1. Problem 1a: Solve the inequality $3x^3 - x^2 - 3x + 1 \le 0$. 2. Strategy: Factor the polynomial to find its real roots and then use a sign chart to determine where the product is $\le 0$. 3. Factor by grouping and simplify. $$3x^3 - x^2 - 3x + 1 = (3x^3 - x^2) - (3x - 1)$$ $$= x^2(3x-1) -1(3x-1)$$ $$= (3x-1)(x^2-1)$$ $$= (3x-1)(x-1)(x+1)$$ 4. The critical points (zeros) are $x = -1$, $x = \tfrac{1}{3}$, and $x = 1$. 5. Sign analysis: test each interval determined by the critical points. For $x<-1$ pick $x=-2$. Then $(3x-1)<0$, $(x-1)<0$, $(x+1)<0$, so the product is negative and the inequality holds on that interval. For $-10$, so the product is positive and the inequality does not hold there. For $\tfrac{1}{3}0$, $(x-1)<0$, $(x+1)>0$, so the product is negative and the inequality holds there. For $x>1$ pick $x=2$. All factors are positive so the product is positive and the inequality does not hold. 6. Include zeros where equality holds. Therefore the solution of 1a is $$(-\infty,-1]\cup[\tfrac{1}{3},1]$$ 7. Problem 1b: Solve the inequality $$\frac{1}{x+2} \le \frac{2}{3x+1}$$ 8. Strategy: bring to one side, combine into a single rational expression, find critical points (zeros and undefined points), and use a sign chart. 9. Bring terms to one side and combine. $$\frac{1}{x+2} - \frac{2}{3x+1} \le 0$$ $$\frac{3x+1 - 2(x+2)}{(x+2)(3x+1)} \le 0$$ $$\frac{x-3}{(x+2)(3x+1)} \le 0$$ 10. Critical points: numerator zero at $x=3$, denominator zeros at $x=-2$ and $x=-\tfrac{1}{3}$ (these two values are excluded from the domain). 11. Sign analysis using intervals $(-\infty,-2)$, $(-2,-\tfrac{1}{3})$, $(-\tfrac{1}{3},3)$, and $(3,\infty)$. For $x<-2$ pick $x=-3$. Numerator $x-3$ is negative and denominator $(x+2)(3x+1)$ is positive, so the fraction is negative and the inequality holds. For $-23$ pick $x=4$. Numerator and denominator are positive, so the fraction is positive and the inequality does not hold. 12. Include $x=3$ because it makes the numerator zero and satisfies the $\le$ condition, but exclude $x=-2$ and $x=-\tfrac{1}{3}$ because they make the expression undefined. Therefore the solution of 1b is $$(-\infty,-2)\cup(-\tfrac{1}{3},3]$$ 13. Problem 2: Find $k$ for $f(x)=\frac{3-k}{2x+k}$ if the graph passes through $(5,-0.35)$. 14. Substitute $x=5$ and $f(5)=-0.35$. $$\frac{3-k}{2\cdot5 + k} = -0.35$$ 15. Convert the decimal to a fraction: $-0.35 = -\tfrac{7}{20}$. $$\frac{3-k}{10+k} = -\frac{7}{20}$$ 16. Multiply both sides by $10+k$ to remove the denominator on the left. $$3-k = -\frac{7}{20}(10+k)$$ 17. Multiply both sides by 20 to clear the fraction. $$20(3-k) = -7(10+k)$$ 18. Expand both sides. $$60 - 20k = -70 -7k$$ 19. Collect like terms to isolate $k$. $$60 +70 = -7k + 20k$$ $$130 = 13k$$ 20. Divide both sides by 13 and show the cancellation of the common factor. $$k = \frac{130}{13} = \frac{\cancel{13}\cdot10}{\cancel{13}} = 10$$ 21. Final answer for problem 2: $k = 10$.