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 + 3x, [1, 8] 88 A) 21 B) 1) C) 11 D) 12 2) y = 9x3 + 5x2 - 8, [-2, 8] A) 492 3) y = C) 615 D) 498 C) D) 10 2x, [2, 8] 3) B) A) 4) y = 2) 1245 B) , [4, 7] x-2 A) - 10 5) y = 4x2 , 0, A) 4) B) C) - B) 7) h(t) = sin (3t), 0, 8) g(t) = + tan t, - 10 D) 6) B) -2 C) - D) -34 π 6 π D) 5) 6) y = -3x2 - x, [5, 6] A) A) - A) - C) 7) B) π C) π D) π π π , 4 8) B) - π C) π D) 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) 0.5 B) C) 9) D) 1.5 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) C) D) 0.5 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) 2 D) 12) x = 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) 12) B) C) D) -8 13) x = 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 13) B) -0.5 C) D) Find the slope of the curve for the given value of x 14) y = x2 + 5x, x = A) slope is 20 15) y = x2 + 11x - 15, x = 1 A) slope is 20 16) y = x3 - 5x, x = A) slope is -3 14) B) slope is -39 C) slope is 25 B) slope is -39 C) slope is 25 D) slope is 13 15) D) slope is 13 16) C) slope is -2 B) slope is D) slope is 17) y = x3 - 2x2 + 4, x = A) slope is B) slope is -15 C) slope is 15 D) slope is 18) y = - x3 , x = -1 A) slope is B) slope is C) slope is -3 D) slope is -1 17) 18) Solve the problem 19) 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 19) III lim f(x) does not exist x→0 A) none 20) Given B) I C) III D) II 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 20) III lim f(x) does not exist x→0 A) I B) III C) II D) none 21) 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) I and IV only C) III and IV only 21) D) II and III only 22) What conditions, when present, are sufficient to conclude that a function f(x) has a limit as x approaches some value of a? A) 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 B) 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 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) 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 22) Use the graph to evaluate the limit 23) lim f(x) x→-1 23) y -6 -5 -4 -3 -2 -1 x -1 A) B) ∞ C) -1 D) - 24) lim f(x) x→0 24) y -4 -3 -2 -1 x -1 -2 -3 -4 A) -2 B) C) D) does not exist 25) lim f(x) x→0 25) y -6 -5 -4 -3 -2 -1 -1 x -2 -3 -4 -5 -6 A) C) -1 B) does not exist D) 26) lim f(x) x→0 26) 12 y 10 -2 -1 x -2 -4 A) -1 B) C) D) does not exist 27) lim f(x) x→0 27) y -4 -3 -2 -1 x -1 -2 -3 -4 A) -1 B) does not exist C) D) ∞ 28) lim f(x) x→0 28) y -4 -3 -2 -1 x -1 -2 -3 -4 A) ∞ C) -1 B) does not exist D) 29) lim f(x) x→0 29) y -4 -3 -2 -1 x -1 -2 -3 -4 A) B) does not exist C) D) -2 30) lim f(x) x→0 30) y -4 -3 -2 -1 x -1 -2 -3 -4 A) does not exist C) -2 B) D) 31) lim f(x) x→0 31) y -4 -3 -2 -1 x -1 -2 -3 -4 B) -1 A) 32) Find C) does not exist D) -2 lim f(x) and lim f(x) x→(-1)x→(-1)+ 32) y -4 -2 x -2 -4 -6 A) -7; -5 B) -7; -2 C) -5; -2 D) -2; -7 Use the table of values of f to estimate the limit 33) Let f(x) = x2 + 8x - 2, find lim f(x) x→2 x f(x) 1.9 1.99 1.999 33) 2.001 2.01 2.1 A) x 1.9 1.99 1.999 2.001 2.01 2.1 ; limit = ∞ f(x) 5.043 5.364 5.396 5.404 5.436 5.763 B) x 1.9 1.99 1.999 2.001 2.01 2.1 ; limit = 5.40 f(x) 5.043 5.364 5.396 5.404 5.436 5.763 C) x 1.9 1.99 1.999 2.001 2.01 2.1 ; limit = 18.0 f(x) 16.810 17.880 17.988 18.012 18.120 19.210 D) x 1.9 1.99 1.999 2.001 2.01 2.1 ; limit = 17.70 f(x) 16.692 17.592 17.689 17.710 17.808 18.789 34) Let f(x) = x f(x) x-4 , find lim f(x) x-2 x→4 3.9 3.99 3.999 34) 4.001 4.01 4.1 A) x 3.9 3.99 3.999 4.001 4.01 4.1 ; limit = 5.10 f(x) 5.07736 5.09775 5.09978 5.10022 5.10225 5.12236 B) x 3.9 3.99 3.999 4.001 