Mathematics, 25.07.2019 11:50 ayeequeen6392

Find a cubic function f(x) = ax3 + bx2 + cx + d that has a local maximum value of 4 at x = −3 and a local minimum value of 0 at x = 1.find a cubic function f(x) = ax3 + bx2 + cx + d that has a local maximum value of 4 at x = −3 and a local minimum value of 0 at x = 1.

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The formula is f(x) = a x ^ 3 + b x ^ 2 + c x + d
f '(x) = 3ax^2 + 2bx + c.
f(- 3) = 3 ==> - 27a + 9b - 3c + d = 3
f '(- 3) = 0 (being a most extreme) ==> 27a - 6b + c = 0.
f(1) = 0 ==> a + b + c + d = 0
f '(1) = 0 (being a base) ==> 3a + 2b + c = 0.
-
Along these lines, we have the four conditions
- 27a + 9b - 3c + d = 3
a + b + c + d = 0
27a - 6b + c = 0
3a + 2b + c = 0
Subtracting the last two conditions yields 24a - 8b = 0 ==> b = 3a.
Along these lines, the last condition yields 3a + 6a + c = 0 ==> c = - 9a.
Consequently, we have from the initial two conditions:
- 27a + 9(3a) - 3(- 9a) + d = 3 ==> 27a + d = 3
a + 3a - 9a + d = 0 ==> d = 5a.
Along these lines, a = 3/32 and d = 15/32.
==> b = 9/32 and c = - 27/32.
That is, f(x) = (1/32)(3x^3 + 9x^2 - 27x + 15).

Answer from: apexdude2020

The required cubic polynomial is $f(x)=\dfrac{1}{8}x^3+\dfrac{3}{8}x^2-\dfrac{9}{8}x+\dfrac{5}{8}$ .

According to the question, for the cubic polynomial f(x)=ax^3+bx^2+cx+d , local maximum value of 4 at x = -3 and a local minimum value of 0 at x = 1 .

So,

$f(-3)=4\\-27a+9b-3c+d=4----(i)$

And,

$f(1)=0\\a+b+c+d=0----(ii)$

Differentiate the given equation with respect to as-

f'(x)=3ax^2+2bx+c

According to law of differentiabilty,

$f'(-3)=0\\27a-6b+c=0----(iii)$

And,

$f'(1)=0\\3a+2b+c=0----(iv)$

Subtract equation (iv) from (iii),

$27a-3a-6b-2b+c-c=0\\24a-8b=0\\b=3a----(v)$

Substitute b=3a in equation ,

$3a+2b+c=0\\3a+6a+c=0\\c=-9a----(vi)$

Substitue equation (iv) and (v) in equations (i) and (ii),

$-27a+9(3a)-3(-9a)+d=4\\27a+d=4$

And,

$a+b+c+d=0\\a+3a-9a+d=0\\-5a+d=0\\d=5a$

So,

$27a+d=4\\27a+5a=4\\a=\dfrac{4}{32}\\a=\dfrac{1}{8}$

And,

$d=5\times \dfrac{1}{8}\\d=\dfrac{5}{8}$

Also,

$c=-9\times \dfrac{1}{8}\\c=\dfrac{-9}{8}$

And,

$b=3\times \dfrac{1}{8}\\b=\dfrac{3}{8}$

Hence, the required cubic polynomial is $f(x)=\dfrac{1}{8}x^3+\dfrac{3}{8}x^2-\dfrac{9}{8}x+\dfrac{5}{8}$

Learn more about cubic polynomial here:

link

Answer from: Quest

here you hope this .

step-by-step explanation:

hypotenuse = √(15² + 8²)

= √(225 +64)

= √(289) = 17

sin(x) = cos(y) = 8/17

they are identical so long as you are doing the sine of one angle and the cosine of the only other angle that isn't the right angle.

Hurry need your use the image below to answer the following question. find the value of sin x° and

Answer from: Quest

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2 = -20

divide by -2 on each side

-2x^2 /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

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