.0 58 1.5 70 2.0 74 2.5 70 3.0 58 3.5 38 4.0 10 A) -8 B) C) D) 13) x = 13) x y 0.900 -0.05263 0.990 -0.00503 0.999 -0.0005 1.000 0.0000 1.001 0.0005 1.010 0.00498 1.100 0.04762 A) 0.5 B) D) -0.5 C) For the given position function, make a table of average velocities and make a conjecture about the instantaneous velocity at the indicated time 14) s(t) = t2 + 8t - at t = 14) t s(t) 1.9 1.99 1.999 2.001 2.01 2.1 A) t 1.9 1.99 1.999 2.001 2.01 2.1 ; instantaneous velocity is 17.70 s(t) 16.692 17.592 17.689 17.710 17.808 18.789 B) t 1.9 1.99 1.999 2.001 2.01 2.1 ; instantaneous velocity is 5.40 s(t) 5.043 5.364 5.396 5.404 5.436 5.763 C) t 1.9 1.99 1.999 2.001 2.01 2.1 ; instantaneous velocity is ∞ s(t) 5.043 5.364 5.396 5.404 5.436 5.763 D) t 1.9 1.99 1.999 2.001 2.01 2.1 ; instantaneous velocity is 18.0 s(t) 16.810 17.880 17.988 18.012 18.120 19.210 15) s(t) = t2 - at t = t s(t) -0.1 15) -0.01 -0.001 0.001 0.01 0.1 A) t -0.1 s(t) -4.9900 -0.01 -4.9999 -0.001 -5.0000 0.001 0.01 0.1 ; instantaneous velocity is -5.0 -5.0000 -4.9999 -4.9900 t -0.1 s(t) -2.9910 -0.01 -2.9999 -0.001 -3.0000 0.001 0.01 0.1 ; instantaneous velocity is -3.0 -3.0000 -2.9999 -2.9910 t -0.1 s(t) -1.4970 -0.01 -1.4999 -0.001 -1.5000 0.001 0.01 0.1 ; instantaneous velocity is -15.0 -1.5000 -1.4999 -1.4970 t -0.1 s(t) -1.4970 -0.01 -1.4999 -0.001 -1.5000 0.001 0.01 0.1 ; instantaneous velocity is ∞ -1.5000 -1.4999 -1.4970 B) C) D) Find the slope of the curve for the given value of x 16) y = x2 + 5x, x = 16) B) slope is 25 C) slope is -39 B) slope is 25 C) slope is 20 D) slope is -39 18) y = x3 - 5x, x = A) slope is B) slope is C) slope is -2 D) slope is -3 19) y = x3 - 3x2 + 4, x = A) slope is B) slope is -9 C) slope is D) slope is 20) y = - x3 , x = -1 A) slope is -1 B) slope is C) slope is -3 D) slope is A) slope is 13 D) slope is 20 17) y = x2 + 11x - 15, x = A) slope is 13 17) 18) 19) 20) Solve the problem 21) Given lim f(x) = Ll, lim f(x) = Lr, and Ll ≠ Lr, which of the following statements is true? x→0 x→0 + I lim f(x) = Ll x→0 II lim f(x) = Lr x→0 III lim f(x) does not exist x→0 A) I B) II C) none D) III 21) 22) Given lim f(x) = Ll, lim f(x) = Lr , and Ll = Lr, which of the following statements is false? x→0 x→0 + I lim f(x) = Ll x→0 II lim f(x) = Lr x→0 22) III lim f(x) does not exist x→0 A) I B) II C) none D) III 23) If lim f(x) = L, which of the following expressions are true? x→0 I lim f(x) does not exist x→0 - II lim f(x) does not exist x→0 + III lim f(x) = L x→0 - IV lim f(x) = L x→0 + A) I and II only B) III and IV only C) I and IV only 23) D) II and III only 24) What conditions, when present, are sufficient to conclude that a function f(x) has a limit as x approaches some value of a? A) The limit of f(x) as x→a from the left exists, the limit of f(x) as x→a from the right exists, and these two limits are the same B) Either the limit of f(x) as x→a from the left exists or the limit of f(x) as x→a from the right exists C) The limit of f(x) as x→a from the left exists, the limit of f(x) as x→a from the right exists, and at least one of these limits is the same as f(a) D) f(a) exists, the limit of f(x) as x→a from the left exists, and the limit of f(x) as x→a from the