Question

Sketch the graph of f(t) = 5/(2+3e^-t), t>=0

Answer 1

Explanation:

The term containing the variable, e^-t has a range from 0 to infinity, as all exponential terms do.

For t → -∞, e^-t → ∞ and the value of the rational expression becomes 5/∞ ≈ 0. That is, there is a horizontal asymptote at f(t)=0 for large negative values of t.

For t → ∞, e^-t → 0 and the value of the rational expression becomes approximately 5/2. That is, there is a horizontal asymptote at f(t) = 5/2 for large positive values of t.

Essentially, the curve is "S" shaped, with a smooth transition between 0 and 5/2 for values of t that make 3e^-t have values within an order of magnitude of the other term in the denominator, 2.

At t=0, 3e^-t = 1 and the denominator is 2+3=5. That is, f(0) = 5/5 = 1. Of course, the curve will cross the line f(t) = 5/4 (halfway between the asymptotes) when 3e^-t = 2, or t=ln(3/2)≈0.405. The curve is symmetrical about that point.

You can sketch the graph by finding values of t that give you points on the transition. Typically, you would choose t such that 3e^-t will be some fraction or multiple of 2, say 1/10, 1/3, 1/2, 1, 2, 3, 10 times 2.

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f(t) is called a "logistic function." It models a situation where growth rate is proportional both to population size and the difference between population size and carrying capacity. In public health terms, it models the spread of disease when that is proportional to the number of people exposed and to the number not yet exposed.