Math 11 Review Packet for Exam #1

Math 11 Review Packet for Exam #1 Limit Practice Problems Evaluate the following limits, including if the limit Does Not Exist, or is +∞ or −∞. Always justify your work: 1. lim 16 = w 2. lim 16 = w w→0+ w→0− 3. lim 16 = w 4. lim 3−t = t−2 w→0 t→2+ 5. lim t→2− 6. lim t→2 3−t = t−2 3−t = t−2 7. lim 3−t = (t − 2)2 8. lim 3−t = (t − 2)2 t→2+ t→2− 3−t = t→2 (t − 2)2 9. lim 10. lim (x + 2)2 = x2 − 3x − 4 11. lim (x + 2)2 = x2 − 3x − 4 x→4+ x→4− (x + 2)2 = x→4 x2 − 3x − 4 12. lim 13. lim x→4+ 14. lim x→4− 15. lim x2 x−4 = − 3x − 4 x−4 = x2 − 3x − 4 x→4 x2 x−4 = − 3x − 4 1 16. lim x2 − 2x − 8 = x2 − 3x − 4 17. lim x2 − 2x − 8 = x2 − 3x − 4 x→4+ x→4− x2 − 2x − 8 = x→4 x2 − 3x − 4 18. lim x2 − 4x − 12 = x→6 x2 − 3x − 18 19. lim x2 − 4x − 12 = x→1 x2 − 3x − 18 20. lim x2 − 4x − 12 = x→0 x2 − 3x − 18 21. lim x2 − 4x − 12 = x→−3 x2 − 3x − 18 22. lim x2 − 4x − 12 = x→−2 x2 − 3x − 18 23. lim x2 − 4x − 12 = x→0+ x2 − 7x 24. lim x2 − 4x − 12 = x→0− x2 − 7x 25. lim x2 − 4x − 12 = x→0 x2 − 7x 26. lim x2 − 4x = x→0 x2 − 7x 27. lim 28. lim x2 − 9 = |x − 3| 29. lim x2 − 9 = |x − 3| x→3+ x→3− 30. lim x3 + 2009x2 + 2000x = |x| 31. lim x3 + 2009x2 + 2000x = |x| x→0+ x→0− 32. 33. lim x2 + 6x + 5 = |x + 5| lim x2 + 6x + 5 = |x + 5| x→(−5)+ x→(−5)− 2 x2 + 6x + 5 = x→−5 |x + 5| 34. lim t2 − 1 = t→1 t2 − 11t + 10 35. lim t2 = t→1 t2 + t − 1 36. lim 2009(t2 + 6t + 5) = t→−1 t2 + t 37. lim x2 − 10x + 9 = x→9 x2 + x − 90 38. lim 39. lim t500 + t400 + t300 + t200 + t100 = t→1 x2 − x − 6 = x→3 x2 − 2x − 3 40. lim 41. lim √ x→1 x−1 = x+3−2 9x − x2 √ = x→9 3 − x √ x2 + 8 − 3 = 43. lim x→1 x−1 42. lim t+4 = + 6t 44. lim t→1 t2 2 = w→0 w + 6 45. lim 2 = w→6 w + 6 46. lim 47. lim x2 − 3x + 6 = x→−5 48. lim w→−2 w 2 49. lim x→2 3 = x−2 50. lim x→−1 51. lim x→1 w+2 = − 3w + 2 5 = 1−x 5x = 1−x 52. lim x→5+ 53. lim x→5− 6x = 5−x 6x = 5−x 3 54. lim x→5 6x = 5−x x2 − 3x − 28 = x→−4 x2 + 4x 55. lim x2 − 3x − 28 = x→0 x2 + 4x 56. lim 57. lim x→3− 58. lim x→3 −4 = x−3 −4 = x−3 59. lim x→3+ −4 = 3−x −4 = x→3 3 − x 60. lim 61. lim |x| + 3 = x→0+ 62. lim |x| + 3 = x→0− 63. lim |x − 1| − 3 = x→1+ 64. lim |x − 1| − 3 = x→1− 65. lim |x − 1| − 3 = x→5 66. lim |x − 1| − 3 = x→−2 67. lim x2 − 1 = |x − 1| 68. lim x2 − 1 = |x − 1| x→1+ x→1− 69. lim x→1 |1 − x| = (1 − x)2 x2 − 4 = x→2 |x − 2| 70. lim 71. lim x→7− 72. lim x→0− 7−x = |x − 7| x = x − |x| 4 73. lim x→2+ 2−x = |x − 2| Use the ε − δ definition for limits to prove each of the following: 74. lim 7x − 6 = 8. x→2 3 75. lim 2 − x = 5. x→−7 7 76. lim 2x + 1 = −3. x→−2 77. lim 1 − 4x = −11. x→3 78. lim 5 − 4t = −3. t→2 79. lim 4 − 3x = 7. x→−1 80. lim −2t − 5 = −9. t→2 81. lim −3x + 17 = 5. x→4 82. lim 1 − 5x = 16. x→−3 83. 