1. Solve the inequalities algebraically. **a) Solve** $3x^3 - x^2 - 3x + 1 \leq 0$ 1. Write the inequality: $$3x^3 - x^2 - 3x + 1 \leq 0$$ 2. Factor the cubic polynomial if possible. Try rational roots using factors of 1: Test $x=1$: $$3(1)^3 - (1)^2 - 3(1) + 1 = 3 - 1 - 3 + 1 = 0$$ So $x=1$ is a root. 3. Use polynomial division or synthetic division to factor out $(x-1)$: Divide $3x^3 - x^2 - 3x + 1$ by $(x-1)$: $$\frac{3x^3 - x^2 - 3x + 1}{x-1} = 3x^2 + 2x - 1$$ 4. Factor the quadratic $3x^2 + 2x - 1$: Find factors of $3 \times (-1) = -3$ that sum to 2: 3 and -1. Rewrite: $$3x^2 + 3x - x - 1 = 3x(x+1) -1(x+1) = (3x - 1)(x + 1)$$ 5. So the full factorization is: $$3x^3 - x^2 - 3x + 1 = (x - 1)(3x - 1)(x + 1)$$ 6. The inequality becomes: $$(x - 1)(3x - 1)(x + 1) \leq 0$$ 7. Find critical points where expression equals zero: $$x = 1, \quad x = \frac{1}{3}, \quad x = -1$$ 8. Test intervals determined by these points: - $(-\infty, -1)$ - $(-1, \frac{1}{3})$ - $(\frac{1}{3}, 1)$ - $(1, \infty)$ 9. Choose test points and check sign: - At $x = -2$: $( -2 - 1)(3(-2) - 1)(-2 + 1) = (-3)(-7)(-1) = (-)(-)(-) = -$ negative - At $x = 0$: $(0 - 1)(3(0) - 1)(0 + 1) = (-1)(-1)(1) = (+)$ positive - At $x = 0.5$: $(0.5 - 1)(3(0.5) - 1)(0.5 + 1) = (-0.5)(0.5)(1.5) = (-)$ negative - At $x = 2$: $(2 - 1)(3(2) - 1)(2 + 1) = (1)(5)(3) = (+)$ positive 10. Inequality $\leq 0$ means where expression is negative or zero: Negative intervals: $(-\infty, -1)$ and $(\frac{1}{3}, 1)$ Include points where expression equals zero: $$x = -1, \frac{1}{3}, 1$$ 11. Final solution: $$(-\infty, -1] \cup \left[\frac{1}{3}, 1\right]$$ --- **b) Solve** $$\frac{1}{x + 2} \leq \frac{2}{3x + 1}$$ 1. Write inequality: $$\frac{1}{x + 2} \leq \frac{2}{3x + 1}$$ 2. Bring all terms to one side: $$\frac{1}{x + 2} - \frac{2}{3x + 1} \leq 0$$ 3. Find common denominator: $$(x + 2)(3x + 1)$$ 4. Combine fractions: $$\frac{(3x + 1) - 2(x + 2)}{(x + 2)(3x + 1)} \leq 0$$ 5. Simplify numerator: $$(3x + 1) - 2x - 4 = 3x + 1 - 2x - 4 = x - 3$$ 6. Inequality becomes: $$\frac{x - 3}{(x + 2)(3x + 1)} \leq 0$$ 7. Find critical points where numerator or denominator is zero: $$x - 3 = 0 \Rightarrow x = 3$$ $$x + 2 = 0 \Rightarrow x = -2$$ $$3x + 1 = 0 \Rightarrow x = -\frac{1}{3}$$ 8. These points divide the number line into intervals: $(-\infty, -2)$, $(-2, -\frac{1}{3})$, $(-\frac{1}{3}, 3)$, $(3, \infty)$ 9. Test each interval: - At $x = -3$: $$\frac{-3 - 3}{(-3 + 2)(3(-3) + 1)} = \frac{-6}{(-1)(-8)} = \frac{-6}{8} = -0.75 < 0$$ - At $x = -1$: $$\frac{-1 - 3}{(-1 + 2)(3(-1) + 1)} = \frac{-4}{(1)(-2)} = \frac{-4}{-2} = 2 > 0$$ - At $x = 0$: $$\frac{0 - 3}{(0 + 2)(3(0) + 1)} = \frac{-3}{(2)(1)} = -1.5 < 0$$ - At $x = 4$: $$\frac{4 - 3}{(4 + 2)(3(4) + 1)} = \frac{1}{(6)(13)} > 0$$ 10. Inequality $\leq 0$ means expression is negative or zero. 11. Check if points where denominator zero are included (they cause division by zero, so excluded): Exclude $x = -2$ and $x = -\frac{1}{3}$ Include $x = 3$ because numerator zero makes fraction zero. 12. Final solution: $$(-\infty, -2) \cup \left(-\frac{1}{3}, 3\right]$$ --- 2. Find $k$ such that $f(x) = \frac{3 - k}{2x + k}$ passes through $(5, -0.35)$ 1. Substitute $x=5$, $f(5) = -0.35$: $$-0.35 = \frac{3 - k}{2(5) + k} = \frac{3 - k}{10 + k}$$ 2. Multiply both sides by denominator: $$-0.35(10 + k) = 3 - k$$ 3. Distribute: $$-3.5 - 0.35k = 3 - k$$ 4. Rearrange terms: $$-0.35k + k = 3 + 3.5$$ $$0.65k = 6.5$$ 5. Solve for $k$: $$k = \frac{6.5}{0.65} = 10$$ --- **Final answers:** 1a) $$(-\infty, -1] \cup \left[\frac{1}{3}, 1\right]$$ 1b) $$(-\infty, -2) \cup \left(-\frac{1}{3}, 3\right]$$ 2) $$k = 10$$