Chapter 3, Problem 25RE

Solution

To find: a formula for the logistic function whose graph is shown in the figure.

  $f\left(x\right)\approx \frac{20}{1+3e^{-0.37x}}$

Given information:

The graph of the required function is:

  

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:

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

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. Now, in the graph of a logistic function the limit to growth is given by the upper horizontal asymptote.

  

On observing the above graph, the upper horizontal asymptote is given by $y=20$ . So,

  $c=20$

So, the function becomes:

  $f\left(x\right)=\frac{20}{1+a\cdot e^{-kx}}$

Now, from the graph, it can be inferred that $f\left(0\right)=5$ . Substitute this in above equation,

  $\begin{array}{c}f\left(0\right)=\frac{20}{1+a\cdot e^{-k\left(0\right)}}\\ 5=\frac{20}{1+a}\\ 1+a=\frac{20}{5}\end{array}$

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

Now, to find the value of the constant k , substitute the given point $\left(3,10\right)$ ,

  $\begin{array}{c}f\left(3\right)=\frac{c}{1+a\cdot e^{-k\left(3\right)}}\\ 10=\frac{20}{1+3e^{-3k}}\\ 1+3e^{-3k}=\frac{20}{10}\end{array}$

  $\begin{array}{c}1+3e^{-3k}=2\\ 3e^{-3k}=1\\ e^{-3k}=\frac{1}{3}\end{array}$

Take natural log on both sides,

  $\begin{array}{c}\ln e^{-3k}=\ln \left(\frac{1}{3}\right)\\ -3k=\ln 1-\ln 3\\ -3k=0-\ln 3\end{array}$

  $\begin{array}{c}k=\frac{\ln 3}{3}\\ k\approx 0.37\end{array}$

Hence, the required function is:

  $f\left(x\right)\approx \frac{20}{1+3e^{-0.37x}}$