4 x + 3 = −5. x→−14 7 Tangent Lines Please use the limit definition for the derivative when computing the derivatives in this section. lim 84. Find an equation for the tangent line to the graph of f (x) = x − 2x2 at the point (1, −1) √ 85. Find an equation for the tangent line to the graph of f (x) = x at x = 4 86. At which point(s) does the graph of f (x) = −x2 + 13 have a horizontal tangent line? 87. At which point(s) of the graph of f (x) = −x3 + 13 is the slope of the tangent line equal to -27? What’s the picture representing this problem? 88. There are two points on the graph of the curve y = −x2 + 7 whose tangent line to the graph at those points passes through the point (0, 11). Find those two points. 89. Find the equation of the line passing through (2, 3) which is perpendicular to the tangent to the curve y = x3 − 3x + 1 at the point (2, 3). 90. Find the equation of the tangent line to the curve y = x3 + x at the point(s) where the slope equals 4. 5 Derivatives Use the limit definition of the derivative to calculate the derivative for each of the following functions: 91. f (x) = 3 − 9x2 92. f (x) = −4x − x2 − 3 93. f (x) = −3 x 94. f (x) = −9x2 + 3 95. f (x) = x3 96. f (x) = x2 − 4x + 3 1 x2 √ 98. f (x) = x − 7 √ 99. f (x) = 3x − 7 97. f (x) = 100. f (x) = 1 x3 1 101. f (x) = √ x Functions and Limit Practice Problems Evaluate the following limits: x−1 = x→1 g(x2 ) − 3 102. Let g(x) = 2x + 1. Compute lim u2 − 2u = u→2 G(u − 3) 103. Let G(u) = u2 + u. Compute lim F (x − 1) = x→4 F (x − 5) 104. Let F (x) = |x| + 1. Compute lim 105. Let h(y) = y 2 − 3. Compute 106. Let g(x) = √ lim x→−2 x+2 = h(2x) − h(x + 6) g(s2 + 8) − 3 = s→1 s−1 x. Compute lim f (t − 1) − 2f (t) 1 = 107. Let f (t) = . Compute lim t→2 t t2 − 4 6 More Tangent Lines Please use the limit definition for the derivative when computing derivatives in this section. 1 at the point (0, −1). 108. Find an equation for the tangent line to the graph of f (x) = x−1   1 1 109. Find an equation for the tangent line to the graph of f (x) = at the point 1, . x+1 2 110. Find an equation for the tangent line to the graph of y = 3 + 1 when x = 1. x 111. Find an equation for the tangent line to the graph of y = x2 − 4x + 2 when x = 1. At what point is the tangent line to this curve horizontal? Functions Please state what the domain is for each of the following functions. x+2 x−1 √ 113. g(x) = x − 2 √ 114. g(x) = 2 − x 112. f (x) = 115. g(x) = √ 1 2−x x−3 x2 + 3 1 117. w(x) = x−4 116. f (x) = x2 + 6x + 8 x+2 More Functions x+1 . Compute (and simplify, if possible) the following: 119. Let g(x) = x 118. f (x) = (a) (b) (c) (d) (e) g(2) = g(0) = g( 12 ) = 1 g( 10 )= g(t − 2) = g(2 + h) − g(2) (f) = h 120. Let f (x) = (a) (b) (c) (d) (e) 1 