right exists 24) Use the graph to evaluate the limit 25) lim f(x) x→-1 25) y -6 -5 -4 -3 -2 -1 B) - x -1 A) ∞ C) -1 D) 26) lim f(x) x→0 26) y -4 -3 -2 -1 x -1 -2 -3 -4 A) C) -1 B) D) does not exist 27) lim f(x) x→0 27) y -6 -5 -4 -3 -2 -1 -1 x -2 -3 -4 -5 -6 A) -1 B) C) does not exist D) 28) lim f(x) x→0 28) 12 y 10 -2 -1 x -2 -4 A) B) -1 C) D) does not exist 29) lim f(x) x→0 29) y -4 -3 -2 -1 x -1 -2 -3 -4 A) ∞ B) -1 C) does not exist D) 30) lim f(x) x→0 30) y -4 -3 -2 -1 x -1 -2 -3 -4 A) -1 B) ∞ C) D) does not exist 31) lim f(x) x→0 31) y -4 -3 -2 -1 x -1 -2 -3 -4 B) -2 A) does not exist C) D) 32) lim f(x) x→0 32) y -4 -3 -2 -1 x -1 -2 -3 -4 A) -2 B) C) D) does not exist 33) lim f(x) x→0 33) y -4 -3 -2 -1 x -1 -2 -3 -4 A) -2 34) Find B) -1 C) does not exist D) lim f(x) and lim f(x) x→(-1)x→(-1)+ 34) y -4 -2 x -2 -4 -6 A) -7; -5 B) -7; -2 C) -5; -2 10 D) -2; -7 193) lim x→14 x2 - B) ± 187 A) 93.5 194) lim x→-8 - lim x→3+ lim t→1 + A) 187 D) Does not exist 194) B) C) D) Does not exist (x - 3)3 x-3 A) 196) C) x2 - 64 A) 195) 193) 195) B) C) D) Does not exist (t + 36)(t - 1) 13t - 13 37 13 196) B) 13 C) D) Does not exist SHORT ANSWER Write the word or phrase that best completes each statement or answers the question Provide an appropriate response 197) Use the Intermediate Value Theorem to prove that 9x3 - 5x2 + 10x + 10 = has a solution 197) between -1 and 198) Use the Intermediate Value Theorem to prove that 8x4 + 4x3 - 7x - = has a solution between -1 and 198) 199) Use the Intermediate Value Theorem to prove that x(x - 6)2 = has a solution between and 199) 200) Use the Intermediate Value Theorem to prove that sin x = x has a solution between π 200) and π MULTIPLE CHOICE Choose the one alternative that best completes the statement or answers the question Find numbers a and b, or k, so that f is continuous at every point 201) x < -4 -4, f(x) = ax + b, -4 ≤ x ≤ -3 5, x > -3 A) a = -4, b = B) a = 9, b = -22 C) a = 9, b = 32 44 201) D) Impossible 202) 202) x2 , x < -4 f(x) = ax + b, -4 ≤ x ≤ x + 20, x > A) a = -1, b = 20 B) a = 1, b = 20 C) a = 1, b = -20 D) Impossible 203) 203) 3x + 9, if x < -6 f(x) = kx + 6, if x ≥ -6 A) k = B) k = C) k = - D) k = 204) 204) x2 , if x ≤ f(x) = x + k, if x > A) k = -4 B) k = 12 C) k = 20 D) Impossible 205) 205) x2 , if x ≤ f(x) = kx, if x > A) k = B) k = C) k = 25 Solve the problem 206) Select the correct statement for the definition of the limit: D) Impossible lim f(x) = L x→x0 206) means that A) if given a number ε > 0, there exists a number δ > 0, such that for all x, < x - x0 < δ implies f(x) - L > ε B) if given any number ε > 0, there exists a number δ > 0, such that for all x, < x - x0 < ε implies f(x) - L < δ C) if given any number ε > 0, there exists a number δ > 0, such that for all x, < x - x0 < ε implies f(x) - L > δ D) if given any number ε > 0, there exists a number δ > 0, such that for all x, < x - x0 < δ implies f(x) - L < ε 207) Identify the incorrect statements about limits I The number L is the limit of f(x) as x approaches x0 if f(x) gets closer to L as x approaches x0 II The number L is the limit of f(x) as x approaches x0 if, for any ε > 0, there corresponds a δ > such that f(x) - L < ε whenever < x - x0 < δ III The number L is the limit of f(x) as x approaches x0 if, given any ε > 0, there exists a value of x for which f(x) - L < ε A) I and II B) II and III C) I and III 45 D) I, II, and III 207) Use the graph to find a δ > such that for all x, < x - x0 < δ ⇒ f(x) - L < ε 208) 208) y y=x+ f(x) = x + x0 = 4.2 L=4 ε = 0.2 3.8 x 0.8 1.2 NOT TO SCALE A) 0.1 B) 0.4 C) D) 0.2 209) 209) y y = 4x - f(x) = 4x - x0 = 5.2 L=5 ε = 0.2 4.8   1.95 2.05 x NOT TO SCALE A) 0.05 B) 0.1 C) 46 D) 0.5 210) 210) y y = -4x - 6.2 f(x) = -4x - x0 = -2 L=6 ε = 0.2 5.8  -2  -2.05 -1.95 x NOT TO SCALE A) 0.5 B) -0.05 C) 12 D) 0.05 211) 211) y y = -2x + 7.2 f(x) = -2x + x0 = -2 L=7 ε = 0.2 6.8 -2.1 -2 -1.9 x NOT TO SCALE A) 0.1 B) -0.1 C) 47 D) 0.2 212) 212) y y= x+2 3.5 f(x) = 3.3 x+2 x0 = 3.1 L = 3.3 ε = 0.2 x 0.8 1.1 NOT TO SCALE A) 2.3 B) 0.1 C) 0.3 D) -0.3 213) 213) f(x) = - y x+3 y=- 4.5 x+3 4.3 x0 = -1 L = 4.3 ε = 0.2 4.1 -1.1 -1 -0.8 x NOT TO SCALE A) -0.3 B) 5.3 C) 0.1 48 D) 0.3 214) 214) y f(x) = x x0 = L= y= ε= x 1.66 1.41 1.16 1.3575 2.7675 x NOT TO SCALE A) -0.59 B) 0.6425 C) 0.7675 D) 1.41 215) 215) y f(x) = x - x0 = L= y= ε= x-2 1.66 1.41 1.16 3.3554 4.7696 x NOT TO SCALE A) 2.59 B) 0.7696 C) 0.6446 49 D) 1.4142 216) 216) y y = 2x2 f(x) = 2x2 x0 = L=2 ε=1   0.71 1.22 x NOT TO SCALE A) 0.29 B) 0.51 C) D) 0.22 217) 217) y y = x2 - f(x) = x2 - x0 = L=3 ε=1   1.73 2.24 x NOT TO SCALE A) 0.51 B) 0.24 C) D) 0.27 A function f(x), a point x , the limit of f(x) as x approaches x , and a positive number ε is given Find a number δ > such that for all x, < x - x < δ ⇒ f(x) - L < ε 218) f(x) = 6x + 5, L = 17, x0 = 2, and ε = 0.01 A) 0.008333 218) B) 0.005 C) 0.001667 D) 0.003333 219) f(x) = 7x - 2, L = 12, x0 = 2, and ε = 0.01 A) 0.000714 219) B) 0.001429 C) 0.002857 D) 0.005 220) f(x) = -2x + 9, L = 1, x0 = 4, and ε = 0.01 A) 0.02 220) B) 0.005 C) 0.01 50 D) -0.0025 221) f(x) = -10x - 1, L = -11, x0 = 1, and ε = 0.01 A) -0.01 221) B) 0.001 C) 0.002 222) f(x) = 6x2, L =150, x0 = 5, and ε = 0.4 A) 4.99333 B) 0.00666 D) 0.0005 222) C) 5.00666 D) 0.00667 SHORT ANSWER Write the word or phrase that best completes each statement or answers the question Prove the limit statement 223) lim (3x - 4) = x→2 223) x2 - 49 224) lim = 14 x→7 x - 225) lim x→9 224) 2x2 - 15x- 27 = 21 x-9 225) 1 226) lim = x→7 x 226) 51 Answer Key Testname: UNTITLED2 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) 19) 20) 21) 22) 23) 24) 25) 26) 27) 28) 29) 30) 31) 32) 33) 34) 35) 36) 37) 38) 39) 40) 41) C C B A A A C B D B A C A D A A A C D C D D B A D D D C C D B A A D D C C C C B C x2 42) Answers may vary One possibility: lim = lim = According to the squeeze theorem, the function x→0 x→0 x sin(x) x2 , which is squeezed between and 1, must also approach as x approaches Thus, - cos(x) x sin(x) lim = x→0 - cos(x) 43) D 52 Answer Key Testname: UNTITLED2 44) 45) 46) 47) 48) 49) 50) 51) 52) 53) 54) 55) 56) 57) 58) 59) 60) 61) 62) 63) 64) 65) 66) 67) 68) 69) 70) 71) 72) 73) 74) 75) 76) 77) 78) 79) 80) 81) 82) 83) 84) 85) 86) 87) 88) 89) 90) 91) 92) 93) A D A C B B D C D D C B A C C D B C A D C B A D D D C C B B A D D D C C A A B B D B C B D B C B C D 53 Answer Key Testname: UNTITLED2 94) 95) 