Chapter 3.2, Problem 26E

Solution

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

  $h\left(x\right)\approx \frac{30}{1+5{\left(0.585\right)}^x}$

Given information:

Initial height = 5, limit to growth = 30, passing through (3, 15).

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:

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

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

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

  $c=30$

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

  $\begin{array}{c}h\left(0\right)=\frac{c}{1+a\cdot b^{\left(0\right)}}\\ =\frac{c}{1+a}\\ =\frac{30}{1+a}\end{array}$

Now, initial value is given to be 5, so,

  $\begin{array}{c}5=\frac{30}{1+a}\\ 1+a=\frac{30}{5}\\ 1+a=6\\ a=5\end{array}$

Now, to find the value of the constant b , substitute the given point $\left(3,15\right)$ ,

  $\begin{array}{c}h\left(3\right)=\frac{c}{1+a\cdot b^{\left(3\right)}}\\ 15=\frac{30}{1+5b^3}\\ 1+5b^3=\frac{30}{15}\end{array}$

  $\begin{array}{c}1+5b^3=2\\ 5b^3=1\\ b=\sqrt[3]{\frac{1}{5}}\\ b\approx 0.585\end{array}$

Hence, the required function is:

  $h\left(x\right)\approx \frac{30}{1+5{\left(0.585\right)}^x}$