Help with maths question please question 4?

a) initial position: x(0) = -0^3/3 + 0^2 + 8(0) + 1 = 1
initial velocity:
x '(t) = -t^2 + 2t + 8
x '(0) = -0^2 + 2(0) + 8 = 8
b) Change in direction occurs when x '(t) = 0.
0 = -t^2 + 2t + 8
0 = t^2 - 2t - 8
0 = (t - 4)(t + 2)
t = 4 or = t = -2
Discard t = -2 because t >= 0 rules out negative times, leaving just
t = 4
distance traveled = x(4) = -4^3/3 + 4^2 + 8(4) + 1 = -64/3 + 16 + 32 + 1 =
-64/3 + 49 = -21 1/3 + 49 = 17 2/3
c) acceleration = x ''(t) = -2t + 2
At time t = 4, x ''(4) = -2(4) + 2 = -6