1. **Problem:** Find the domain and critical values of the function $$y = x^3 + 6x^2 + 12x - 1$$. 2. **Domain:** Since this is a polynomial, the domain is all real numbers, $$\mathbb{R}$$. 3. **Find the derivative:** $$y' = 3x^2 + 12x + 12$$ 4. **Set derivative equal to zero to find critical points:** $$3x^2 + 12x + 12 = 0$$ 5. **Divide entire equation by 3:** $$\cancel{3}x^2 + \cancel{3}4x + \cancel{3}4 = 0 \Rightarrow x^2 + 4x + 4 = 0$$ 6. **Factor the quadratic:** $$ (x + 2)^2 = 0 $$ 7. **Solve for x:** $$ x = -2 $$ 8. **Critical value:** Substitute back into original function: $$ y(-2) = (-2)^3 + 6(-2)^2 + 12(-2) - 1 = -8 + 24 - 24 - 1 = -9 $$ --- 9. **Problem:** Find the domain and critical values of $$y = x^5 + 5x^4 - 5x^3 - 35x^2 - 40x - 1$$. 10. **Domain:** All real numbers, $$\mathbb{R}$$. 11. **Derivative:** $$ y' = 5x^4 + 20x^3 - 15x^2 - 70x - 40 $$ 12. **Set derivative to zero:** $$ 5x^4 + 20x^3 - 15x^2 - 70x - 40 = 0 $$ 13. **Divide by 5:** $$ \cancel{5}x^4 + \cancel{5}4x^3 - \cancel{5}3x^2 - \cancel{5}14x - \cancel{5}8 = 0 \Rightarrow x^4 + 4x^3 - 3x^2 - 14x - 8 = 0 $$ 14. **Solve quartic equation (numerical or factoring methods needed).** --- 15. **Problem:** Find the domain and critical values of $$y = 4x^3 - 21x^2 + 48x - 7$$. 16. **Domain:** All real numbers, $$\mathbb{R}$$. 17. **Derivative:** $$ y' = 12x^2 - 42x + 48 $$ 18. **Set derivative to zero:** $$ 12x^2 - 42x + 48 = 0 $$ 19. **Divide by 6:** $$ \cancel{6}2x^2 - \cancel{6}7x + \cancel{6}8 = 0 \Rightarrow 2x^2 - 7x + 8 = 0 $$ 20. **Calculate discriminant:** $$ \Delta = (-7)^2 - 4 \cdot 2 \cdot 8 = 49 - 64 = -15 < 0 $$ 21. **No real roots, so no critical points.** --- 22. **Problem:** Find the domain and critical values of $$y = x^4 - 32x^2 + 12$$. 23. **Domain:** All real numbers, $$\mathbb{R}$$. 24. **Derivative:** $$ y' = 4x^3 - 64x $$ 25. **Set derivative to zero:** $$ 4x^3 - 64x = 0 $$ 26. **Factor out 4x:** $$ 4x(x^2 - 16) = 0 $$ 27. **Set each factor to zero:** $$ 4x = 0 \Rightarrow x = 0 $$ $$ x^2 - 16 = 0 \Rightarrow x = \pm 4 $$ 28. **Critical points:** $$ x = -4, 0, 4 $$ 29. **Find corresponding y-values:** $$ y(-4) = (-4)^4 - 32(-4)^2 + 12 = 256 - 512 + 12 = -244 $$ $$ y(0) = 0 - 0 + 12 = 12 $$ $$ y(4) = 256 - 512 + 12 = -244 $$ --- 30. **Problem:** Find the domain and critical values of $$y = 3x^5 - 40x^3 + 180x - 71$$. 31. **Domain:** All real numbers, $$\mathbb{R}$$. 32. **Derivative:** $$ y' = 15x^4 - 120x^2 + 180 $$ 33. **Set derivative to zero:** $$ 15x^4 - 120x^2 + 180 = 0 $$ 34. **Divide by 15:** $$ \cancel{15}x^4 - \cancel{15}8x^2 + \cancel{15}12 = 0 \Rightarrow x^4 - 8x^2 + 12 = 0 $$ 35. **Substitute $$u = x^2$$:** $$ u^2 - 8u + 12 = 0 $$ 36. **Solve quadratic:** $$ u = \frac{8 \pm \sqrt{64 - 48}}{2} = \frac{8 \pm \sqrt{16}}{2} = \frac{8 \pm 4}{2} $$ 37. **Solutions for u:** $$ u_1 = 6, u_2 = 2 $$ 38. **Back-substitute:** $$ x^2 = 6 \Rightarrow x = \pm \sqrt{6} $$ $$ x^2 = 2 \Rightarrow x = \pm \sqrt{2} $$ 39. **Critical points:** $$ x = \pm \sqrt{6}, \pm \sqrt{2} $$ 40. **Find y-values by substitution (optional).