Consider the function f(x) whose second derivative is f''(x)=9x+10sin(x). If f(0)=4 and f'(0)=4, what is f(3)?

Answer 1

∫(9x + 10sin x)dx goes to 4.5x² − 10cos x +C₁ as f'(x).

∫(4.5x² − 10cos x +C₁)dx goes to 1.5x³ − 10sin x +C₁x + C₂ as f(x).

1.5(0)³ − 10sin (0) + C₁(0) + C₂ = 4 gives C₁(0) + C₂ = 4 or C₂ = 4.

4.5(0)² − 10cos (0) + C₁ = 4 gives -10 + C₁ = 4 or C₁ = 14.

Now construct f(x) = 1.5x³ − 10sin x +14x + 4 and confirm that f(0) is 4.

Also confirm that f'(x) goes to 4.5x² − 10cos x + 14 and that f'(0) is 4.

Finally f(3) is given by 1.5(3)³ − 10sin (3 Radians) +14(3) + 4 equal to 85.08879992.