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:** $$y' = \frac{d}{dx}(x^3 + 6x^2 + 12x - 1) = 3x^2 + 12x + 12$$ 4. **Set derivative equal to zero to find critical points:** $$3x^2 + 12x + 12 = 0$$ 5. **Divide entire equation by 3:** $$\cancel{3}x^2 + \cancel{3} \cdot 4x + \cancel{3} \cdot 4 = 0 \Rightarrow x^2 + 4x + 4 = 0$$ 6. **Factor the quadratic:** $$x^2 + 4x + 4 = (x + 2)^2 = 0$$ 7. **Solve for x:** $$x + 2 = 0 \Rightarrow x = -2$$ 8. **Find the y-value at critical point:** $$y(-2) = (-2)^3 + 6(-2)^2 + 12(-2) - 1 = -8 + 24 - 24 - 1 = -9$$ 9. **Critical value:** The critical point is at $$(-2, -9)$$. --- 1. **Problem:** Find the domain and critical values of the function $$y = x^5 + 5x^4 - 5x^3 - 35x^2 - 40x - 1$$. 2. **Domain:** Polynomial, so domain is $$\mathbb{R}$$. 3. **Derivative:** $$y' = 5x^4 + 20x^3 - 15x^2 - 70x - 40$$ 4. **Set derivative to zero:** $$5x^4 + 20x^3 - 15x^2 - 70x - 40 = 0$$ 5. **Divide by 5:** $$\cancel{5}x^4 + \cancel{5} \cdot 4x^3 - \cancel{5} \cdot 3x^2 - \cancel{5} \cdot 14x - \cancel{5} \cdot 8 = 0 \Rightarrow x^4 + 4x^3 - 3x^2 - 14x - 8 = 0$$ 6. **Solve quartic equation (factoring or numerical methods):** Try rational roots: test $x = -2$: $$(-2)^4 + 4(-2)^3 - 3(-2)^2 - 14(-2) - 8 = 16 - 32 - 12 + 28 - 8 = -8$$ (not zero) Try $x = 1$: $$1 + 4 - 3 - 14 - 8 = -20$$ (not zero) Try $x = 2$: $$16 + 32 - 12 - 28 - 8 = 0$$ (root found) 7. **Divide polynomial by $(x - 2)$:** Use synthetic division: Coefficients: 1, 4, -3, -14, -8 Bring down 1, multiply by 2, add: 1 1*2=2 +4=6 6*2=12 -3=9 9*2=18 -14=4 4*2=8 -8=0 Quotient: $$x^3 + 6x^2 + 9x + 4$$ 8. **Solve cubic $$x^3 + 6x^2 + 9x + 4 = 0$$:** Try rational roots: $x = -1$: $$-1 + 6 - 9 + 4 = 0$$ (root found) 9. **Divide cubic by $(x + 1)$:** Coefficients: 1, 6, 9, 4 Bring down 1, multiply by -1, add: 1 1*(-1)=-1 +6=5 5*(-1)=-5 +9=4 4*(-1)=-4 +4=0 Quotient: $$x^2 + 5x + 4$$ 10. **Factor quadratic:** $$x^2 + 5x + 4 = (x + 4)(x + 1)$$ 11. **Roots:** $$x = 2, -1, -4, -1$$ (note $-1$ repeated) 12. **Critical points:** Evaluate $$y$$ at each critical $$x$$: - $$y(2) = 32 + 80 - 40 - 140 - 80 - 1 = -149$$ - $$y(-1) = -1 + 5 + 5 - 35 + 40 - 1 = 13$$ - $$y(-4) = -1024 + 1280 + 320 - 560 + 160 - 1 = 175$$ --- 1. **Problem:** Find the domain and critical values of $$y = 4x^3 - 21x^2 + 48x - 7$$. 2. **Domain:** $$\mathbb{R}$$. 3. **Derivative:** $$y' = 12x^2 - 42x + 48$$ 4. **Set derivative to zero:** $$12x^2 - 42x + 48 = 0$$ 5. **Divide by 6:** $$2x^2 - 7x + 8 = 0$$ 6. **Calculate discriminant:** $$\Delta = (-7)^2 - 4 \cdot 2 \cdot 8 = 49 - 64 = -15 < 0$$ 7. **No real roots, so no critical points.