9. The curve passes through the point (-1, -3), which means
[tex]-3 = a(-1) + \dfrac b{-1} \implies a + b = 3[/tex]
Compute the derivative.
[tex]y = ax + \dfrac bx \implies \dfrac{dy}{dx} = a - \dfrac b{x^2}[/tex]
At the given point, the gradient is -7 so that
[tex]-7 = a - \dfrac b{(-1)^2} \implies a-b = -7[/tex]
Eliminating [tex]b[/tex], we find
[tex](a+b) + (a-b) = 3+(-7) \implies 2a = -4 \implies \boxed{a=-2}[/tex]
Solve for [tex]b[/tex].
[tex]a+b=3 \implies b=3-a \implies \boxed{b = 5}[/tex]
10. Compute the derivative.
[tex]y = \dfrac{x^3}3 - \dfrac{5x^2}2 + 6x - 1 \implies \dfrac{dy}{dx} = x^2 - 5x + 6[/tex]
Solve for [tex]x[/tex] when the gradient is 2.
[tex]x^2 - 5x + 6 = 2[/tex]
[tex]x^2 - 5x + 4 = 0[/tex]
[tex](x - 1) (x - 4) = 0[/tex]
[tex]\implies x=1 \text{ or } x=4[/tex]
Evaluate [tex]y[/tex] at each of these.
[tex]\boxed{x=1} \implies y = \dfrac{1^3}3 - \dfrac{5\cdot1^2}2 + 6\cdot1 - 1 = \boxed{y = \dfrac{17}6}[/tex]
[tex]\boxed{x = 4} \implies y = \dfrac{4^3}3 - \dfrac{5\cdot4^2}2 + 6\cdot4 - 1 \implies \boxed{y = \dfrac{13}3}[/tex]
11. a. Solve for [tex]x[/tex] where both curves meet.
[tex]\dfrac{x^3}3 - 2x^2 - 8x + 5 = x + 5[/tex]
[tex]\dfrac{x^3}3 - 2x^2 - 9x = 0[/tex]
[tex]\dfrac x3 (x^2 - 6x - 27) = 0[/tex]
[tex]\dfrac x3 (x - 9) (x + 3) = 0[/tex]
[tex]\implies x = 0 \text{ or }x = 9 \text{ or } x = -3[/tex]
Evaluate [tex]y[/tex] at each of these.
[tex]A:~~~~ \boxed{x=0} \implies y=0+5 \implies \boxed{y=5}[/tex]
[tex]B:~~~~ \boxed{x=9} \implies y=9+5 \implies \boxed{y=14}[/tex]
[tex]C:~~~~ \boxed{x=-3} \implies y=-3+5 \implies \boxed{y=2}[/tex]
11. b. Compute the derivative for the curve.
[tex]y = \dfrac{x^3}3 - 2x^2 - 8x + 5 \implies \dfrac{dy}{dx} = x^2 - 4x - 8[/tex]
Evaluate the derivative at the [tex]x[/tex]-coordinates of A, B, and C.
[tex]A: ~~~~ x=0 \implies \dfrac{dy}{dx} = 0^2-4\cdot0-8 \implies \boxed{\dfrac{dy}{dx} = -8}[/tex]
[tex]B:~~~~ x=9 \implies \dfrac{dy}{dx} = 9^2-4\cdot9-8 \implies \boxed{\dfrac{dy}{dx} = 37}[/tex]
[tex]C:~~~~ x=-3 \implies \dfrac{dy}{dx} = (-3)^2-4\cdot(-3)-8 \implies \boxed{\dfrac{dy}{dx} = 13}[/tex]
12. a. Compute the derivative.
[tex]y = 4x^3 + 3x^2 - 6x - 1 \implies \boxed{\dfrac{dy}{dx} = 12x^2 + 6x - 6}[/tex]
12. b. By completing the square, we have
[tex]12x^2 + 6x - 6 = 12 \left(x^2 + \dfrac x2\right) - 6 \\\\ ~~~~~~~~ = 12 \left(x^2 + \dfrac x2 + \dfrac1{4^2}\right) - 6 - \dfrac{12}{4^2} \\\\ ~~~~~~~~ = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4[/tex]
so that
[tex]\dfrac{dy}{dx} = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4 \ge 0 \\\\ ~~~~ \implies 12 \left(x + \dfrac14\right)^2 \ge \dfrac{27}4 \\\\ ~~~~ \implies \left(x + \dfrac14\right)^2 \ge \dfrac{27}{48} = \dfrac9{16} \\\\ ~~~~ \implies \left|x + \dfrac14\right| \ge \sqrt{\dfrac9{16}} = \dfrac34 \\\\ ~~~~ \implies x+\dfrac14 \ge \dfrac34 \text{ or } -\left(x+\dfrac14\right) \ge \dfrac34 \\\\ ~~~~ \implies \boxed{x \ge \dfrac12 \text{ or } x \le -1}[/tex]
13. a. Compute the derivative.
[tex]y = x^3 + x^2 - 16x - 16 \implies \boxed{\dfrac{dy}{dx} = 3x^2 - 2x - 16}[/tex]
13. b. Complete the square.
[tex]3x^2 - 2x - 16 = 3 \left(x^2 - \dfrac{2x}3\right) - 16 \\\\ ~~~~~~~~ = 3 \left(x^2 - \dfrac{2x}3 + \dfrac1{3^2}\right) - 16 - \dfrac13 \\\\ ~~~~~~~~ = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3[/tex]
Then
[tex]\dfrac{dy}{dx} = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3 \le 0 \\\\ ~~~~ \implies 3 \left(x - \dfrac13\right)^2 \le \dfrac{49}3 \\\\ ~~~~ \implies \left(x - \dfrac13\right)^2 \le \dfrac{49}9 \\\\ ~~~~ \implies \left|x - \dfrac13\right| \le \sqrt{\dfrac{49}9} = \dfrac73 \\\\ ~~~~ \implies x - \dfrac13 \le \dfrac73 \text{ or } -\left(x-\dfrac13\right) \le \dfrac73 \\\\ ~~~~ \implies \boxed{x \le 2 \text{ or } x \ge \dfrac83}[/tex]