96) 97) 98) 99) 100) 101) 102) 103) 104) 105) 106) 107) 108) 109) 110) 111) 112) 113) 114) 115) 116) 117) 118) 119) 120) 121) 122) 123) 124) 125) 126) 127) 128) 129) 130) 131) 132) 133) 134) 135) 136) 137) 138) 139) 140) 141) 142) 143) D A C B B C B D D A D A D B B D B C A D A D B B A B D B B D D B D C D A B A D C B B D A D D C C C D 54 Answer Key Testname: UNTITLED2 144) 145) 146) 147) 148) 149) 150) 151) 152) C A C D A B D D Answers may vary One possible answer: y -8 -6 -4 -2 x -2 -4 -6 -8 153) Answers may vary One possible answer: y -8 -6 -4 -2 x -2 -4 -6 -8 55 Answer Key Testname: UNTITLED2 154) Answers may vary One possible answer: y 12 10 -12 -10 -8 -6 -4 -2-2 -4 -6 10 12 x -8 -10 -12 155) Answers may vary One possible answer: y -8 -6 -4 -2 x -2 156) 157) 158) 159) 160) 161) 162) 163) 164) 165) 166) 167) 168) 169) 170) 171) 172) 173) 174) 175) 176) B C B C A D D C A C C D D A A B B B A A B 56 Answer Key Testname: UNTITLED2 177) 178) 179) 180) 181) 182) 183) 184) 185) 186) 187) 188) 189) 190) 191) 192) 193) 194) 195) 196) C C D D C D D A D C C A A C B C C C B A 197) Let f(x) = 9x3 - 5x2 + 10x + 10 and let y0 = f(-1) = -14 and f(0) = 10 Since f is continuous on [-1, 0] and since y0 = is between f(-1) and f(0), by the Intermediate Value Theorem, there exists a c in the interval (-1 , 0) with the property that f(c) = Such a c is a solution to the equation 9x3 - 5x2 + 10x + 10 = 198) Let f(x) = 8x4 + 4x3 - 7x - and let y0 = f(-1) = and f(0) = -5 Since f is continuous on [-1, 0] and since y0 = is between f(-1) and f(0), by the Intermediate Value Theorem, there exists a c in the interval (-1, 0) with the property that f(c) = Such a c is a solution to the equation 8x4 + 4x3 - 7x - = 199) Let f(x) = x(x - 6)2 and let y0 = f(5) = and f(7) = Since f is continuous on [5, 7] and since y0 = is between f(5) and f(7), by the Intermediate Value Theorem, there exists a c in the interval (5, 7) with the property that f(c) = Such a c is a solution to the equation x(x - 6)2 = 200) Let f(x) = sin x π π and let y0 = f , π and since y0 = is between f ≈ 0.6366 and f(π) = Since f is continuous on x 2 π π and f(π), by the Intermediate Value Theorem, there exists a c in the interval , π , with the property that f(c) = 2 201) 202) 203) 204) 205) 206) 207) 208) 209) 210) 211) 212) 213) 214) Such a c is a solution to the equation sin x = x C B D B A D C D A D A B C B 57 Answer Key Testname: UNTITLED2 215) 216) 217) 218) 219) 220) 221) 222) 223) C D B C B B B B Let ε > be given Choose δ = ε/3 Then < x - < δ implies that (3x - 4) - = 3x - = 3(x - 2) = x - < 3δ = ε Thus, < x - < δ implies that (3x - 4) - < ε 224) Let ε > be given Choose δ = ε Then < x - < δ implies that x2 - 49 (x - 7)(x + 7) - 14 = - 14 x-7 x-7 = (x + 7) - 14 for x ≠ = x -7 < δ=ε x2 - 49 Thus, < x - < δ implies that - 14 < ε x-7 225) Let ε > be given Choose δ = ε/2 Then < x - < δ implies that 2x2 - 15x- 27 (x - 9)(2x + 3) - 21 = - 21 x-9 x-9 = (2x + 3) - 21 for x ≠ = 2x - 18 = 2(x - 9) = x - < 2δ = ε 2x2 - 15x- 27 Thus, < x - < δ implies that - 21 < ε x-9 226) Let ε > be given Choose δ = min{7/2, 49ε/2} Then < x - < δ implies that 1 7-x = x 7x = 1 ∙ ∙ x-7 x < 1 49ε ∙ ∙ =ε 7/2 Thus, < x - < δ implies that 1