Chapter 3.4, Problem 148E

a.

To determine

To use : the graphing utility to graph the two functions

Explanation of Solution

Given information : The question is

The function is

  $\begin{array}{l}m\left(x\right)=\frac{100}{1+e^{-0.6114\left(x-69.71\right)}}\\ f\left(x\right)=\frac{100}{1+e^{-0.66607\left(x-64.51\right)}}\end{array}$

Graph : Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression $\frac{100}{1+e^{-0.6114\left(x-69.71\right)}}\&\frac{100}{1+e^{-0.66607\left(x-64.51\right)}}$ which is required to graph.

Step 4: Press

  $\boxed{\text{GRAPH}}$ button to graph the function.

The graph is obtained as:

  

Interpretation :

From the above graph the blue line shows the function $\frac{100}{1+e^{-0.66607\left(x-64.51\right)}}$ and the red line shows the function $\frac{100}{1+e^{-0.6114\left(x-69.71\right)}}$

b.

To determine

To calculate : the horizontal asymptote of the function

Answer to Problem 148E

The horizontal asymptote is x-axis component to $\infty$

Explanation of Solution

Given information : The question is

The function is

  $\begin{array}{l}m\left(x\right)=\frac{100}{1+e^{-0.6114\left(x-69.71\right)}}\\ f\left(x\right)=\frac{100}{1+e^{-0.66607\left(x-64.51\right)}}\end{array}$

Calculation :

The horizontal asymptote

  $\begin{array}{l}f\left(x\right)=\&f\left(x\right)=100\\ g\left(x\right)=0\&g\left(x\right)=100\end{array}$

So, the horizontal asymptote of the function is given by substituting the value of x-axis component to $\infty$

c.

To determine

To calculate : the average height of each sex

Answer to Problem 148E

The average height of male is $69.71$ and female is $64.51$

Explanation of Solution

Given information : The question is

The function is

  $\begin{array}{l}m\left(x\right)=\frac{100}{1+e^{-0.6114\left(x-69.71\right)}}\\ f\left(x\right)=\frac{100}{1+e^{-0.66607\left(x-64.51\right)}}\end{array}$

Calculation :

The average height of male

  $\begin{array}{l}m\left(x\right)=\frac{100}{1+e^{-0.6114\left(x-69.71\right)}}\\ m\left(x\right)=69.71\end{array}$

The average height of female

  $\begin{array}{l}f\left(x\right)=\frac{100}{1+e^{-0.66607\left(x-64.51\right)}}\\ f\left(x\right)=64.51\end{array}$