Question

I honestly need help with these

Answer 1

9. The curve passes through the point (-1, -3), which means

`-3 = a(-1) + \frac b{-1} \implies a + b = 3`

Compute the derivative.

`y = ax + \frac bx \implies (dy)/(dx) = a - \frac b{x^2}`

At the given point, the gradient is -7 so that

`-7 = a - \frac b{(-1)^2} \implies a-b = -7`

Eliminating
`b`, we find

`(a+b) + (a-b) = 3+(-7) \implies 2a = -4 \implies \boxed{a=-2}`

Solve for
`b`.

`a+b=3 \implies b=3-a \implies \boxed{b = 5}`

10. Compute the derivative.

`y = \frac{x^3}3 - \frac{5x^2}2 + 6x - 1 \implies (dy)/(dx) = x^2 - 5x + 6`

Solve for
`x` when the gradient is 2.

`x^2 - 5x + 6 = 2`

`x^2 - 5x + 4 = 0`

`(x - 1) (x - 4) = 0`

`\implies x=1 \text{ or } x=4`

Evaluate
`y` at each of these.

`\boxed{x=1} \implies y = \frac{1^3}3 - \frac{5\cdot1^2}2 + 6\cdot1 - 1 = \boxed{y = \frac{17}6}`

`\boxed{x = 4} \implies y = \frac{4^3}3 - \frac{5\cdot4^2}2 + 6\cdot4 - 1 \implies \boxed{y = \frac{13}3}`

11. a. Solve for
`x` where both curves meet.

`\frac{x^3}3 - 2x^2 - 8x + 5 = x + 5`

`\frac{x^3}3 - 2x^2 - 9x = 0`

`\frac x3 (x^2 - 6x - 27) = 0`

`\frac x3 (x - 9) (x + 3) = 0`

`\implies x = 0 \text{ or }x = 9 \text{ or } x = -3`

Evaluate
`y` at each of these.

`A:~~~~ \boxed{x=0} \implies y=0+5 \implies \boxed{y=5}`

`B:~~~~ \boxed{x=9} \implies y=9+5 \implies \boxed{y=14}`

`C:~~~~ \boxed{x=-3} \implies y=-3+5 \implies \boxed{y=2}`

11. b. Compute the derivative for the curve.

`y = \frac{x^3}3 - 2x^2 - 8x + 5 \implies (dy)/(dx) = x^2 - 4x - 8`

Evaluate the derivative at the
`x`-coordinates of A, B, and C.

`A: ~~~~ x=0 \implies (dy)/(dx) = 0^2-4\cdot0-8 \implies \boxed{(dy)/(dx) = -8}`

`B:~~~~ x=9 \implies (dy)/(dx) = 9^2-4\cdot9-8 \implies \boxed{(dy)/(dx) = 37}`

`C:~~~~ x=-3 \implies (dy)/(dx) = (-3)^2-4\cdot(-3)-8 \implies \boxed{(dy)/(dx) = 13}`

12. a. Compute the derivative.

`y = 4x^3 + 3x^2 - 6x - 1 \implies \boxed{(dy)/(dx) = 12x^2 + 6x - 6}`

12. b. By completing the square, we have

`12x^2 + 6x - 6 = 12 \left(x^2 + \frac x2\right) - 6 \\\\ ~~~~~~~~ = 12 \left(x^2 + \frac x2 + \frac1{4^2}\right) - 6 - (12)/(4^2) \\\\ ~~~~~~~~ = 12 \left(x + \frac14\right)^2 - \frac{27}4`

so that

`(dy)/(dx) = 12 \left(x + \frac14\right)^2 - \frac{27}4 \ge 0 \\\\ ~~~~ \implies 12 \left(x + \frac14\right)^2 \ge \frac{27}4 \\\\ ~~~~ \implies \left(x + \frac14\right)^2 \ge (27)/(48) = \frac9{16} \\\\ ~~~~ \implies \left|x + \frac14\right| \ge \sqrt{\frac9{16}} = \frac34 \\\\ ~~~~ \implies x+\frac14 \ge \frac34 \text{ or } -\left(x+\frac14\right) \ge \frac34 \\\\ ~~~~ \implies \boxed{x \ge \frac12 \text{ or } x \le -1}`

13. a. Compute the derivative.

`y = x^3 + x^2 - 16x - 16 \implies \boxed{(dy)/(dx) = 3x^2 - 2x - 16}`

13. b. Complete the square.

`3x^2 - 2x - 16 = 3 \left(x^2 - \frac{2x}3\right) - 16 \\\\ ~~~~~~~~ = 3 \left(x^2 - \frac{2x}3 + \frac1{3^2}\right) - 16 - \frac13 \\\\ ~~~~~~~~ = 3 \left(x - \frac13\right)^2 - \frac{49}3`

Then

`(dy)/(dx) = 3 \left(x - \frac13\right)^2 - \frac{49}3 \le 0 \\\\ ~~~~ \implies 3 \left(x - \frac13\right)^2 \le \frac{49}3 \\\\ ~~~~ \implies \left(x - \frac13\right)^2 \le \frac{49}9 \\\\ ~~~~ \implies \left|x - \frac13\right| \le \sqrt{\frac{49}9} = \frac73 \\\\ ~~~~ \implies x - \frac13 \le \frac73 \text{ or } -\left(x-\frac13\right) \le \frac73 \\\\ ~~~~ \implies \boxed{x \le 2 \text{ or } x \ge \frac83}`