Identify equilibrium values and determine which are stable and which are unstable.
-y = (y - 4)(y - 6)(y - 7)
A) y = 7 is a stable equilibrium value and y = 6 and y = 4 are unstable equilibria.
B) y = 4 is a stable equilibrium value and y = 6 and y = 7 are unstable equilibria.
C) y = 6 is a stable equilibrium value and y = 4 and y = 7 are unstable equilibria.
D) y = 5, y = 4 and y = 7 are unstable equilibria.

Question

Quizplus · Accepted Answer

To determine the stability of the equilibrium points of the function $y = (y - 4)(y - 6)(y - 7)$, we look at the sign changes of the function around these points. The equilibrium values are $y = 4$, $y = 6$, and $y = 7$, where the function equals zero. For values of $y$ just less than 4, the expression is negative, and for values just greater than 4, it is positive, indicating a sign change from negative to positive, making $y = 4$ an unstable equilibrium. Between $y = 4$ and $y = 6$, the function is positive, and just after $y = 6$, it becomes negative, indicating a stable equilibrium at $y = 6$ due to the sign change from positive to negative. Finally, for $y = 7$, the sign changes from negative just before it to positive just after, indicating an unstable equilibrium. Thus, $y = 6$ is a stable equilibrium, while $y = 4$ and $y = 7$ are unstable equilibria.

Identify Equilibrium Values and Determine Which Are Stable and Which