Figure out this calculus problem?

Question

a particle moves along the x-axis and its position at any time t>or equal to 0 is given by x(t)=(t-1)^3(2t-3)

for wat values of t is the velocity of the particle less than zero

Answer 1

First take the derivative of x(t) to get v(t):

v(t) = 3*(t-1)^2 * (2t-3) + (t-1)^3 * 2

the term (t-1)^2 is always positive, so the sign of this expression depends on the (2t - 3)*(t-1)^3 terms. (t-1)^3 is negative for t<1. (2t -3) is negative for t < 1.5. When both of these terms are negative (t<1) the velocity is positive. From 1 < t < 1.5, the term (t - 1) is positive and (3t - 2) is negative, so the expression is negative in that region. Above that, the expression is always positive. Therefore the velocity is negative when 1 < t < 1.5.