Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 52

(a) f is defined and infinitely differential everywhere, so it hasn't vertical asymptotes.
When x->-oo, arctan(x)->-pi/2 and f(x)->e^(-pi/2) approx 0.21.
When x->+oo, arctan(x)->+pi/2 and f(x)->e^(pi/2) approx 4.81.
So f has horisontal asymptotes y=e^(-pi/2) and y=e^(pi/2).

(b, c) arctan(x) monotonely increases, and e^x monotonely increases, so f(x) also monotonely increases on RR and has no maximums or minimums.
(d) now we have to compute f''(x). Recall that (arctan(x))' = 1/(1+x^2).
f'(x) = f(x)*1/(1+x^2),
f''(x) = f'(x)*1/(1+x^2) + f(x)*((-2x)/(1+x^2)^2) =(f(x))/(1+x^2)*(1-2x).
f''(x)=0 only at x=1/2.f''(x) is positive for x<1/2, so f is concave upward on (-oo, 1/2). f''(x) is negative for x>1/2, so f is concave downward on (1/2, +oo). And x=1/2 is the only inflection point of f.
(e) please look at the picture attached