Science:Math Exam Resources/Courses/MATH215/December 2011/Question 03 (b)/Solution 1

From part (a) we have that the equilibrium points are $\pm 1,2$ and therefore these points divide our derivative space into four regions where the derivative can be positive or negative, $(-\infty ,-1)\cup (-1,1)\cup (1,2)\cup (2,\infty )$. We have that

${\frac {dy}{dt}}=(y^{2}-1)(y-2)^{2}$

and we first note that $\displaystyle (y-2)^{2}$ 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

${\begin{aligned}y^{2}-1<0&,\quad (-1,1)\\y^{2}-1>0&,\quad (-\infty ,-1)\cup (1,\infty ).\end{aligned}}$

Therefore

${\begin{aligned}{\frac {dy}{dt}}>0&,\quad (-\infty ,-1)\cup (1,2)\cup (2,\infty )\\{\frac {dy}{dt}}<0&,\quad (-1,1).\end{aligned}}$

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.