What is the instantaneous rate of change at t = 4 ? Is there a maximum, minimum, or neither at t = 4?

Question

1. A particle moves along the y axis with the relationship where s(t) represents the displacement and t represents the time.

a) What is the instantaneous rate of change at t = 4

b) Is there a maximum, minimum, or neither at t = 4? Give reasoning.

Answer 1

s(t) = -5t^3 + 9t^2 - 4t + 64

rate of change = the derivative = velocity

s'(t) = v(t) = -15t^2 + 18t - 4

s'(4) = -15(4)^2 + 18(4) -4 = -240 +72 -4 = -172

v(4) = -172 meters per second

-15t^2 + 18t -4 = 0

t = 18/30 +/- (1/30)sqr(18^2 -15(4)(4))

t = 0.6 + or minus about 0.306

t = 0.906 or 0.294 seconds at the local extrema

plug that t value into the original equation to see if it's a max or min, or take the 2nd derivative to determine if the extremum was a max or min.

there is no global max or min at t=4 or at any t value, but there are local extrema, one local max and one local min