Question

PLEASE I'm DESPERATE!. Find a, b, c, and d such that the cubic . f(x) = ax3 + bx2 + cx + d. satisfies the given conditions.. Relative maximum: (3, 11). Relative minimum: (5, 9). Inflection point: (4, 10)

Answer 1

We are given with the equation f(x) = ax3 + bx2 + cx + d

Substituting, (3,11) 
11= 27a + 9b + 3c + d
@(5, 9) 
9 = 125 a + 25 b + 5c + d
@(4, 10)
10 = 64 a + 16 b + 4c + d

@inflection point, second derivative is equal to zero
f'(x) = 3ax2 + 2bx + c 
f''(x) = 6ax + 2b = 0

when x is 4, 24 a + 2b = 0 or 12a + b = 0. 

There are 4 equations, 4 unknowns: answer is 
0.5 x^3 - 6x^2 + 22.5 - 24 = 0

Answer 2

Answer:

f(x) = (1/2)x^3 - 6x^2 + (45/2)x - 16

Step-by-step explanation:

Given: f(x) = ax^3 + bx^2 + cx + d

Both maximum and minimum satisfied the function, then:

f(3) = 11 => 11 = a(3)^3 + b(3)^2 + c(3) + d

11 = 27a + 9b + 3c + d    (eq. 1)

f(5) = 9 => 9 = a(5)^3 + b(5)^2 + c(5) + d

9 = 125a + 25b +  5c +d      (eq. 2)

The first derivative of the function is:

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

And the second derivative is:

f''(x) = 6ax + 2b

In the relative minimum the first derivative is equal to zero, then:

f'(3) = 0 => 0 = 3a(3)^2 + 2b(3) + c

0 = 27a + 6b + c     (eq. 3)

In the inflection point the second derivative is equal to zero, then:

f''(4) = 0 => 0 = 6a4 + 2b  <=> b = -12a

Replacing b = -12a in eq. 3:

0 = 27a + 6(-12)a + c <=> c = 45a

Replacing b = -12a and c = 45a in eq. 1:

11 = 27a + 9(-12)a + 3(45)a + d

11 - 54a = d     (eq. 4)

Replacing b = -12a and c = 45a in eq. 2:

9 = 125a + 25(-12)a +  5(45)a +d

9 - 50a = d      (eq. 5)

Equating eq. 4 and eq. 5:

11 - 54a = 9 - 50a

2 = 4a

a = 1/2

So, b = -6, c = 45/2 and d = -16