`f(x) = e^(-x^2)` (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`.
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and of course `+-1/sqrt(2)` are inflection points
(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.
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