1. **Determine the average rate of change of the function** $F(x) = -\frac{3}{x^2} - 2x$ over the interval $[-1, \frac{1}{3}]$. The average rate of change of a function $F(x)$ over an interval $[a,b]$ is given by the formula: $$\text{Average rate of change} = \frac{F(b) - F(a)}{b - a}$$ **Step 1:** Identify $a = -1$ and $b = \frac{1}{3}$. **Step 2:** Calculate $F(a) = F(-1)$: $$F(-1) = -\frac{3}{(-1)^2} - 2(-1) = -\frac{3}{1} + 2 = -3 + 2 = -1$$ **Step 3:** Calculate $F(b) = F\left(\frac{1}{3}\right)$: $$F\left(\frac{1}{3}\right) = -\frac{3}{\left(\frac{1}{3}\right)^2} - 2\left(\frac{1}{3}\right) = -\frac{3}{\frac{1}{9}} - \frac{2}{3} = -3 \times 9 - \frac{2}{3} = -27 - \frac{2}{3} = -\frac{81}{3} - \frac{2}{3} = -\frac{83}{3}$$ **Step 4:** Calculate the average rate of change: $$\frac{F\left(\frac{1}{3}\right) - F(-1)}{\frac{1}{3} - (-1)} = \frac{-\frac{83}{3} - (-1)}{\frac{1}{3} + 1} = \frac{-\frac{83}{3} + 1}{\frac{4}{3}} = \frac{-\frac{83}{3} + \frac{3}{3}}{\frac{4}{3}} = \frac{-\frac{80}{3}}{\frac{4}{3}}$$ **Step 5:** Simplify the fraction: $$= -\frac{80}{3} \times \frac{3}{4} = -\frac{80}{\cancel{3}} \times \frac{\cancel{3}}{4} = -\frac{80}{4} = -20$$ **Final answer:** The average rate of change is $-20$. 2. **Estimate the instantaneous rate of change of** $F(x) = \frac{2x - 1}{x + 2}$ at $x = -3$, to 2 decimal places. The instantaneous rate of change at $x = a$ is the derivative $F'(a)$. **Step 1:** Use the quotient rule for derivatives: If $F(x) = \frac{u(x)}{v(x)}$, then $$F'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$$ Here, $u(x) = 2x - 1$, $v(x) = x + 2$. **Step 2:** Compute derivatives: $$u'(x) = 2, \quad v'(x) = 1$$ **Step 3:** Substitute into the quotient rule: $$F'(x) = \frac{2(x + 2) - (2x - 1)(1)}{(x + 2)^2} = \frac{2x + 4 - 2x + 1}{(x + 2)^2} = \frac{5}{(x + 2)^2}$$ **Step 4:** Evaluate at $x = -3$: $$F'(-3) = \frac{5}{(-3 + 2)^2} = \frac{5}{(-1)^2} = \frac{5}{1} = 5$$ **Final answer:** The instantaneous rate of change at $x = -3$ is $5.00$. 3. **Find the value of** $k$ when $2x^4 - kx^2 + kx + 2$ is divided by $x + 2$ and the remainder is 10. By the Remainder Theorem, the remainder when a polynomial $P(x)$ is divided by $x - a$ is $P(a)$. Here, divisor is $x + 2$, so $a = -2$. **Step 1:** Evaluate $P(-2)$: $$P(-2) = 2(-2)^4 - k(-2)^2 + k(-2) + 2 = 2(16) - k(4) - 2k + 2 = 32 - 4k - 2k + 2 = 34 - 6k$$ **Step 2:** Set equal to remainder 10: $$34 - 6k = 10$$ **Step 3:** Solve for $k$: $$-6k = 10 - 34 = -24$$ $$k = \frac{-24}{-6} = 4$$ **Final answer:** $k = 4$. 4. **Find values of** $a$ and $b$ when $ax^3 - 2x^2 + bx - 4$ is divided by $(x + 2)$ with remainder 4, and divided by $(x - 1)$ with remainder -11. By the Remainder Theorem: $$P(-2) = 4, \quad P(1) = -11$$ **Step 1:** Write $P(x) = ax^3 - 2x^2 + bx - 4$. Evaluate at $x = -2$: $$P(-2) = a(-2)^3 - 2(-2)^2 + b(-2) - 4 = -8a - 8 - 2b - 4 = -8a - 2b - 12$$ Set equal to 4: $$-8a - 2b - 12 = 4 \implies -8a - 2b = 16$$ Divide both sides by -2: $$\cancel{-8a} \div \cancel{-2} = 4a, \quad \cancel{-2b} \div \cancel{-2} = b, \quad 16 \div -2 = -8$$ So: $$4a + b = -8 \quad (1)$$ **Step 2:** Evaluate at $x = 1$: $$P(1) = a(1)^3 - 2(1)^2 + b(1) - 4 = a - 2 + b - 4 = a + b - 6$$ Set equal to -11: $$a + b - 6 = -11 \implies a + b = -5 \quad (2)$$ **Step 3:** Solve the system: From (2): $b = -5 - a$ Substitute into (1): $$4a + (-5 - a) = -8 \implies 4a - 5 - a = -8 \implies 3a - 5 = -8$$ $$3a = -3 \implies a = -1$$ Then: $$b = -5 - (-1) = -5 + 1 = -4$$ **Final answer:** $a = -1$, $b = -4$.