Chapter 4.3, Problem 40E

(a)

To determine

To find: The vertical and the horizontal asymptotes.

Answer to Problem 40E

The horizontal asymptotes are $y=e^{\frac{\pi }{2}}$ and $y=e^{-\frac{\pi }{2}}$ . No vertical asymptote exists.

Explanation of Solution

Given:

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

Calculation:

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

  $\begin{array}{c}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }{e}^{\arctan x}\\ =e^{\arctan \left(\infty \right)}\\ =e^{\frac{\pi }{2}}\end{array}$

So, $y=e^{\frac{\pi }{2}}$ and $y=e^{-\frac{\pi }{2}}$ are the horizontal asymptote.

Since the function is a simple exponential function, no vertical asymptote exists.

(b)

To determine

To find: The intervals of increase and decrease.

Answer to Problem 40E

The function is increasing for all x .

Explanation of Solution

Given:

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

Calculation:

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

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

Since $f^{\prime }\left(x\right)$ is always greater than 0, $f\left(x\right)$ is an always increasing function.

(c)

To determine

To find: The local maximum and minimum values.

Answer to Problem 40E

No local minimum or maximum value.

Explanation of Solution

Given:

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

Calculation:

Since $f^{\prime }\left(x\right)=\frac{e^{\arctan x}}{1+x^2}$ 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 40E

The function $f\left(x\right)$ is concave upwards in the intervals $\left(-\infty ,\frac{1}{2}\right)$ and concave downwards in the interval $\left(\frac{1}{2},\infty \right)$ . The point $x=\frac{1}{2}$ becomes the point of inflection.

Explanation of Solution

Given:

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

Calculation:

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

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

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

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

So, $f\left(x\right)$ is concave upwards in the intervals $\left(-\infty ,\frac{1}{2}\right)$ and concave downwards in the interval $\left(\frac{1}{2},\infty \right)$ . The point $x=\frac{1}{2}$ becomes the point of inflection.

(e)

To determine

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

Explanation of Solution

Given:

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

Calculation:

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

  

Figure 1