Determine f(t) when f''(t)= 6(3t-1) and f'(1)=1, f(1)=3?

Answer 1

f " (t) = 18 t - 6
f `(t) = 9 t ² - 6 t + C
1 = 9 - 6 + C
C = - 2
f `(t) = 9 t ² - 6 t - 2
f (t) = 3 t ³ - 3 t ² - 2 t + k
3 = 3 - 3 - 2 + k
k = 5
f (t) = 3 t ³ - 3 t ² - 2 t + 5

Answer 2

3t^3-3t^2-3t+6=f(t)