Answer
(a) $y =4.81$ is a horizontal asymptote.
$y =0.21$ is a horizontal asymptote.
(b) $f$ is increasing on the interval $(-\infty, \infty)$
(c) There is no local maximum or local minimum.
(d) The graph is concave down on this interval: $(\frac{1}{2}, \infty)$
The graph is concave up on this interval: $(-\infty, \frac{1}{2})$
The point of inflection is $(0.5, 1.59)$
(e) We can see a sketch of the graph below.
Work Step by Step
(a) $f(x) = e^{arctan~x}$
$f(x)$ is defined for all values of $x$
There are no vertical asymptotes.
$\lim\limits_{x \to \infty}arctan~x = \frac{\pi}{2}$
$\lim\limits_{x \to \infty}f(x) = e^{\pi/2} = 4.81$
$y =4.81$ is a horizontal asymptote.
$\lim\limits_{x \to -\infty}arctan~x = -\frac{\pi}{2}$
$\lim\limits_{x \to -\infty}f(x) = e^{-\pi/2} = 0.21$
$y =0.21$ is a horizontal asymptote.
(b) We can find the points where $f'(x) = 0$:
$f'(x) =e^{arctan~x}~(\frac{1}{x^2+1})$
$f'(x) = \frac{e^{arctan~x}}{x^2+1} = 0$
$e^{arctan~x} = 0$
There are no values of $x$ such that $f'(x) = 0$
$f$ is not decreasing on any interval.
For all values of $x$, $f'(x) \gt 0$
$f$ is increasing on the interval $(-\infty, \infty)$
(c) There are no values of $x$ such that $f'(x) = 0$
There is no local maximum or local minimum.
(d) We can find the points where $f''(x) = 0$:
$f''(x) = \frac{\frac{e^{arctan~x}}{x^2+1}(x^2+1)-(e^{arctan~x})(2x)}{(x^2+1)^2}$
$f''(x) = \frac{e^{arctan~x-2x~e^{arctan~x}}}{(x^2+1)^2}$
$f''(x) = \frac{e^{arctan~x~(1-2x)}}{(x^2+1)^2} = 0$
$1-2x = 0$
$x = \frac{1}{2}$
The graph is concave down when $f''(x) \lt 0$
The graph is concave down on this interval: $(\frac{1}{2}, \infty)$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on this interval: $(-\infty, \frac{1}{2})$
$f(\frac{1}{2}) = e^{arctan~0.5} = e^{0.4636} = 1.59$
The point of inflection is $(0.5, 1.59)$
(e) We can see a sketch of the graph below.
