From part (a) we have that the equilibrium points are
and therefore these points divide our derivative space into four regions where the derivative can be positive or negative,
. We have that
![{\displaystyle {\frac {dy}{dt}}=(y^{2}-1)(y-2)^{2}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d5a91a7da360631385c4eea2bad7b6005c0caa1f)
and we first note that
is always positive (unless y = 2) and so the sign of the derivative will be determined by the first term only. For this term we get
![{\displaystyle {\begin{aligned}y^{2}-1<0&,\quad (-1,1)\\y^{2}-1>0&,\quad (-\infty ,-1)\cup (1,\infty ).\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/6fd480158b64c60b0bc2d12b97363e6b7e1e7b2e)
Therefore
![{\displaystyle {\begin{aligned}{\frac {dy}{dt}}>0&,\quad (-\infty ,-1)\cup (1,2)\cup (2,\infty )\\{\frac {dy}{dt}}<0&,\quad (-1,1).\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/8bf7fd46a78b60eea5caa420ce50f42de6864523)
Note: We separate the domain of increase for the derivative at the point y=2 because strictly speaking the derivative is not increasing at this point.