(a) there is no vertical asymptotes, f is defined and differentiable everywhere.
When x->+-oo, -x^2->-oo and f(x)->+0, so there is one horizontal asymptote y=0.
(b) f'(x) = -2x*e^(-x^2). It is >0 for x<0 and <0 for x>0,
so f(x) is increases on (-oo, 0) and decreases on (0, +oo).
(c) therefore there is one local maximum x=0. f(0)=1.
(d) f''(x) = e^(-x^2)*(-2 + 4x^2).
This is <0 for x on (-1/sqrt(2), 1/sqrt(2)) , f is concave downward there. And f''>0 on (-oo, -1/sqrt(2)) and on (1/sqrt(2), +oo), f is concave upward there.
(e)
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