Chapter 5.3, Problem 62E

a.

To determine

To show: f is increasing in the interval $\left(-\infty ,\infty \right)$ , if $abc>0$ , and decreasing if $abc<0$

Explanation of Solution

Given information:let $f\left(x\right)=c/\left(1+a{e}^{-bx}\right)$ with $a>0,abc\ne 0.$

Proof:

Determine $f\text{'}\left(x\right)$ ,

  $\begin{array}{l}f\text{'}\left(x\right)=\frac{-c}{{\left(1+a{e}^{-bx}\right)}^2}\left(1+a{e}^{-bs}\right)\\ =\frac{-c}{{\left(1+a{e}^{-bx}\right)}^2}\left(-ab\right)e^{-bx}\\ =\frac{abc\left(e^{-bx}\right)}{{\left(1+a{e}^{-bx}\right)}^2}\end{array}$

  $e^{-bx}>0$ and $\left(1+a{e}^{-bx}\right)>0$ , thus, when $abc>0$ , f is increasing and when $abc<0$ , f is decreasing in the interval $\left(-\infty ,\infty \right)$ .

Hence, proved.

b.

To determine

To show: point of inflection of f occurs at $x=\frac{\ln \left|a\right|}{b}$

Explanation of Solution

Given information: let $f\left(x\right)=c/\left(1+a{e}^{-bx}\right)$ with $a>0,abc\ne 0.$

Proof:

From part a, find the $f\text{'}\text{'}\left(x\right)$

The possible points of inflection occur where $f\text{'}\text{'}\left(x\right)=0$ or undefined, since, $\left(1+a{e}^{-bx}\right)>0$ , $f\text{'}\text{'}\left(x\right)$ can never be undefined for all x . The possible point of inflection can occur at $\left(a{e}^{-bx}-1\right)=0$ , solve for x.

  $\begin{array}{l}\left(a{e}^{-bx}-1\right)=0\\ a{e}^{-bx}=1\\ a=e^{bx}\\ \ln \left|a\right|=\ln e^{bx}\\ \ln \left|a\right|=bx\\ x=\frac{\ln \left|a\right|}{b}\end{array}$

Hence, proved.