For the function 𝑓(𝑥) = 𝑎𝑥^4+ 𝑏𝑥^3− 4𝑥^2+ 2𝑐𝑥 + 14, determine the values of a, b, and c so that 𝑓(−2) = 2, 𝑓′(−2) = 16 and 𝑓′′(−2) = −8

Answer 1

f(x) = ax⁴ + bx³ – 4x² + 2cx + 14

First derivative:

f’(x) = 4ax³ + 3bx² – 8x + 2c

Second derivative:

f’’(x) = 12ax² + 6bx – 8

Given:

f(-2) = a(-2)⁴ + b(-2)³ – 4(-2)² + 2c(-2) + 14 = 2

16a - 8b – 16 – 4c + 14 = 2

16a -8b -4c =4 (Eqn. 1)

f’(-2) = 4a(-2)³ + 3b(-2)² – 8(-2) + 2c = 16

-32a + 12b + 16 + 2c = 16

-32a + 12b +2c = 0 (Eqn. 2)

f’’(x) = 12a(-2)² + 6b(-2) – 8 = -8

48a – 12b – 8 = -8

48a – 12b = 0 (Eqn. 3)

Solving for b (Eqn. 3):

b = 4a

Substitute this value in Eqn. 2:

-32a + 12(4a) + 2c = 0

16a + 2c = 0

c = -8a

Substitute these values in Eqn. 1:

16a -8(4a) -4(-8a) =4

16a = 4

a=1/4

b = 4a = 1

c = -8a = -2

Original function is therefore:

f(x) = (¼)x⁴ + x³ – 4x² -4x + 14