Consider the function f(x) whose second derivative is f''(x) = 3x + 8 sin(x). If f(0) = 3 and f'(0) = 3 , what is f(4)?

Answer 1

f''(x) = 3x + 8 sin(x)

∫f''(x)dx = ∫[3x+8sin(x)]dx

f'(x) = (3/2)x²-8cos(x)+C

f'(0) = (3/2)(0)²-8cos(0)+C

3 = -8 + C

11 = C

f'(x) = (3/2)x²-8cos(x)+11

∫f'(x)dx = ∫[(3/2)x²-8cos(x)+11]dx

f(x) = (1/2)x³-8sin(x)+11x+C

f(0) = (1/2)(0)³-8sin(0)+11(0)+C

3 = C

f(x) = (1/2)x³-8sin(x)+11x+3

f(4) = (1/2)(4)³-8sin(4)+11(4)+3

f(4) = (64/2)-8sin(4)+44+3

f(4) = 32-8sin(4)+47

f(4) = 79-8sin(4)

f(4) ≈ 85.05442

Hope this helped! The key to solving this problem is integrating each time from f''(x) to f(x) while getting your values for C using the given initial conditions.

Answer 2

f"(x) = 3x+8sinx

f'(x) = 3x^2/2 -8cosx + C

f'(0) = 3 = 0- 8 +C

C = 3+8 =11

f'(x) = 3x^2/2 - 8cosx + 11

f(x) = 3x^3/2(3) - 8sinx + 11x+ C2

f(0) = 3 = C2

f(x) = .5x^3 - 8sinx +11x + 3

f(4) = .5(64) -8sin4 +11(4) +3

= 32+44+3 -8sin4

= 79 -8sin4

= 79-.56 = 78.44