26. If 4h = kM², show that typical solution curves look as illustrated in Fig. 2.2.14. Thus if xo 2 M/2, then x(1) → M/2 as t →+o. But if xo < M/2, then x(t) = 0 after a finite period of time, so the lake is fished out. The critical point x = M/2 might be called semistable, because it looks stable from one side, unstable from the other. x= M/2 x = 0 FIGURE 2.2.14. Solution curves for harvesting a logistic population with 4h = kM2.

Calculus: Early Transcendentals
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Chapter1: Functions And Models
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26. If 4h = kM², show that typical solution curves look
as illustrated in Fig. 2.2.14. Thus if xo 2 M/2, then
x(1) → M/2 as t →+o. But if xo < M/2, then x(t) = 0
after a finite period of time, so the lake is fished out. The
critical point x = M/2 might be called semistable, because
it looks stable from one side, unstable from the other.
x= M/2
x = 0
FIGURE 2.2.14. Solution curves for
harvesting a logistic population with
4h = kM2.
Transcribed Image Text:26. If 4h = kM², show that typical solution curves look as illustrated in Fig. 2.2.14. Thus if xo 2 M/2, then x(1) → M/2 as t →+o. But if xo < M/2, then x(t) = 0 after a finite period of time, so the lake is fished out. The critical point x = M/2 might be called semistable, because it looks stable from one side, unstable from the other. x= M/2 x = 0 FIGURE 2.2.14. Solution curves for harvesting a logistic population with 4h = kM2.
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