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Homework 8 SolutionsTotal points: 16March 11, 2019Chapter 7.1Solution:When the slope is zero, the equilibrium point can have any stability (stable, unstable, semistable). Thus,the fact thatdX0dXX*= 0tells us nothing about the stability ofX*. In other words, linear stability analysis fails in this case.Solution:As you learned in LS 30A, you find the equilibrium points by setting the change equation equal to zeroand solving forX:X0=X3-X= 0X(X2-1) = 0X(X-1)(X+ 1) = 0X=-1,0,1Thus, the equilibrium points are-1,0,and 1. On the other hand,dX0dX= 2X2-1.Thus, linear stability analysis tells us thatdX0dXX=-1= 1>0 =⇒ -1 is unstabledX0dXX=0=-1<0 =⇒0 is stabledX0dXX=1= 1>0 =⇒1 is unstable1
Chapter 7.2Solution:By definition, a function is a rule that takes each element of its domain to exactly one element of itscodomain. Hence, for any (X, Y)∈R2, there is single valuef(X, Y). This means that the graph off:R2→Rhasjust a single point above (or below) each (X, Y). Otherwise said, the graph of a function cannot have two points