** --- 1. **Problem:** Find the domain and critical values of $$y = x^4 - 32x^2 + 12$$. 2. **Domain:** $$\mathbb{R}$$. 3. **Derivative:** $$y' = 4x^3 - 64x$$ 4. **Set derivative to zero:** $$4x^3 - 64x = 0$$ 5. **Factor out $$4x$$:** $$4x(x^2 - 16) = 0$$ 6. **Set each factor to zero:** $$4x = 0 \Rightarrow x = 0$$ $$x^2 - 16 = 0 \Rightarrow x = \pm 4$$ 7. **Critical points:** Evaluate $$y$$ at $$x = -4, 0, 4$$: - $$y(-4) = 256 - 512 + 12 = -244$$ - $$y(0) = 0 - 0 + 12 = 12$$ - $$y(4) = 256 - 512 + 12 = -244$$ --- 1. **Problem:** Find the domain and critical values of $$y = 3x^5 - 40x^3 + 180x - 71$$. 2. **Domain:** $$\mathbb{R}$$. 3. **Derivative:** $$y' = 15x^4 - 120x^2 + 180$$ 4. **Set derivative to zero:** $$15x^4 - 120x^2 + 180 = 0$$ 5. **Divide by 15:** $$x^4 - 8x^2 + 12 = 0$$ 6. **Substitute $$u = x^2$$:** $$u^2 - 8u + 12 = 0$$ 7. **Solve quadratic:** $$u = \frac{8 \pm \sqrt{64 - 48}}{2} = \frac{8 \pm \sqrt{16}}{2} = \frac{8 \pm 4}{2}$$ 8. **Roots for $$u$$:** $$u_1 = 6, u_2 = 2$$ 9. **Back to $$x$$:** $$x^2 = 6 \Rightarrow x = \pm \sqrt{6}$$ $$x^2 = 2 \Rightarrow x = \pm \sqrt{2}$$ 10. **Critical points:** Evaluate $$y$$ at these points: - $$y(\sqrt{6}) = 3(6\sqrt{6}) - 40(6\sqrt{6}) + 180\sqrt{6} - 71$$ (approximate) - $$y(-\sqrt{6})$$ same as above - $$y(\sqrt{2})$$ and $$y(-\sqrt{2})$$ similarly --- 1. **Problem:** Find the domain and critical values of $$y = x^3 + x^2 - 5x - 3$$ on $$x \in [-3, 1]$$. 2. **Domain:** $$[-3, 1]$$. 3. **Derivative:** $$y' = 3x^2 + 2x - 5$$ 4. **Set derivative to zero:** $$3x^2 + 2x - 5 = 0$$ 5. **Calculate discriminant:** $$\Delta = 2^2 - 4 \cdot 3 \cdot (-5) = 4 + 60 = 64$$ 6. **Roots:** $$x = \frac{-2 \pm 8}{6}$$ 7. **Calculate roots:** $$x_1 = \frac{-2 + 8}{6} = 1$$ $$x_2 = \frac{-2 - 8}{6} = -\frac{10}{6} = -\frac{5}{3} \approx -1.6667$$ 8. **Check if roots are in domain:** Both $$1$$ and $$-\frac{5}{3}$$ are in $$[-3, 1]$$. 9. **Evaluate $$y$$ at critical points and endpoints:** - $$y(1) = 1 + 1 - 5 - 3 = -6$$ - $$y(-\frac{5}{3}) = (-\frac{5}{3})^3 + (-\frac{5}{3})^2 - 5(-\frac{5}{3}) - 3$$ Calculate stepwise: $$= -\frac{125}{27} + \frac{25}{9} + \frac{25}{3} - 3 = -4.6296 + 2.7778 + 8.3333 - 3 = 3.4815$$ - $$y(-3) = -27 + 9 + 15 - 3 = -6$$ --- 1. **Problem:** Find the domain and critical values of $$y = 2x^3 + x^2 - 4x + 8$$ on $$x \in [0, 7]$$. 2. **Domain:** $$[0, 7]$$. 