Interval in which $x=2t^3-3t^2$ is negative

Question

A particle move along x-axis so that its position is given by $x=2t^3-3t^2$ at time $t$ sec. What is the time interval during which the particle will be on the negative half of the axis ?

$$ x=2t^3-3t^2=t^2(2t-3)<0\implies\boxed{t<\dfrac{3}{2}}\\ \frac{dx}{dt}=6t^2-6t=6t(t-1)=0\implies t=0\text{ or } t=1\\ \frac{d^2x}{dt^2}=12t-6\\ \frac{d^2x}{dt^2}\Big|_{t=0}=-6<0\;;\quad\;\frac{d^2x}{dt^2}\Big|_{t=1}=6>0\\ x(t=0)=0\;;\quad\;x(t=1)=-1\\ x(t=3/2)=0 $$

My reference gives the solution $0<t<3/2$, so where does the lower limit $0$ comes from ?

Answer 1

$t$ is time, hence $t \ge 0.$ Furthermore we have $x(0)=0.$

Hence:

the particle will be on the negative half of the axis $ \iff 0<t<3/2.$

Answer 2

Hint (for avoiding differentiation) : \begin{align*} \mathrm{sgn}(2t^3-3t^2) &= \mathrm{sgn}(t^2(2t-3))\\ &= \mathrm{sgn}(2t-3)\\ &= \mathrm{sgn}\left(t-\frac 32\right) \end{align*}