Chapter 3.2, Problem 23E

Solution

To find: the logistic function that satisfies the given conditions.

  $f\left(x\right)=\frac{40}{1+3{\left(\frac{1}{3}\right)}^x}$

Given information:

Initial value = 10, limit to growth = 40, passing through $\left(1,20\right)$ .

Definition Used:

Logistic Growth Function:

Let a , b , c , and k be positive constants, with $b<1$ . A logistic growth function in x is a function that can be written in the form

  $f\left(x\right)=\frac{c}{1+a\cdot b^x}\text{ or }f\left(x\right)=\frac{c}{1+a\cdot e^{-kx}}$

Where the constant c is the limit to growth.

Explanation:

Now, the required function will be of the form:

  $f\left(x\right)=\frac{c}{1+a\cdot b^x}$

Here, as x becomes larger, $e^{-kx}$ will approach to zero. So,

  $\underset{x\to \infty }{\lim }f\left(x\right)=c$ thus, c is the limit to growth.

  $c=40$

Now, let’s find the initial value by substituting $x=0$ ,

  $\begin{array}{c}f\left(0\right)=\frac{c}{1+a\cdot b^{\left(0\right)}}\\ =\frac{c}{1+a}\\ =\frac{40}{1+a}\end{array}$

Now, initial value is given to be 10, so,

  $\begin{array}{c}10=\frac{40}{1+a}\\ 1+a=\frac{40}{10}\\ 1+a=4\\ a=3\end{array}$

Now, to find the value of the constant b , substitute the given point $\left(1,20\right)$ ,

  $\begin{array}{c}f\left(1\right)=\frac{c}{1+a\cdot b^{\left(1\right)}}\\ 20=\frac{40}{1+3b}\\ 1+3b=\frac{40}{20}\end{array}$

  $\begin{array}{c}1+3b=2\\ 3b=1\\ b=\frac{1}{3}\end{array}$

Hence, the required function is:

  $f\left(x\right)=\frac{40}{1+3{\left(\frac{1}{3}\right)}^x}$