Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 47

The function is defined everywhere and is infinitely differentiable.
(a) so no vertical asymptotes
At right asymptote y=0:
lim_(x->+oo)f(x) = lim_(x->+oo)(sqrt(x^2+1)-x) =

lim_(x->+oo)((x^2+1-x^2)/(sqrt(x^2+1)+x)) =lim_(x->+oo)(1/(sqrt(x^2+1)+x)) = 0.
When x->-oo, f(x)->+oo, no horizontal asymptote at left (slanted is).

(b, c) f'(x)=1/(2*sqrt(x^2+1))*2x-1 = x/sqrt(x^2+1) - 1
which is <0, so f decreases on (-oo, +oo) and has no minimums and maximums.

d) f''(x)=(1* sqrt(x^2+1) - x*(1/(2*sqrt(x^2+1))*2x))/(x^2+1)=( x^2+1- x^2)/( sqrt(x^2+1)* (x^2+1))=1/( sqrt(x^2+1)* (x^2+1)).

which is >0 on (-oo, +oo) so f is concave upward on (-oo, +oo) .

e)