** --- 41. **Problem:** Find the domain and critical values of $$y = x^3 + x^2 - 5x - 3$$ on $$x \in [-3,1]$$. 42. **Domain:** Closed interval $$[-3,1]$$. 43. **Derivative:** $$ y' = 3x^2 + 2x - 5 $$ 44. **Set derivative to zero:** $$ 3x^2 + 2x - 5 = 0 $$ 45. **Calculate discriminant:** $$ \Delta = 2^2 - 4 \cdot 3 \cdot (-5) = 4 + 60 = 64 $$ 46. **Solve for x:** $$ x = \frac{-2 \pm 8}{6} $$ 47. **Solutions:** $$ x_1 = 1, x_2 = -\frac{5}{3} \approx -1.6667 $$ 48. **Both critical points are in the domain.** 49. **Evaluate y at critical points and endpoints:** $$ y(1) = 1 + 1 - 5 - 3 = -6 $$ $$ y(-\frac{5}{3}) = (-\frac{5}{3})^3 + (-\frac{5}{3})^2 - 5(-\frac{5}{3}) - 3 = -\frac{125}{27} + \frac{25}{9} + \frac{25}{3} - 3 = \frac{256}{27} \approx 9.48 $$ $$ y(-3) = -27 + 9 + 15 - 3 = -6 $$ --- 50. **Problem:** Find the domain and critical values of $$y = 2x^3 + x^2 - 4x + 8$$ on $$x \in [0,7]$$. 51. **Domain:** Closed interval $$[0,7]$$. 52. **Derivative:** $$ y' = 6x^2 + 2x - 4 $$ 53. **Set derivative to zero:** $$ 6x^2 + 2x - 4 = 0 $$ 54. **Divide by 2:** $$ 3x^2 + x - 2 = 0 $$ 55. **Calculate discriminant:** $$ \Delta = 1^2 - 4 \cdot 3 \cdot (-2) = 1 + 24 = 25 $$ 56. **Solve for x:** $$ x = \frac{-1 \pm 5}{6} $$ 57. **Solutions:** $$ x_1 = \frac{4}{6} = \frac{2}{3} \approx 0.6667, x_2 = -1 $$ 58. **Only $$x=\frac{2}{3}$$ is in the domain.** 59. **Evaluate y at critical point and endpoints:** $$ y(0) = 8 $$ $$ y(\frac{2}{3}) = 2(\frac{2}{3})^3 + (\frac{2}{3})^2 - 4(\frac{2}{3}) + 8 = 2(\frac{8}{27}) + \frac{4}{9} - \frac{8}{3} + 8 = \frac{16}{27} + \frac{12}{27} - \frac{72}{27} + \frac{216}{27} = \frac{172}{27} \approx 6.37 $$ $$ y(7) = 2(343) + 49 - 28 + 8 = 686 + 49 - 28 + 8 = 715 $$ --- 60. **Problem:** Find the domain and critical values of $$y = -3x^4 + 24x^2 + 5$$ on $$x \in [-3,5]$$. 61. **Domain:** Closed interval $$[-3,5]$$. 62. **Derivative:** $$ y' = -12x^3 + 48x $$ 63. **Set derivative to zero:** $$ -12x^3 + 48x = 0 $$ 64. **Factor:** $$ -12x(x^2 - 4) = 0 $$ 65. **Solve:** $$ x = 0, x = \pm 2 $$ 66. **All critical points are in the domain.** 67. **Evaluate y at critical points and endpoints:** $$ y(-3) = -3(81) + 24(9) + 5 = -243 + 216 + 5 = -22 $$ $$ y(-2) = -3(16) + 24(4) + 5 = -48 + 96 + 5 = 53 $$ $$ y(0) = 5 $$ $$ y(2) = 53 $$ $$ y(5) = -3(625) + 24(25) + 5 = -1875 + 600 + 5 = -1270 $$ --- 68. **Problem:** Find the domain and critical values of $$y = 2x^5 - 35x^3 + 135x - 29$$ on $$x \in [-2,6]$$. 69. **Domain:** Closed interval $$[-2,6]$$. 70. **Derivative:** $$ y' = 10x^4 - 105x^2 + 135 $$ 71. **Set derivative to zero:** $$ 10x^4 - 105x^2 + 135 = 0 $$ 72. **Divide by 5:** $$ 2x^4 - 21x^2 + 27 = 0 $$ 73. **Substitute $$u = x^2$$:** $$ 2u^2 - 21u + 27 = 0 $$ 74. **Calculate discriminant:** $$ \Delta = (-21)^2 - 4 \cdot 2 \cdot 27 = 441 - 216 = 225 $$ 75. **Solve for u:** $$ u = \frac{21 \pm 15}{4} $$ 76. **Solutions:** $$ u_1 = 9, u_2 = 1.5 $$ 77. **Back-substitute:** $$ x^2 = 9 \Rightarrow x = \pm 3 $$ $$ x^2 = 1.5 \Rightarrow x = \pm \sqrt{1.5} \approx \pm 1.2247 $$ 78. **Critical points in domain:** $$-2 \leq x \leq 6$$ includes $$-3$$ is outside domain, so discard. 