An object moves along the x-axis, its position at each time t> 0 given by \"x(t) = 81/2*t^2-3/4*t^4\" determine the time interval(s), if any, during which the object slows down.

You can say an object slow down when the acceleration is negative which can be observed on a negative slope during the graph of acceleration, from a position equation lets derive 2 times to attain the acceleration equation

x(t) = 81/2*t^2-3/4*t^4
V(t) = 81X - 3t^3
A(t) = 81 - 9t^2

81 - 9t^2 < 0
t^2 > 9
-3 > t > 3

The final answer is 3x3+6x2+8x+24+47x−2mark me brainliest