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Find a cubic function f(x) = ax^3 + bx^2 + cx + d that has a local maximum value of 3 at x = −3 and a local minimum value of 0 a

Ahat [919]

Answer:

f(x) = (3x³ + 9x² - 27x + 15)/32

Step-by-step explanation:

f'(x) = 3ax² + 2bx + c

Maximum or minimum: f'(x) = 0

Maximum: f'(-3) = 0 and f(-3) = 3

Minimum: f'(1) = 0 and f(1) = 0

f'(-3) = 0 leads 27a - 6b + c = 0 (1)

f'(1) = 0 leads 3a + 2b + c = 0 (2)

f(-3) = 3 leads -27a + 9b -3c + d = 3 (3)

f(1) = 0 leads a + b + c + d = 0 (4)

(1) - (2): 24a - 8b = 0, or 3a - b = 0, b = 3a

(3) - (4) -28a + 8b - 4c = 3 (5)

(1) * 4 + (5): 80a - 16b = 3,

use b = 3a, get 80a - 16*3a = 3, 32a = 3, a = 3/32, b = 3a = 9/32

From (2): c = -3a - 2b = -9/32 - 18/32 = -27/32

From (4): d = -a - b - c = -3/32 - 9/32 + 27/32 = 15/32

So

f(x) = (3x³ + 9x² - 27x + 15)/32

3 0

2 years ago

Points J and K are midpoints of the sides of triangle FGH. Triangle F G H is cut by lines segment J K. Point J is the midpoint o

anastassius [24]

Answer:

Value of y = 7

Step-by-step explanation:

By the midpoint theorem of the triangles,

"Segment joining midpoints of the two sides of a triangle is parallel to the third side and measure half the length of the third side."

Therefore, from the figure attached,

In ΔFGH,

J and K are the mid points of the two sides GH and FH.

By theorem, segment JK║GF and m(JK) = $\frac{1}{2}m(FG)$

$(2y+5)=\frac{1}{2}(5y+3)$

(5y + 3) = 2(2y + 5)

5y + 3 = 4y + 10

5y - 4y = 10 - 3

y = 7

7 0

3 years ago

Read 2 more answers

If y varies directly with x, the constant of variation is . The equation representing this relationship is . What is the value o

Julli [10]

Answer:

Y~x

Y = k × v

Y = 22

Step-by-step explanation:

Y~ x

Y = k × X

Since the question isn't complete,we are going to assign figures to complete this very question and move on to other things and you can as well solve questions that are similar to this ones.

Now let's assume that the question says that is 6 when x is 3,what is y when x is 11.

To solve this question above,we need to find the constant of variation which will then enable us to know what y is when x is 11

Y~ x

Y = k × V

When y is 6 and x is 3,k will be

6 = 3k

K= 2

Now when k is 2,x is 11,y Will then be

Y = kx

Y = 2 × 11

Y = 22.

With this procedure,one can find any value of x or y when they have the constant of variation.

If you have a figure assigned to x and y is to be found,the constant of variation will help to get y and vice versa

8 0

3 years ago

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the

devlian [24]

Answer:

The rocket will reach its maximum height after 6.13 seconds

Step-by-step explanation:

To find the time of the maximum height of the rocket differentiate the equation of the height with respect to the time and then equate the differentiation by 0 to find the time of the maximum height

∵ y is the height of the rocket after launch, x seconds

∵ y = -16x² + 196x + 126

- Differentiate y with respect to x

∴ y' = -16(2)x + 196

∴ y' = -32x + 196

- Equate y' by 0

∴ 0 = -32x + 196

- Add 32x to both sides

∴ 32x = 196

- Divide both sides by 32

∴ x = 6.125 seconds

- Round it to the nearest hundredth

∴ x = 6.13 seconds

∴ The rocket will reach its maximum height after 6.13 seconds

There is another solution you can find the vertex point (h , k) of the graph of the quadratic equation y = ax² + bx + c, where h = $-\frac{b}{2a}$ and k is the value of y at x = h and k is the maximum/minimum value