SOLUTION: Find the logistic function that satisfies the given conditions. Initial value=12​, limit to growth=36​, passing through (5,22). What is the correct expression for x

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Question 1126020: Find the logistic function that satisfies the given conditions.
Initial value=12​, limit to growth=36​, passing through (5,22).
What is the correct expression for x?

Answer by greenestamps(13209)   (Show Source): You can put this solution on YOUR website!


I have worked very little with logistic functions; and I have never tried to work a problem where the goal is to find the logistic function when the initial value, the limiting value, and the value at one particular point are given.

I was curious about the methods for doing that; since no other tutor has responded to your question, I decided to look into it.

My methods may be far more complicated then necessary; however, having no training in this kind of problem, the following is what I came up with.

The general logistic function is of the form



Our logistic function has initial value 12; so f(0) = 12:

(1)

Our logistic function has limiting value 36; so f(100) = 36 (100 is an arbitrary "large" number):

(2)

The value of the logistic function is 22 at x=5:

(3)

We can subtract (1) from (2) to eliminate a, allowing us to get an expression for c in terms of b:







And we can get a in terms of b from (2):

Now we can substitute these expressions for a and c in terms of b into (3) and solve for b:



Solving this equation with a graphing calculator yields the solution b = 24.116857.

Then a = 36-b = 11.883143
and c = 24/0.116857 = 205.379

The logistic function with the given requirements is



A graph....

The window is x=0 to x=10, so that x=5 is the middle of the domain. The constant lines y=12, y=22, and y=36 are also shown, to show the initial value, the limiting value, and the value at x=5.



I enjoyed the mental exercise I got from figuring this out....

Now perhaps another tutor who knows logistic functions will respond and show me (and you) that there is a much easier path to the answer....

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