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3 years ago

I need help with this! Quick ASAP?

Mathematics

1 answer:

sp2606 [1]3 years ago

6 0

Answer:

a) F

b) B, E, D

Step-by-step explanation:

a) The segment with the greatest gradient has the largest change in y-values per unit change in x-values

From the given option, the rate of change of the <em>y </em>to the<em> </em>x-values of B = the gradient = (4 units)/(2 units) = 2

The gradient of F = (-3units)/(1 unit) = -3

The gradient of A = 4/4 = 1

The gradient of C = -2/5

The gradient of D = 2/6 = 1/3

The gradient of E = 3/4

The segment with the greatest gradient is F

b) The steepest segment has the higher gradient

From their calculated we have;

The gradient of segment B = 2 therefore, B is steeper than E that has a gradient of 3/4, and E is steeper than D, as the gradient of D = 1/3

Therefore, we have;

B, E, D.

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I honestly need help with these

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

$-3 = a(-1) + \dfrac b{-1} \implies a + b = 3$

Compute the derivative.

$y = ax + \dfrac bx \implies \dfrac{dy}{dx} = a - \dfrac b{x^2}$

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

$-7 = a - \dfrac b{(-1)^2} \implies a-b = -7$

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.

$y = \dfrac{x^3}3 - \dfrac{5x^2}2 + 6x - 1 \implies \dfrac{dy}{dx} = x^2 - 5x + 6$

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.

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

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

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

$\dfrac{x^3}3 - 2x^2 - 8x + 5 = x + 5$

$\dfrac{x^3}3 - 2x^2 - 9x = 0$

$\dfrac x3 (x^2 - 6x - 27) = 0$

$\dfrac x3 (x - 9) (x + 3) = 0$

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

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.

$y = \dfrac{x^3}3 - 2x^2 - 8x + 5 \implies \dfrac{dy}{dx} = x^2 - 4x - 8$

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

$A: ~~~~ x=0 \implies \dfrac{dy}{dx} = 0^2-4\cdot0-8 \implies \boxed{\dfrac{dy}{dx} = -8}$

$B:~~~~ x=9 \implies \dfrac{dy}{dx} = 9^2-4\cdot9-8 \implies \boxed{\dfrac{dy}{dx} = 37}$

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

12. a. Compute the derivative.

$y = 4x^3 + 3x^2 - 6x - 1 \implies \boxed{\dfrac{dy}{dx} = 12x^2 + 6x - 6}$

12. b. By completing the square, we have

$12x^2 + 6x - 6 = 12 \left(x^2 + \dfrac x2\right) - 6 \\\\ ~~~~~~~~ = 12 \left(x^2 + \dfrac x2 + \dfrac1{4^2}\right) - 6 - \dfrac{12}{4^2} \\\\ ~~~~~~~~ = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4$

so that

$\dfrac{dy}{dx} = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4 \ge 0 \\\\ ~~~~ \implies 12 \left(x + \dfrac14\right)^2 \ge \dfrac{27}4 \\\\ ~~~~ \implies \left(x + \dfrac14\right)^2 \ge \dfrac{27}{48} = \dfrac9{16} \\\\ ~~~~ \implies \left|x + \dfrac14\right| \ge \sqrt{\dfrac9{16}} = \dfrac34 \\\\ ~~~~ \implies x+\dfrac14 \ge \dfrac34 \text{ or } -\left(x+\dfrac14\right) \ge \dfrac34 \\\\ ~~~~ \implies \boxed{x \ge \dfrac12 \text{ or } x \le -1}$

13. a. Compute the derivative.

$y = x^3 + x^2 - 16x - 16 \implies \boxed{\dfrac{dy}{dx} = 3x^2 - 2x - 16}$

13. b. Complete the square.

$3x^2 - 2x - 16 = 3 \left(x^2 - \dfrac{2x}3\right) - 16 \\\\ ~~~~~~~~ = 3 \left(x^2 - \dfrac{2x}3 + \dfrac1{3^2}\right) - 16 - \dfrac13 \\\\ ~~~~~~~~ = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3$

Then

$\dfrac{dy}{dx} = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3 \le 0 \\\\ ~~~~ \implies 3 \left(x - \dfrac13\right)^2 \le \dfrac{49}3 \\\\ ~~~~ \implies \left(x - \dfrac13\right)^2 \le \dfrac{49}9 \\\\ ~~~~ \implies \left|x - \dfrac13\right| \le \sqrt{\dfrac{49}9} = \dfrac73 \\\\ ~~~~ \implies x - \dfrac13 \le \dfrac73 \text{ or } -\left(x-\dfrac13\right) \le \dfrac73 \\\\ ~~~~ \implies \boxed{x \le 2 \text{ or } x \ge \dfrac83}$

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Answer:

This question needs to be written better

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