4.01 4.1 ; limit = 4.0 f(x) 3.97484 3.99750 3.99975 4.00025 4.00250 4.02485 C) x 3.9 3.99 3.999 4.001 4.01 4.1 ; limit = 1.20 f(x) 1.19245 1.19925 1.19993 1.20007 1.20075 1.20745 D) x 3.9 3.99 3.999 4.001 4.01 4.1 ; limit = ∞ f(x) 1.19245 1.19925 1.19993 1.20007 1.20075 1.20745 10 SHORT ANSWER Write the word or phrase that best completes each statement or answers the question Provide an appropriate response 190) Use the Intermediate Value Theorem to prove that 5x3 + 4x2 + 10x + 10 = has a solution 190) between -1 and 191) Use the Intermediate Value Theorem to prove that 6x4 + 9x3 - 5x - = has a solution 191) between -2 and -1 192) Use the Intermediate Value Theorem to prove that x(x - 9)2 = has a solution between and 10 193) Use the Intermediate Value Theorem to prove that sin x = x has a solution between π 192) 193) 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 194) x < -3 -12, f(x) = ax + b, -3 ≤ x ≤ 8, x>1 A) a = 5, b = 13 B) a = -12, b = C) a = 5, b = 194) D) Impossible 195) 195) x2 , x < -3 f(x) = ax + b, -3 ≤ x ≤ x + 12, x > A) a = -1, b = 12 B) a = 1, b = 12 C) a = 1, b = -12 D) Impossible 196) 196) 3x + 8, if x < -5 f(x) = kx + 7, if x ≥ -5 A) k = B) k = C) k = - D) k = 14 197) 197) x2 , if x ≤ f(x) = x + k, if x > A) k = 56 B) k = 42 C) k = -7 45 D) Impossible 198) 198) x2 , if x ≤ f(x) = kx, if x > A) k = 64 B) k = C) k = Solve the problem 199) Select the correct statement for the definition of the limit: D) Impossible lim f(x) = L x→x0 199) means that A) 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 B) 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 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 a number ε > 0, there exists a number δ > 0, such that for all x, < x - x0 < δ implies f(x) - L > ε 200) 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 200) 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) II and III B) I and III C) I and II D) I, II, and III Use the graph to find a δ > such that for all x, < x - x < δ ⇒ f(x) - L < ε 201) 201) y y=x+2 f(x) = x + x0 = 4.2 L=4 ε = 0.2 3.8 x 1.8 2.2 NOT TO SCALE A) 0.1 B) 0.2 C) 46 D) 0.4 202) 202) y y = 4x - f(x) = 4x - x0 = 7.2 L=7 ε = 0.2 6.8 2  1.95 x 2.05 NOT TO SCALE A) 0.1 B) 0.05 C) D) 0.5 203) 203) y y = -4x - 5.2 f(x) = -4x - x0 = -2 L=5 ε = 0.2 4.8  -2  -2.05 x -1.95 NOT TO SCALE A) 0.5 B) 0.05 C) 13 47 D) -0.05 204) 204) y y = -x + 4.2 f(x) = -x + x0 = -2 L=4 ε = 0.2 3.8 -2.2 -2 -1.8 x NOT TO SCALE A) -0.2 B) 0.2 C) 0.4 D) 205) 205) y y= x+3 4.7 f(x) = 4.5 x0 = L = 4.5 ε = 0.2 4.3 x+3 x 0.9 1.1 NOT TO SCALE A) -0.2 B) 0.2 C) 3.5 48 D) 0.1 206) 206) y =- y x+2 3.7 f(x) = - x + 2 3.5 x0 = -1 L = 3.5 ε = 0.2 3.3 -1.1 -1 -0.9 x NOT TO SCALE A) 4.5 B) 0.1 C) 0.2 D) -0.2 207) 207) y f(x) = x x0 = L= y= ε= x 1.66 1.41 1.16 1.3575 2.7675 x NOT TO SCALE A) 0.7675 C) -0.59 B) 0.6425 49 D) 1.41 208) 208) y f(x) = x - x0 = y= L=1 ε= x-2 1.25 0.75 2.5625 3.5625 x NOT TO SCALE A) B) C) 0.4375 D) 0.5625 209) 209) y y = x2 f(x) = x2 x0 = L=4 ε=1   1.73 x 2.24 NOT TO SCALE A) 0.51 B) 0.27 C) 50 D) 0.24 210) 210) y y = x2 - f(x) = x2 - x0 = L=7 ε=1 3  2.83 x 3.16 NOT TO SCALE A) 0.17 B) C) 0.33 D) 0.16 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 211) f(x) = 3x + 10, L = 13, x0 = 1, and ε = 0.01 A) 0.003333 211) B) 0.01 C) 0.016667 D) 0.006667 C) 0.000556 D) 0.01 C) 0.013333 D) 0.006667 212) f(x) = 9x - 7, L = 2, x0 = 1, and ε = 0.01 A) 0.001111 212) B) 0.002222 213) f(x) = -3x + 9, L = 6, x0 = 1, and ε = 0.01 A) 0.003333 213) B) -0.01 214) f(x) = -9x - 7, L = -34, x0 = 3, and ε = 0.01 A) 0.000556 214) B) -0.003333 