3. **Derivative:** $$y' = 6x^2 + 2x - 4$$ 4. **Set derivative to zero:** $$6x^2 + 2x - 4 = 0$$ 5. **Divide by 2:** $$3x^2 + x - 2 = 0$$ 6. **Calculate discriminant:** $$\Delta = 1^2 - 4 \cdot 3 \cdot (-2) = 1 + 24 = 25$$ 7. **Roots:** $$x = \frac{-1 \pm 5}{6}$$ 8. **Calculate roots:** $$x_1 = \frac{4}{6} = \frac{2}{3} \approx 0.6667$$ $$x_2 = \frac{-6}{6} = -1$$ 9. **Check domain:** Only $$x = \frac{2}{3}$$ is in $$[0,7]$$. 10. **Evaluate $$y$$ at critical point and endpoints:** - $$y(\frac{2}{3}) = 2(\frac{8}{27}) + (\frac{4}{9}) - 4(\frac{2}{3}) + 8 = \frac{16}{27} + \frac{4}{9} - \frac{8}{3} + 8 = 0.5926 + 0.4444 - 2.6667 + 8 = 6.3703$$ - $$y(0) = 8$$ - $$y(7) = 2(343) + 49 - 28 + 8 = 686 + 49 - 28 + 8 = 715$$ --- 1. **Problem:** Find the domain and critical values of $$y = -3x^4 + 24x^2 + 5$$ on $$x \in [-3, 5]$$. 2. **Domain:** $$[-3, 5]$$. 3. **Derivative:** $$y' = -12x^3 + 48x$$ 4. **Set derivative to zero:** $$-12x^3 + 48x = 0$$ 5. **Factor:** $$-12x(x^2 - 4) = 0$$ 6. **Solve:** $$x = 0, x^2 - 4 = 0 \Rightarrow x = \pm 2$$ 7. **Check domain:** All $$x = -2, 0, 2$$ are in $$[-3, 5]$$. 8. **Evaluate $$y$$ at critical points and endpoints:** - $$y(-3) = -3(81) + 24(9) + 5 = -243 + 216 + 5 = -22$$ - $$y(-2) = -3(16) + 24(4) + 5 = -48 + 96 + 5 = 53$$ - $$y(0) = 5$$ - $$y(2) = 53$$ - $$y(5) = -3(625) + 24(25) + 5 = -1875 + 600 + 5 = -1270$$ --- 1. **Problem:** Find the domain and critical values of $$y = 2x^5 - 35x^3 + 135x - 29$$ on $$x \in [-2, 6]$$. 2. **Domain:** $$[-2, 6]$$. 3. **Derivative:** $$y' = 10x^4 - 105x^2 + 135$$ 4. **Set derivative to zero:** $$10x^4 - 105x^2 + 135 = 0$$ 5. **Divide by 5:** $$2x^4 - 21x^2 + 27 = 0$$ 6. **Substitute $$u = x^2$$:** $$2u^2 - 21u + 27 = 0$$ 7. **Calculate discriminant:** $$\Delta = (-21)^2 - 4 \cdot 2 \cdot 27 = 441 - 216 = 225$$ 8. **Roots for $$u$$:** $$u = \frac{21 \pm 15}{4}$$ 9. **Calculate roots:** $$u_1 = \frac{36}{4} = 9$$ $$u_2 = \frac{6}{4} = 1.5$$ 10. **Back to $$x$$:** $$x^2 = 9 \Rightarrow x = \pm 3$$ $$x^2 = 1.5 \Rightarrow x = \pm \sqrt{1.5} \approx \pm 1.2247$$ 11. **Check domain:** Only $$x = -2, -1.2247, 1.2247, 3, 6$$ are in $$[-2, 6]$$ (exclude $$-3$$). 12. **Evaluate $$y$$ at critical points and endpoints:** - $$y(-2) = 2(-32) - 35(-8) + 135(-2) - 29 = -64 + 280 - 270 - 29 = -83$$ - $$y(-1.2247)$$ approx - $$y(1.2247)$$ approx - $$y(3) = 2(243) - 35(27) + 135(3) - 29 = 486 - 945 + 405 - 29 = -83$$ - $$y(6) = 2(7776) - 35(216) + 135(6) - 29 = 15552 - 7560 + 810 - 29 = 10773$$ --- 1. **Problem:** Find the domain and extreme points of $$y = x^5 - 16x + 12$$. 