1. Problem: Find $(f+g)(1)$ where $f(x) = \frac{3x+1}{x+1}$ and $g(x) = \frac{x^{-5}}{x}$.\n\n2. Formula: $(f+g)(x) = f(x) + g(x)$. We evaluate each function at $x=1$ and then add.\n\n3. Calculate $f(1)$:\n$$f(1) = \frac{3(1)+1}{1+1} = \frac{3+1}{2} = \frac{4}{2} = 2.$$\n\n4. Calculate $g(1)$:\n$$g(1) = \frac{1^{-5}}{1} = \frac{1}{1} = 1.$$\n\n5. Add the results:\n$$(f+g)(1) = f(1) + g(1) = 2 + 1 = 3.$$\n\n---\n\n6. Problem: Find degree, leading coefficient, and constant term of $f(x) = \frac{12x^3 + 4x^2}{8} + x^7$.\n\n7. Simplify the fraction:\n$$\frac{12x^3 + 4x^2}{8} = \frac{12}{8}x^3 + \frac{4}{8}x^2 = \frac{3}{2}x^3 + \frac{1}{2}x^2.$$\n\n8. So, $f(x) = x^7 + \frac{3}{2}x^3 + \frac{1}{2}x^2$.\n\n9. Degree is the highest power of $x$, which is $7$.\n\n10. Leading coefficient is the coefficient of $x^7$, which is $1$.\n\n11. Constant term is the term without $x$, here none, so $0$.\n\n---\n\n12. Problem: Simplify $\frac{2^{x+2} + 3(2^{x+3})}{7(2^{x+1})}$.\n\n13. Factor $2^x$ from numerator:\n$$2^{x+2} + 3(2^{x+3}) = 2^x 2^2 + 3 \cdot 2^x 2^3 = 2^x (4 + 24) = 2^x \cdot 28.$$\n\n14. Denominator is $7 \cdot 2^{x+1} = 7 \cdot 2^x \cdot 2^1 = 7 \cdot 2^x \cdot 2 = 14 \cdot 2^x$.\n\n15. So the expression is:\n$$\frac{2^x \cdot 28}{14 \cdot 2^x}.$$\n\n16. Cancel $2^x$:\n$$\frac{\cancel{2^x} \cdot 28}{14 \cdot \cancel{2^x}} = \frac{28}{14} = 2.$$\n\n---\n\n17. Problem: Find quotient $q(x)$ and remainder $r(x)$ when $p(x) = 3x^4 - 2x^3 + 2x - 1$ is divided by $d(x) = x + 1$.\n\n18. Use polynomial long division or synthetic division with root $-1$.\n\n19. Synthetic division setup:\nCoefficients: 3, -2, 0 (for $x^2$), 2, -1\nDivide by $x+1$ means use $-1$.\n\n20. Perform synthetic division:\n- Bring down 3\n- Multiply: 3 * (-1) = -3; add to -2 = -5\n- Multiply: -5 * (-1) = 5; add to 0 = 5\n- Multiply: 5 * (-1) = -5; add to 2 = -3\n- Multiply: -3 * (-1) = 3; add to -1 = 2 (remainder)\n\n21. Quotient is $3x^3 - 5x^2 + 5x - 3$, remainder $2$.\n\n---\n\n22. Problem: Find $b$ if $x+1$ is a factor of $p(x) = x^3 + 3x^4 + x + b$.\n\n23. Rewrite $p(x)$ in standard order:\n$$p(x) = 3x^4 + x^3 + x + b.$$\n\n24. Since $x+1$ is a factor, $p(-1) = 0$.\n\n25. Calculate $p(-1)$:\n$$3(-1)^4 + (-1)^3 + (-1) + b = 3(1) -1 -1 + b = 3 - 1 - 1 + b = 1 + b.$$\n\n26. Set equal to zero:\n$$1 + b = 0 \Rightarrow b = -1.