1 − . Compute (and simplify, if possible) the following: x+1 x f (1) = f (−1) = f (− 21 ) = f (t − 1) = f ( 1t ) = 7 More Functions 121. Let f (x) = following: √ x, g(x) = x2 + 4, and h(x) = 1 . Compute (and simplify, if possible) the x (a) f ◦ g(x) = (b) g ◦ f (x) = (c) g ◦ h(x) − h ◦ g(x) = (d) f ◦ h(x) − h ◦ f (x) = (e) h ◦ g ◦ f (x) = (f) g ◦ f ◦ f (x) = (g) g ◦ g(x) = 122. Suppose lim f (x) = 3 and lim g(x) = −2. Assume g is continuous at x = 3. Find each of the x→3 x→3 following values: (a) lim 2f (x) − 4g(x) = x→3 (b) lim g(x) · x→3 (c) g(3) = x2 − 9 = x−3 (d) lim g(f (x)) = x→3 p (e) lim (f (x))2 − 8g(x) = x→3 Consider each of the following piecewise defined functions. Answer the related questions. Justify your answers please.  if x < 0  x 2 x if 0 ≤ x ≤ 2 123. Let f (x) =  8 − x if x > 2 Sketch the graph. Find the numbers at which f is discontinuous. Evaluate: lim f (x) = x→2 lim f (x) = x→0   x + 2 if x < 0 2x2 if 0 ≤ x ≤ 1 124. Let f (x) =  3 − x if x > 1 Sketch the graph. Find the numbers at which f is discontinuous. Evaluate: lim f (x) = x→2 lim f (x) = x→1 lim f (x) = x→0 8 125. Let f (x) =      1 x−4  −3x + 4 if x ≤ 3 −2 if x > 3 if x < 2     1 if x ≥ 2 x Sketch the graph. Find the numbers at which f is discontinuous. Evaluate: lim f (x) = x→1 lim f (x) = x→2 126. Let f (x) = Sketch the graph. Find the numbers at which f is discontinuous. Evaluate: lim f (x) = x→3 lim f (x) = x→−2  t−3    3−t 127. Let f (t) = 1    3−t if if if if t≤3 35 Sketch the graph. Find the numbers at which f is discontinuous. Evaluate: lim f (t) = t→3 lim f (t) = t→0 lim f (t) = t→5   −2x if x < 0 x2 if 0 ≤ x ≤ 2 128. Let f (x) =  6 − x if x > 2 Sketch the graph. Find the numbers at which f is discontinuous. Evaluate: lim f (x) = x→−2 lim f (x) = x→0 lim f (x) = x→2 lim f (x) = x→6  3 if x < −1  x x if − 1 < x < 1 129. Let f (x) =  1 − x if x ≥ 1 Sketch the graph. Find the numbers at which f is discontinuous. Evaluate: lim f (x) = x→−1 lim f (x) = x→1 9  x−1    1 130. Let f (x) = 3    √ x if if if if x<2 24 Sketch the graph. Find the numbers at which f is discontinuous. Evaluate: lim f (x) = x→0 lim f (x) = x→2 lim f (x) = x→4 f (4) =  8       x+2 2 131. Let h(x) = 1   x−4   2   √ x if x < 0 if x = 0 if 0 < x ≤ 16 if x > 16 Sketch the graph. Find the numbers at which h is discontinuous. Evaluate: lim h(x) = x→−2 lim h(x) = x→0 lim h(x) = x→16  8   if x < 0   x−2 132. Let h(x) = 2 if x = 0   1  x − 4 if x > 0  2 Sketch the graph. Find the numbers at which h is discontinuous. Evaluate: lim h(x) = x→0 lim h(x) = x→2 133. Let h(x) =      8 x−2 if x ≤ 0     1 x − 4 if x > 0 2 Sketch the graph. Find the numbers at which h is discontinuous. Evaluate: lim h(x) = x→0 lim h(x) = x→1 h(0) = 10