Posted by paul on .
a particle moves on the x-axis in such a way that its position at time t is given by x(t)=3t^5-25^3+60t. for what values of t is the particle moving to the left.
b.-2<t<-1 and 1<t<2
d.-1<t<1 and t>2
e.t<-2,-1<t<1, and t>2
CALCULUS BC -
Assuming there is a typo and the original function is:
The derivative is:
Factor x'(t) and solve for the zeroes of x'(t)=0.
Determine on which interval(s) x'(t) is negative, which means that the particle is moving "backwards", or to the left.
There are two such intervals, and two of the zeroes of x'(t) are t=-1 and +1.