Use the second derivative to determine where the maximum and minimum rates of change occur for a vehicle’s speed modeled by the function
v(t) = 0.0591t3 - 0.3399t2 + 0.5019t + 0.00594, where t is the time in minutes after leaving a stoplight.
v'=3*.059t^2+ 2*.3399t + 0.05019 set =0, and solve for the two t using the quadrataic equaiton.
v"=6*.059t+ 2*.3399
now for each t, find the sign of v"
if negative, you were at a maximum.
if positive, you were at a minimum