Chapter 4.3, Problem 38E

(a)

To determine

To find: The vertical and the horizontal asymptotes.

Answer to Problem 38E

The horizontal asymptote is line $y=1$ and there exists no horizontal asymptote.

Explanation of Solution

Given:

The function is $f\left(x\right)=\frac{e^x}{1+e^x}$ .

Calculation:

To find the horizontal asymptote, determine $\underset{x\to \infty }{\lim }f\left(x\right)$ .

  $\begin{array}{c}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{e^x}{1+e^x}\\ =1\left[\text{By standard result}\right]\end{array}$

So, $y=1$ is the horizontal asymptote.

To find the vertical asymptote, put the denominator $1+e^x=0$ and determine the values of $x$ for which $e^x=-1$ hold true. Since the exponential curve is never negative, there exist no vertical asymptotes.

(b)

To determine

To find: The intervals of increase and decrease.

Answer to Problem 38E

The function is increasing everywhere.

Explanation of Solution

Given:

The function is $f\left(x\right)=\frac{e^x}{1+e^x}$ .

Calculation:

To find the intervals of increase and decrease, determine the derivative of $f\left(x\right)$ .

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{e^x}{1+e^x}\right)\\ =\frac{e^x\left(1+e^x\right)-e^x\left(e^x\right)}{{\left(1+e^x\right)}^2}\\ =\frac{e^x+e^{2x}-e^{2x}}{{\left(1+e^x\right)}^2}\\ =\frac{e^x}{{\left(1+e^x\right)}^2}\end{array}$

Since $\frac{e^x}{{\left(1+e^x\right)}^2}$ is an always increasing function, $f\left(x\right)$ is always increasing.

(c)

To determine

To find: The local maximum and minimum values.

Answer to Problem 38E

No local minimum or maximum value.

Explanation of Solution

Given:

The function is $f\left(x\right)=\frac{e^x}{1+e^x}$ .

Calculation:

Since $f^{\prime }\left(x\right)=\frac{e^x}{{\left(1+e^x\right)}^2}$ is never zero anywhere, there exist no local minimum or maximum value.

(d)

To determine

To find: The intervals of concavity and the inflection points.

Answer to Problem 38E

The function $f\left(x\right)$ is concave upwards in the interval $\left(-\infty ,0\right)$ and concave downwards in the interval $\left(0,\infty \right)$ . The point $x=0$ is the point of inflection.

Explanation of Solution

Given:

The function is $f\left(x\right)=\frac{e^x}{1+e^x}$ .

Calculation:

To find the intervals of concavity, determine the second derivative of $f\left(x\right)$ .

  $\begin{array}{c}{f}^{″}\left(x\right)=\frac{d}{dx}\left(\frac{e^x}{{\left(1+e^x\right)}^2}\right)\\ =\frac{e^x{\left(1+e^x\right)}^2-2e^x\left(1+e^x\right)e^x}{{\left(1+e^x\right)}^4}\\ =\frac{e^x\left(1+e^x\right)-2e^{2x}}{{\left(1+e^x\right)}^3}\\ =\frac{e^x\left(1-e^x\right)}{{\left(1+e^x\right)}^2}\end{array}$

Now to determine the points where $f^{″}\left(x\right)$ is 0, put $\frac{e^x\left(1-e^x\right)}{{\left(1+e^x\right)}^2}$ equal to 0.

  $\begin{array}{c}\frac{e^x\left(1-e^x\right)}{{\left(1+e^x\right)}^2}=0\\ e^x\left(1-e^x\right)=0\\ 1-e^x=0\\ e^x=1\end{array}$

Since $e^0=1$ , $f\left(x\right)$ is concave upwards in the interval $\left(-\infty ,0\right)$ and concave downwards in the interval $\left(0,\infty \right)$ . The point $x=0$ serves as the point of inflection.

(e)

To determine

To sketch: The graph of $f\left(x\right)$ .

Explanation of Solution

Given:

The function is $f\left(x\right)=\frac{e^x}{1+e^x}$ .

Calculation:

To graph of $f\left(x\right)$ is shown below:

  

Figure 1