Chapter 3.2, Problem 28E

Solution

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

  $f\left(x\right)=\frac{60}{1+3{\left(0.87\right)}^x}$

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:

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

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. 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 $y=60$ . So,

  $c=60$

So, the function becomes:

  $f\left(x\right)=\frac{60}{1+a\cdot b^x}$

Now, from the graph, it can be inferred that $f\left(0\right)=15$ . Substitute this in above equation,

  $\begin{array}{c}f\left(0\right)=\frac{60}{1+a\cdot b^{\left(0\right)}}\\ 15=\frac{60}{1+a}\\ 1+a=\frac{60}{15}\end{array}$

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

Now, to find the value of the constant b , substitute the given point $\left(8,30\right)$ ,

  $\begin{array}{c}f\left(8\right)=\frac{c}{1+a\cdot b^{\left(8\right)}}\\ 30=\frac{60}{1+3b^8}\\ 1+3b^8=\frac{60}{30}\end{array}$

  $\begin{array}{c}1+3b^8=2\\ 3b^8=1\\ b=\sqrt[8]{\frac{1}{3}}\\ b=0.87\end{array}$

Hence, the required function is:

  $f\left(x\right)=\frac{60}{1+3{\left(0.87\right)}^x}$