f ''(x) = 4 + 6x + 24x2, f(0) = 3, f (1) = 12

Question

f ''(x) = 4 + 6x + 24x²,    f(0) = 3,    f (1) = 12

Answer 1

I assume you want to find f(x).

Integrate twice

f "(x) = 4 + 6x+24x²

f '(x) = 4x + 3x²+8x³ + A Where A is an integration constant

f(x) = 2x² + x³ + 2x⁴ + Ax + B B is another integration constant

Need to solve for A and B

f(0) = 0 + 0 + 0+ 0 + B = 3 --> B = 3

f(1) = 2 + 1 + 2 + A + 3 = 12 --> A = 4

f(x) = 2x⁴+ x³ + 2x² + 4x + 3

Answer 2

f'(x) = ∫(4 +6x + 24x²)dx = 4x + 3x² + 8x³ + C₁;

f(x) = ∫(4x + 3x² + 8x³ + C₁)dx = 2x² + x³ + 2x⁴ + C₁x + C₂;

f(0) = C₂ = 3;

f(1) = 2 + 1 + 2 + C₁ + C₂ = 12; C₁ = 12 - 5 - 3 = 4.

So, f(x) = 2x² + x³ + 2x⁴ + 4x + 3.