The velocity  $v$ of a particle is given by the equation $v=6t^2-6t^3$ , where $v$ is in  $m{s}^{-1},t$  is the instant of time in seconds while 6 and 6 are suitable dimensional constants. 

Given  $v=6l^2-6t^3$  Differentiating  $v$  w.r.t   $t$ , we have  $\frac{dv}{dt}=12t-18t^2$
Putting  $\frac{dv}{dt}=0$ , we will get the values of  $t$ at which  $v$  is maximum or minimum. Therefore,
 $12t-18t^2=0\Rightarrow t=0,2/3s$
To the distinguish between points of maxima and minima, we need the second derivative of  $v$
 $\frac{d^{2}v}{d{t}^2}=12-36t$
Now  ${\frac{d^{2}v}{d{t}^2}|}_{t=0}=12>0$
So, $t=0$  is a point of minima
 ${\frac{d^{2}v}{d{t}_2}|}_{t=2/3s}=12-36\times \frac{2}{3}=-12<0$
So,  $t=2/3s$  is a point of maxima
Hence, the minimum value of  $v$  is  $0m{s}^{-1}$ (by putting $t=0sinv$ )
The maximum value of $v$  is
$6\times \frac{4}{9}-6\times \frac{8}{27}=\frac{8}{3}-\frac{16}{9}=\frac{8}{9}m{s}^{-1}$  (by putting  $t=\frac{2}{3}\sin v$ )