Find a cubic function, in the form below,

Question

max value of 4 at x= -1 
min value of 89/108 at x=1/6

Answer 1

  

The cubic function f(x) = ax³ + bx² + cx + d

Maximum value 4 at x = - 1

f(- 1) = a(- 1)³ + b(- 1)² + c(- 1) + d

 - a + b - c + d = 4 ---> (1)

Minimum value 89/108 at x = 1/6

f(1/6) = a(1/6)³ + b(1/6)² + c(1/6) + d

89/108 = a/216 + b/36 + c/6 + d

89/108 = (a + 6b + 36c + 216d)/216

89(216)/108 = a + 6b + 36c + 216d

a + 6b + 36c + 216d = 178 ---> (2)

f(x) = ax³ + bx² + cx + d

f'(x) = 3ax² + 2bx + c

First derivative is zero for the function at maximum and minimum.

f'(x) = 3ax² + 2bx + c

f'(-1) = 0 and f'(1/6) = 0

3a(- 1)² + 2b(- 1) + c = 0

3a - 2b + c = 0 ---> (3)

3a(1/6)² + 2b(1/6) + c = 0

3a(1/36) + (b/3) + c = 0

a/12 + b/3 + c = 0

(a + 4b + 12c)/12 = 0

a + 4b + 12c = 0 ---> (4)

 

Now solve above four equations for four unknowns.

To eliminate d, eqn(1)(- 216) + eqn(2)⇒

(216a - 216b + 216c - 216d) + (a + 6b + 36c + 216d) = - 864 + 178

217a - 210b + 252c = - 686 ---> (5)

To eliminate a, eqn(5) + eqn(4)(- 217)⇒

(217a - 210b + 252c) + (- 217a - 868b - 2604c) = - 686 + 0

- 1078b - 2352c = - 686 ---> (7)

To eliminate a, eqn(3) + eqn(4)(- 3)⇒

(3a - 2b + c)+ (- 3a - 12b - 36c) = 0 + 0

- 14b - 35c = 0 ---> (8)

To eliminate b, eqn(7) + eqn(8)(-77)⇒

( - 1078b - 2352c) + (1078b + 2695c) = - 686 + 0

343c = - 686

⇒ c = - 2

Substitute c in - 14b - 35c = 0 .

 - 14b - 35(- 2) = 0

 - 14b = - 70

⇒ b = 5

Substitute b,c in a + 4b + 12c = 0.

a + 20 - 24 = 0

⇒ a = 4

Substitute a, b, c in - a + b - c + d = 4.

- 4 + 5 + 2 + d = 4

⇒ d = 1

Substitute the values of a, b, c, d in f(x) = ax³ + bx² + cx + d.

The cubic function is f(x) = 4x³ + 5x² - 2x + 1.