Question

(a) Find the vertical and horizontal asymptotes.

(b) Find the intervals of increase or decrease.

(c) Find the local maximum and minimum values.

(d) Find the intervals of concavity and the inflection points.

(e) Use the information from parts (a)–(d) to sketch the graph of \(f\).

\(f(x) = {e^{\arctan x}}\)

Short answer

(a) The function has no vertical asymptote and the horizontal asymptotes are and \(y = {e^{ - \frac{\pi }{2}}}\).

(b) The given function is decreasing on the interval \(( - \infty ,\infty )\).

(c) The function has no local maximum and minimum value.

(d) The function \(f(x)\) is concave upward on \(\left( { - \infty ,\frac{1}{2}} \right)\) and concave downward on \(\left( {\frac{1}{2},\infty } \right)\) and the inflection point are \(\left( {\frac{1}{2},1.5899} \right)\).

(e) The graph satisfies the expression.

Step-by-step solution

Given data

The given function is \(f(x) = {e^{\arctan x}}\).

Concept of Differentiation

Differentiation is a method of finding the derivative of a function. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables.

Simplify the given function

(a)

Vertical asymptote is obtained at \(x = a\), if \(\mathop {\lim }\limits_{x \to a} f(x) = - \infty \) or \(\mathop {\lim }\limits_{x \to a} f(x) = \infty \).

There is no vertical asymptote since the function is defined for any value.

As \(x\) approaches \( - \infty \),

\(\begin{aligned}{l}\mathop {\lim }\limits_{x \to - \infty } f(x) = \mathop {\lim }\limits_{x \to - \infty } \left( {{e^{\arctan x}}} \right)\\\mathop {\lim }\limits_{x \to - \infty } f(x) = {e^{\arctan ( - \infty )}}\\\mathop {\lim }\limits_{x \to - \infty } f(x) = {e^{ - \frac{x}{2}}}\end{aligned}\)

As \(x\) approaches \(\infty \),

\(\begin{aligned}{l}\mathop {\lim }\limits_{x \to \infty } f(x) = \mathop {\lim }\limits_{x \to \infty } \left( {{e^{\arctan x}}} \right)\\\mathop {\lim }\limits_{x \to \infty } f(x) = {e^{\arctan (\infty )}}\\\mathop {\lim }\limits_{x \to \infty } f(x) = {e^{\frac{\pi }{2}}}\end{aligned}\)

As \(x \to - \infty \), the function approaches \(y = {e^{ - \frac{x}{2}}}\) and \(x \to \infty \), the function approaches . Therefore, the horizontal asymptotes are \(y = {e^{ - \frac{x}{2}}}\) and .

Determine the derivative of the function

(b)

Obtain the derivative of \(f(x)\):

\(\begin{aligned}{l}f(x) = {e^{\arctan x}}\\f(x) = {e^{\arctan x}} \cdot \frac{1}{{1 + {x^2}}}\end{aligned}\)

For any value of \(x\) that \({f^\prime }(x) > 0\) for all \(x\) on the interval \(( - \infty ,\infty )\). Therefore, \(f(x)\) is decreasing on the interval \(( - \infty ,\infty )\).

Determine the first Derivative test

(c)

"Suppose that \(c\) is a critical number of a continuous function \(f\).

(a) If \({f^\prime }\) changes from positive to negative at \(c\), then \(f\) have a local maximum at \(c\).

(b) If \({f^\prime }\) changes from negative to positive at \(c\), then \(f\) has a local minimum at \(c\).

(c) If \({f^\prime }\) is positive to the left and right of \(c\), or negative to the left and right of \(c\), then \(f\) has no local maximum or minimum at \(c\).

From the above table observe that \(f\) changes from increasing to decreasing nor from decreasing to increasing.

Hence, the function has no local maximum and minimum value.

Determine the derivative of the function

(d)

Obtain the derivative of \({f^\prime }(x)\).

\(\begin{aligned}{l}{f^{\prime \prime }}(x) = \frac{d}{{dx}}\left( {{f^\prime }(x)} \right)\\{f^{\prime \prime }}(x) = \frac{d}{{dx}}\left( {\frac{{{e^{{\mathop{\rm arclan}\nolimits} x}}}}{{1 + {x^2}}}} \right)\end{aligned}\)

Apply the Quotient rule:

\(\begin{aligned}{l}{f^{\prime \prime }}(x) = \frac{{\left( {1 + {x^2}} \right){e^{\arctan x}}\left( {\frac{1}{{1 + {x^2}}}} \right) - {e^{\arctan x}}(2x)}}{{{{\left( {1 + {x^2}} \right)}^2}}}\\{f^{\prime \prime }}(x) = \frac{{{e^{\arctan x}}(1 - 2x)}}{{{{\left( {1 + {x^2}} \right)}^2}}}\end{aligned}\)

Set:

\(\begin{aligned}{c}{f^{\prime \prime }}(x) = 0\\{e^{\arctan x}}(1 - 2x) = 0\\1 - 2x = 0\\2x = 1\\x = \frac{1}{2}\end{aligned}\)

Simplify the expression

Use the below table and find on which the function \(f(x)\) is concave upward and downward.

Interval	\(f''\left( x \right)\)	\(f\left( x \right)\)
\( - \infty < x < \frac{1}{2}\)	\(f''\left( x \right) > 0\)	Concave upward on \(\left( { - \infty ,\frac{1}{2}} \right)\)
\(\frac{1}{2} < x < \infty \)	\(f''\left( x \right) < 0\)	Concave downward on \(\left( {\frac{1}{2},\infty } \right)\)

Hence, \(f(x)\) is concave upward on \(\left( { - \infty ,\frac{1}{2}} \right)\) and concave downward on \(\left( {\frac{1}{2},\infty } \right)\).

"A point \(P\) on a curve under the function \(f(x)\) is called an inflection point if \(f\) is continuous and curve changes from concave upward to concave downward or from concave downward to concave upward at P ".

Hence, the curve changes from concave upward to concave downward at \(x = \frac{1}{2}\).

Substitute \(x = \frac{1}{2}\) in \(f(x)\),

\(\begin{aligned}{l}f\left( {\frac{1}{2}} \right) = {e^{\arctan \left( {\frac{1}{2}} \right)}}\\f\left( {\frac{1}{2}} \right) \approx 1.5899\end{aligned}\)

Thus, the inflection point is \( \approx 1.5899\)

Simplify the expression

(e)

From part (a), the function has no vertical asymptote and horizontal asymptote at and \(y = {e^{ - \frac{\pi }{2}}}\).

From part (b), the function \(f(x)\) does not increases on and decreases on \(( - \infty ,\infty )\).

From part (c), the function has no local maximum and minimum value.

From part (d), the function \(f(x)\) is concave upward on \(\left( { - \infty ,\frac{1}{2}} \right)\) and concave downward on \(\left( {\frac{1}{2},\infty } \right)\) and the inflection point is \(\left( {\frac{1}{2},1.5899} \right)\).

Use online calculator and the information collected from part (a) to part (d), sketch the graph as shown below in Figure 1.

The graph satisfies the expression.