2. **Domain:** $$\mathbb{R}$$. 3. **Derivative:** $$y' = 5x^4 - 16$$ 4. **Set derivative to zero:** $$5x^4 - 16 = 0 \Rightarrow 5x^4 = 16 \Rightarrow x^4 = \frac{16}{5}$$ 5. **Solve for $$x$$:** $$x = \pm \sqrt[4]{\frac{16}{5}} = \pm \left(\frac{16}{5}\right)^{1/4}$$ 6. **Approximate:** $$\left(\frac{16}{5}\right)^{1/4} \approx 1.34$$ 7. **Extreme points:** Evaluate $$y$$ at $$x = \pm 1.34$$: - $$y(1.34) = (1.34)^5 - 16(1.34) + 12 \approx 4.3 - 21.44 + 12 = -5.14$$ - $$y(-1.34) = -4.3 + 21.44 + 12 = 29.14$$ --- 1. **Problem:** Find the domain and extreme points of $$y = 2x^3 + 9x^2 - 168x$$. 2. **Domain:** $$\mathbb{R}$$. 3. **Derivative:** $$y' = 6x^2 + 18x - 168$$ 4. **Set derivative to zero:** $$6x^2 + 18x - 168 = 0$$ 5. **Divide by 6:** $$x^2 + 3x - 28 = 0$$ 6. **Calculate discriminant:** $$\Delta = 3^2 - 4 \cdot 1 \cdot (-28) = 9 + 112 = 121$$ 7. **Roots:** $$x = \frac{-3 \pm 11}{2}$$ 8. **Calculate roots:** $$x_1 = \frac{8}{2} = 4$$ $$x_2 = \frac{-14}{2} = -7$$ 9. **Extreme points:** Evaluate $$y$$ at $$x = 4, -7$$: - $$y(4) = 2(64) + 9(16) - 168(4) = 128 + 144 - 672 = -400$$ - $$y(-7) = 2(-343) + 9(49) - 168(-7) = -686 + 441 + 1176 = 931$$ --- 1. **Problem:** Find the domain and extreme points of $$y = x^4 - 6x^2 - 12$$ on $$x \in [-1, 2]$$. 2. **Domain:** $$[-1, 2]$$. 3. **Derivative:** $$y' = 4x^3 - 12x$$ 4. **Set derivative to zero:** $$4x^3 - 12x = 0$$ 5. **Factor:** $$4x(x^2 - 3) = 0$$ 6. **Solve:** $$x = 0, x = \pm \sqrt{3}$$ 7. **Check domain:** Only $$x = 0$$ is in $$[-1, 2]$$. 8. **Evaluate $$y$$ at critical point and endpoints:** - $$y(0) = -12$$ - $$y(-1) = 1 - 6 - 12 = -17$$ - $$y(2) = 16 - 24 - 12 = -20$$ --- 1. **Problem:** Find the domain and extreme points of $$y = 3x^3 + 2x^2 - 27x - 18$$ on $$x \in [-3, 3)$$. 2. **Domain:** $$[-3, 3)$$. 3. **Derivative:** $$y' = 9x^2 + 4x - 27$$ 4. **Set derivative to zero:** $$9x^2 + 4x - 27 = 0$$ 5. **Calculate discriminant:** $$\Delta = 4^2 - 4 \cdot 9 \cdot (-27) = 16 + 972 = 988$$ 6. **Roots:** $$x = \frac{-4 \pm \sqrt{988}}{18}$$ 7. **Approximate roots:** $$\sqrt{988} \approx 31.44$$ $$x_1 = \frac{-4 + 31.44}{18} = 1.52$$ $$x_2 = \frac{-4 - 31.44}{18} = -1.97$$ 8. **Check domain:** Both roots are in $$[-3, 3)$$. 9. **Evaluate $$y$$ at critical points and endpoint $$-3$$:** - $$y(1.52) \approx 3(3.51) + 2(2.31) - 27(1.52) - 18 = 10.53 + 4.62 - 41.04 - 18 = -43.89$$ - $$y(-1.97) \approx 3(-7.64) + 2(3.88) + 53.19 - 18 = -22.92 + 7.76 + 53.19 - 18 = 20.03$$ - $$y(-3) = -81 + 18 + 81 - 18 = 0$$