$$\n\n---\n\n27. Problem: Does $f(x) = x^4 - 5x^3 + x^2 + 12x - 5$ have a zero between 0 and 1?\n\n28. Use Intermediate Value Theorem: check $f(0)$ and $f(1)$.\n\n29. Calculate $f(0)$:\n$$0 - 0 + 0 + 0 - 5 = -5.$$\n\n30. Calculate $f(1)$:\n$$1 - 5 + 1 + 12 - 5 = 4.$$\n\n31. Since $f(0) = -5 < 0$ and $f(1) = 4 > 0$, function changes sign, so there is at least one zero between 0 and 1.\n\n---\n\n32. Problem: If $\log_a 18 = y$ and $\log_3 a = y$, find $\log_a 2$.\n\n33. From $\log_3 a = y$, rewrite as $a = 3^y$.\n\n34. From $\log_a 18 = y$, rewrite as $18 = a^y = (3^y)^y = 3^{y^2}$.\n\n35. Take $\log_3$ of both sides:\n$$\log_3 18 = y^2.$$\n\n36. But $\log_3 18 = \log_3 (2 \cdot 9) = \log_3 2 + \log_3 9 = \log_3 2 + 2.$\n\n37. So $\log_3 2 + 2 = y^2$.\n\n38. We want $\log_a 2 = \frac{\log_3 2}{\log_3 a} = \frac{\log_3 2}{y}$.\n\n39. From step 37, $y^2 = \log_3 2 + 2 \Rightarrow \log_3 2 = y^2 - 2$.\n\n40. Substitute:\n$$\log_a 2 = \frac{y^2 - 2}{y} = y - \frac{2}{y}.$$\n\n---\n\n41. Problem: Simplify $\ln(e^{\sqrt{3}})$.\n\n42. Use property $\ln(e^x) = x$.\n\n43. So $\ln(e^{\sqrt{3}}) = \sqrt{3}$.\n\n---\n\n44. Problem: Solve $8^{2x-1} = \left(\frac{1}{4}\right)^{3 - 2x}$.\n\n45. Rewrite bases as powers of 2:\n$$8 = 2^3, \quad \frac{1}{4} = 4^{-1} = (2^2)^{-1} = 2^{-2}.$$\n\n46. So equation becomes:\n$$2^{3(2x-1)} = 2^{-2(3 - 2x)}.$$\n\n47. Simplify exponents:\n$$2^{6x - 3} = 2^{-6 + 4x}.$$\n\n48. Since bases equal, set exponents equal:\n$$6x - 3 = -6 + 4x.$$\n\n49. Solve for $x$:\n$$6x - 4x = -6 + 3 \Rightarrow 2x = -3 \Rightarrow x = -\frac{3}{2}.$$\n\n---\n\n50. Problem: Find vertex of parabola $f(x) = x^2 - 6$.\n\n51. Standard form is $f(x) = x^2 + 0x - 6$. Vertex formula:\n$$x = -\frac{b}{2a} = -\frac{0}{2 \cdot 1} = 0.$$\n\n52. Find $f(0)$:\n$$0^2 - 6 = -6.$$\n\n53. Vertex is at $(0, -6)$.\n\n---\n\n54. Problem: Find maximum value of $f(x) = -x^2 + 4x + 6$.\n\n55. Since coefficient of $x^2$ is negative, parabola opens downward, so vertex is maximum.\n\n56. Vertex $x$ coordinate:\n$$x = -\frac{b}{2a} = -\frac{4}{2 \cdot (-1)} = 2.$$\n\n57. Calculate $f(2)$:\n$$-(2)^2 + 4(2) + 6 = -4 + 8 + 6 = 10.$$\n\n58. Maximum value is $10$.\n\n---\n\n59. Problem: Solve $\log_3 (x+2) - \log_{1/3} (x-2) = 1$.