Let f′′(x)=36x^(2)+10 f(1)=4, f′(1)=8. Then f(x) =

Answer 1

Since f"(x) = 36x² + 10, f'(x) = ∫(36x²+10)dx = 12x³ + 10x + C_1.

Since f'(1) = 8, we have 12 + 10 + C₁ = 8. So, C₁ = -14

Therefore, f'(x) = 12x³ + 10x - 14.

f(x) = ∫f'(x)dx = 3x⁴ + 5x² - 14x + C₂

Since f(1) = 4, -6 + C₂ = 4. So, C₂ = 10

f(x) = 3x⁴ + 5x² - 14x + 10