CALCULUS BC

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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.
a.-2<t<1 only
b.-2<t<-1 and 1<t<2
c.1<t<2 only
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:
    x(t)=3t^5-25t^3+60t
    The derivative is:
    x'(t)=15*t^4-75*t^2+60
    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.

    Hint:
    There are two such intervals, and two of the zeroes of x'(t) are t=-1 and +1.

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