The position at time t of an insect crawling along the x-axis is given by x(t) = t^3 - 3t^2 - 9t + 4. What is the maximum velocity on the closed interval 0 <= t <= 2?

Answer 1

find v(t), and then just take its derivative to find its maximum:

v(t) = 3t^2-6t-9
v'(t) = 6t-6
clearly, v'=0 at t=1
However, since v' is just a parabola, its vertex is a minimum, not a maximum.

That means that its maximum is either at t=0 or t=2. So, just compare v(0) and v(2).