contestada

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.

Respuesta :

shinmin
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).
ernikhil

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

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

So,

[tex]f(-3)=4\\-27a+9b-3c+d=4----(i)[/tex]

And,

[tex]f(1)=0\\a+b+c+d=0----(ii)[/tex]

Differentiate the given equation with respect to [tex]x[/tex] as-

[tex]f'(x)=3ax^2+2bx+c[/tex]

According to law of differentiabilty,

[tex]f'(-3)=0\\27a-6b+c=0----(iii)[/tex]

And,

[tex]f'(1)=0\\3a+2b+c=0----(iv)[/tex]

Subtract equation (iv) from (iii),

[tex]27a-3a-6b-2b+c-c=0\\24a-8b=0\\b=3a----(v)[/tex]

Substitute [tex]b=3a[/tex] in equation [tex]iv[/tex],

[tex]3a+2b+c=0\\3a+6a+c=0\\c=-9a----(vi)[/tex]

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

[tex]-27a+9(3a)-3(-9a)+d=4\\27a+d=4[/tex]

And,

[tex]a+b+c+d=0\\a+3a-9a+d=0\\-5a+d=0\\d=5a[/tex]

So,

[tex]27a+d=4\\27a+5a=4\\a=\dfrac{4}{32}\\a=\dfrac{1}{8}[/tex]

And,

[tex]d=5\times \dfrac{1}{8}\\d=\dfrac{5}{8}[/tex]

Also,

[tex]c=-9\times \dfrac{1}{8}\\c=\dfrac{-9}{8}[/tex]

And,

[tex]b=3\times \dfrac{1}{8}\\b=\dfrac{3}{8}[/tex]

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

Learn more about cubic polynomial here:

https://brainly.com/question/23638618?referrer=searchResults

Otras preguntas