79. **Valid critical points:** $$x = -1.2247, 1.2247, 3$$ --- 80. **Problem:** Find the domain and extreme points of $$y = x^5 - 16x + 12$$. 81. **Domain:** All real numbers, $$\mathbb{R}$$. 82. **Derivative:** $$ y' = 5x^4 - 16 $$ 83. **Set derivative to zero:** $$ 5x^4 - 16 = 0 \Rightarrow 5x^4 = 16 \Rightarrow x^4 = \frac{16}{5} $$ 84. **Solve for x:** $$ x = \pm \sqrt[4]{\frac{16}{5}} = \pm \left(\frac{16}{5}\right)^{1/4} $$ 85. **Extreme points at these x-values.** --- 86. **Problem:** Find the domain and extreme points of $$y = 2x^3 + 9x^2 - 168x$$. 87. **Domain:** All real numbers, $$\mathbb{R}$$. 88. **Derivative:** $$ y' = 6x^2 + 18x - 168 $$ 89. **Set derivative to zero:** $$ 6x^2 + 18x - 168 = 0 $$ 90. **Divide by 6:** $$ x^2 + 3x - 28 = 0 $$ 91. **Calculate discriminant:** $$ \Delta = 3^2 - 4 \cdot 1 \cdot (-28) = 9 + 112 = 121 $$ 92. **Solve for x:** $$ x = \frac{-3 \pm 11}{2} $$ 93. **Solutions:** $$ x_1 = 4, x_2 = -7 $$ --- 94. **Problem:** Find the domain and extreme points of $$y = x^4 - 6x^2 - 12$$ on $$x \in [-1,2]$$. 95. **Domain:** Closed interval $$[-1,2]$$. 96. **Derivative:** $$ y' = 4x^3 - 12x $$ 97. **Set derivative to zero:** $$ 4x^3 - 12x = 0 $$ 98. **Factor:** $$ 4x(x^2 - 3) = 0 $$ 99. **Solve:** $$ x = 0, x = \pm \sqrt{3} $$ 100. **Only $$x=0$$ is in domain.** 101. **Evaluate y at critical point and endpoints:** $$ y(-1) = 1 - 6 - 12 = -17 $$ $$ y(0) = 0 - 0 - 12 = -12 $$ $$ y(2) = 16 - 24 - 12 = -20 $$ --- 102. **Problem:** Find the domain and extreme points of $$y = 3x^3 + 2x^2 - 27x - 18$$ on $$x \in [-3,3)$$. 103. **Domain:** Half-open interval $$[-3,3)$$. 104. **Derivative:** $$ y' = 9x^2 + 4x - 27 $$ 105. **Set derivative to zero:** $$ 9x^2 + 4x - 27 = 0 $$ 106. **Calculate discriminant:** $$ \Delta = 4^2 - 4 \cdot 9 \cdot (-27) = 16 + 972 = 988 $$ 107. **Solve for x:** $$ x = \frac{-4 \pm \sqrt{988}}{18} $$ 108. **Approximate roots:** $$ x_1 \approx 1.56, x_2 \approx -1.92 $$ 109. **Both roots are in domain.** --- 110. **Problem:** Find the domain and range of $$y = x^2 + 5x - 7$$. 111. **Domain:** All real numbers, $$\mathbb{R}$$. 112. **Find vertex (minimum or maximum):** $$ x = -\frac{b}{2a} = -\frac{5}{2} = -2.5 $$ 113. **Evaluate y at vertex:** $$ y(-2.5) = (-2.5)^2 + 5(-2.5) - 7 = 6.25 - 12.5 - 7 = -13.25 $$ 114. **Since $$a=1 > 0$$, parabola opens upward, so minimum is $$-13.25$$. Range is $$[-13.25, \infty)$$. --- 115. **Problem:** Find the domain and range of $$y = 3x^4 + 2x^3 + 12x^2 + 12x - 42$$. 116. **Domain:** All real numbers, $$\mathbb{R}$$. 117. **Since $$x^4$$ dominates and coefficient is positive, $$y \to \infty$$ as $$|x| \to \infty$$. 118. **Find critical points by derivative:** $$ y' = 12x^3 + 6x^2 + 24x + 12 $$ 119. **Set derivative to zero:** $$ 12x^3 + 6x^2 + 24x + 12 = 0 $$ 120. **Divide by 6:** $$ 2x^3 + x^2 + 4x + 2 = 0 $$ 121. **Solve cubic (numerical or factoring needed) to find critical points and evaluate y to find minimum/maximum for range.**