Derivative as a rate of change

$\displaystyle s = -t^3 +3t^2 -3t$ with the interval $\displaystyle 0\leq t \leq 3$

a. Find the body's displacement and average velocity for the given time interval.

$\displaystyle V=-3t^2 +6t - 3$

$\displaystyle V(0) = - 3$

$\displaystyle V(3) = -32$

Is this all thats needed on this part or do I need to do $\displaystyle V(3)-V(0)$ to find the average velocity?

b. Find the body's speed and acceleration at the endpoints of the interval.

$\displaystyle s = -t^3 +3t^2 - 3t$

$\displaystyle s(0)= 0$

$\displaystyle s(3)=9$ is positive because speed is always absolute value

$\displaystyle A = -6t + 6$

$\displaystyle A(0) = 6$

$\displaystyle A(3) = -12$

c. When, if ever, during the interval does the body change direction.

Use the Velocity equation for this I believe and when I plug in values between the interval, the output (v) is never changed into a positive number so can I conclude it does not change direction?