\n\n60. Use change of base for second term:\n$$\log_{1/3} (x-2) = \frac{\log_3 (x-2)}{\log_3 (1/3)} = \frac{\log_3 (x-2)}{-1} = -\log_3 (x-2).$$\n\n61. Substitute back:\n$$\log_3 (x+2) - (-\log_3 (x-2)) = 1 \Rightarrow \log_3 (x+2) + \log_3 (x-2) = 1.$$\n\n62. Use log addition rule:\n$$\log_3 ((x+2)(x-2)) = 1.$$\n\n63. Rewrite as exponential:\n$$(x+2)(x-2) = 3^1 = 3.$$\n\n64. Expand:\n$$x^2 - 4 = 3 \Rightarrow x^2 = 7 \Rightarrow x = \pm \sqrt{7}.$$\n\n65. Check domain: $x+2 > 0 \Rightarrow x > -2$, $x-2 > 0 \Rightarrow x > 2$. So only $x = \sqrt{7}$ is valid.\n\n---\n\n66. Problem: Use remainder theorem to find remainder when $f(x) = -x^4 - 2x^3 - x^2 - 4$ is divided by $x - 1$.\n\n67. Remainder theorem: remainder is $f(1)$.\n\n68. Calculate $f(1)$:\n$$-1 - 2 - 1 - 4 = -8.$$\n\n69. Remainder is $-8$.\n\n---\n\n70. Problem: Given $p(x) = ax^4 + x^2 - bx + 1$, divided by $x+1$ and $x-2$ with remainders 4 and 9 respectively, find $a$ and $b$.\n\n71. By remainder theorem:\n$$p(-1) = 4, \quad p(2) = 9.$$\n\n72. Calculate $p(-1)$:\n$$a(-1)^4 + (-1)^2 - b(-1) + 1 = a + 1 + b + 1 = a + b + 2 = 4.$$\n\n73. Calculate $p(2)$:\n$$a(2)^4 + (2)^2 - b(2) + 1 = 16a + 4 - 2b + 1 = 16a - 2b + 5 = 9.$$\n\n74. From first equation:\n$$a + b = 2.$$\n\n75. From second equation:\n$$16a - 2b = 4.$$\n\n76. Solve system:\nMultiply first by 2:\n$$2a + 2b = 4.$$\n\nAdd to second:\n$$16a - 2b = 4.$$\n\nAdd equations:\n$$(16a - 2b) + (2a + 2b) = 4 + 4 \Rightarrow 18a = 8 \Rightarrow a = \frac{8}{18} = \frac{4}{9}.$$\n\n77. Substitute $a$ back:\n$$\frac{4}{9} + b = 2 \Rightarrow b = 2 - \frac{4}{9} = \frac{18}{9} - \frac{4}{9} = \frac{14}{9}.$$\n\n---\n\n78. Problem: Find zeros of $f(x) = x^3 + 4x^2 + 2x - 1$.\n\n79. Try rational root theorem candidates: $\pm1$.\n\n80. Test $x=1$:\n$$1 + 4 + 2 - 1 = 6 \neq 0.$$\n\n81. Test $x=-1$:\n$$-1 + 4 - 2 - 1 = 0.$$\n\n82. So $x = -1$ is a root. Divide polynomial by $x+1$.\n\n83. Synthetic division with root $-1$:\nCoefficients: 1, 4, 2, -1\n- Bring down 1\n- Multiply: 1*(-1) = -1; add 4 + (-1) = 3\n- Multiply: 3*(-1) = -3; add 2 + (-3) = -1\n- Multiply: -1*(-1) = 1; add -1 + 1 = 0\n\n84. Quotient: $x^2 + 3x - 1$.\n\n85. Solve quadratic $x^2 + 3x - 1 = 0$:\n$$x = \frac{-3 \pm \sqrt{9 + 4}}{2} = \frac{-3 \pm \sqrt{13}}{2}.$$\n\n86. Zeros are $x = -1$, $x = \frac{-3 + \sqrt{13}}{2}$, $x = \frac{-3 - \sqrt{13}}{2}$.