C) 0.002222 D) 0.001111 215) f(x) = 2x2, L =98, x0 = 7, and ε = 0.4 A) 7.01427 215) B) 0.01427 C) 0.0143 D) 6.9857 SHORT ANSWER Write the word or phrase that best completes each statement or answers the question Prove the limit statement 216) lim (2x - 1) = x→5 216) x2 - 25 217) lim = 10 x→5 x - 218) lim x→7 217) 3x2 - 19x- 14 = 23 x-7 218) 51 1 219) lim = x→5 x 219) 52 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) D D B A D D D C C D B A A D D C C C C B C D A D A C B B D C D D C B A C C D B x2 40) 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 = 2 cos(x) x→0 41) C 42) A 43) D 53 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) 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 D A C B B C B D D B C A D B B D B C A D 54 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) A D B B A B D B B D D B C D D A B A D C B B D A D D C D D B A D A D A B D A B C C D A D D C A C C D 55 Answer Key Testname: UNTITLED2 144) 145) 146) 147) 148) 149) 150) 151) 152) 153) 154) 155) C B D A B A D B C A C Answers may vary One possible answer: y -8 -6 -4 -2 x -2 -4 -6 -8 156) Answers may vary One possible answer: y -8 -6 -4 -2 x -2 -4 -6 -8 56 Answer Key Testname: UNTITLED2 157) Answers may vary One possible answer: y 12 10 -12 -10 -8 -6 -4 -2-2 -4 -6 10 12 x -8 -10 -12 158) Answers may vary One possible answer: y -8 -6 -4 -2 x -2 159) 160) 161) 162) 163) 164) 165) 166) 167) 168) 169) 170) 171) 172) 173) 174) 175) 176) 177) 178) 179) D C D D A D C C A A A B A A C D D B A D B 57 Answer Key Testname: UNTITLED2 180) 181) 182) 183) 184) 185) 186) 187) 188) 189) C D A D D A C B B A 190) Let f(x) = 5x3 + 4x2 + 10x + 10 and let y0 = f(-1) = -1 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 5x3 + 4x2 + 10x + 10 = 191) Let f(x) = 6x4 + 9x3 - 5x - and let y0 = f(-2) = 26 and f(-1) = -6 Since f is continuous on [-2, -1] and since y0 = is between f(-2) and f(-1), by the Intermediate Value Theorem, there exists a c in the interval (-2, -1) with the property that f(c) = Such a c is a solution to the equation 6x4 + 9x3 - 5x - = 192) Let f(x) = x(x - 9)2 and let y0 = f(8) = and f(10) = 10 Since f is continuous on [8, 10] and since y0 = is between f(8) and f(10), by the Intermediate Value Theorem, there exists a c in the interval (8, 10) with the property that f(c) = Such a c is a solution to the equation x(x - 9)2 = 193) Let f(x) = f π π and f(π), by the Intermediate Value Theorem, there exists a c in the interval , π , with the property that 2 f(c) = 194) 195) 196) 197) 198) 199) 200) 201) 202) 203) 204) 205) 206) 207) 208) 209) 210) 211) 212) 213) 214) 215) sin x π π and let y0 = f , π and since y0 = is between ≈ 0.6366 and f(π) = Since f is continuous on x 2 Such a c is a solution to the equation sin x = x C B D B C B B B B B B D B B C D D A A A D B 58 Answer Key Testname: UNTITLED2 216) Let ε > be given Choose δ = ε/2 Then < x - < δ implies that (2x - 1) - = 2x - 10 = 2(x - 5) = x - < 2δ = ε Thus, < x - < δ implies that (2x - 1) - < ε 217) Let ε > be given Choose δ = ε Then < x - < δ implies that x2 - 25 (x - 5)(x + 5) - 10 = - 10 x-5 x-5 for x ≠ = (x + 5) - 10 = x -5 < δ=ε x2 - 25 Thus, < x - < δ implies that - 10 < ε x-5 218) Let ε > be given Choose δ = ε/3 Then < x - < δ implies that 3x2 - 19x- 14 (x - 7)(3x + 2) - 23 = - 23 x-7 x-7 for x ≠ = (3x + 2) - 23 = 3x - 21 = 3(x - 7) = x - < 3δ = ε 3x2 - 19x- 14 Thus, < x - < δ implies that - 23 < ε x-7 219) Let ε > be given Choose δ = min{5/2, 25ε/2} Then < x - < δ implies that 1 5-x = x 5x = 1 ∙ ∙ x-5 x < 1 25ε ∙ ∙ =ε 5/2 Thus, < x - < δ implies that 1